My number is bigger!
Moderators: jestingrabbit, Moderators General, Prelates
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Okay, so where exactly is EliezerYudkowsky's number in the fast growing hierarchy? Because, there should be some function in the fast growing hierarchy that he would never reach, too.

 Posts: 563
 Joined: Tue Jul 27, 2010 8:48 am UTC
Re: My number is bigger!
Vytron wrote:Okay, so where exactly is EliezerYudkowsky's number in the fast growing hierarchy? Because, there should be some function in the fast growing hierarchy that he would never reach, too.
I don't know, but there is an ordinal that passes his function, it is at most W1CK if his function is computable (W1CK is the level of the busy beaver function), or further if it is not, but there is an ordinal that surpasses it. I believe that his function is computable, because in theory you can check all the proofs (up to any fixed finite length) for the turning machines with enough processing time and power. Also, if you somehow managed to come up with a function that is not surpassed by any ordinal, then it would be cheating, because it wouldn't be computable.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Thanks tomtom. I find it really odd that the biggest number of the thread is agreed on to be much bigger than the second one, and people agree it's computable, yet, nobody has been able to tell exactly how big it is, and not even have some idea about what value of the functions the people have posted it's around? (the creator even accept the structure of his number is unknowable.)
I mean, at least my number has some bounds of reaching F_omega^^n at as soon as xnth symbol (for a blue number without brackets) or as late as 1[1]...n...[1] (for n brackets), but there's no lower bound or upper bound on EliezerYudkowsky's number? And when you say, "almost W1CK", what is the distance's magnitude and, do people just think the number is big because it's reaching the limit of computability, and, how do they know computable numbers can't be bigger in some other way without having to reference theories, axioms, and Turing machines?
Because, his, does look like cheating (he went and said "a larger number than the fastgrowing hierarchy for any ordinal for which we could directly write a computable ordering".)
Would like someone to take a closer look at his number and explain what kind of values are passed at every step, because something like "using Turning machines produce mind bogglingly huge numbers" doesn't cut it for me.
I mean, at least my number has some bounds of reaching F_omega^^n at as soon as xnth symbol (for a blue number without brackets) or as late as 1[1]...n...[1] (for n brackets), but there's no lower bound or upper bound on EliezerYudkowsky's number? And when you say, "almost W1CK", what is the distance's magnitude and, do people just think the number is big because it's reaching the limit of computability, and, how do they know computable numbers can't be bigger in some other way without having to reference theories, axioms, and Turing machines?
Because, his, does look like cheating (he went and said "a larger number than the fastgrowing hierarchy for any ordinal for which we could directly write a computable ordering".)
Would like someone to take a closer look at his number and explain what kind of values are passed at every step, because something like "using Turning machines produce mind bogglingly huge numbers" doesn't cut it for me.

 Posts: 563
 Joined: Tue Jul 27, 2010 8:48 am UTC
Re: My number is bigger!
Vytron wrote:I mean, at least my number has some bounds of reaching F_omega^^n at as soon as xnth symbol (for a blue number without brackets) or as late as 1[1]...n...[1] (for n brackets), but there's no lower bound or upper bound on EliezerYudkowsky's number? And when you say, "almost W1CK", what is the distance's magnitude and, do people just think the number is big because it's reaching the limit of computability, and, how do they know computable numbers can't be bigger in some other way without having to reference theories, axioms, and Turing machines?
It would be cheating if he did not put in the clause "that are provably halting in less than 2^^p symbols." Then his function would be able to solve the halting problem in its definition, and so it would be uncomputable, which I think is against the rules (unless it has been changed, there is a rule about all functions being computable). I'm also not sure if his function grows as fast as he says, because it probably takes a lot of symbols to prove halting if it only halts in 10^36534 steps (the current record for 6 state machines). Also, for the value of the function at 5 and above, the function will not be able to practically calculated, because S(5) is not known, and searching through all proofs with up to 2^^5=2^65536 characters is not feasible.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
tomtom2357 wrote: Also, for the value of the function at 5 and above, the function will not be able to practically calculated, because S(5) is not known
Thank you. So then, if his number is "unknowningly huge", how do we know exactly at what point some other number surpasses it? Does it become a "battle of the unknowns" if someone posts another number that we don't know how big is, and we don't know which one is bigger?
(By "cheating" I don't mean his number is uncomputable or invalid, just that, with all the previous numbers in the thread, we can build a number line, or a system of lines, or a systems of systems, to see how big those numbers are [they fall somewhere, it's clear that one is bigger than another and roughly by how much], but with this number, we can't, so how does it counts as "well defined"?)

 Posts: 563
 Joined: Tue Jul 27, 2010 8:48 am UTC
Re: My number is bigger!
Vytron wrote:tomtom2357 wrote: Also, for the value of the function at 5 and above, the function will not be able to practically calculated, because S(5) is not known
Thank you. So then, if his number is "unknowningly huge", how do we know exactly at what point some other number surpasses it? Does it become a "battle of the unknowns" if someone posts another number that we don't know how big is, and we don't know which one is bigger?
(By "cheating" I don't mean his number is uncomputable or invalid, just that, with all the previous numbers in the thread, we can build a number line, or a system of lines, or a systems of systems, to see how big those numbers are [they fall somewhere, it's clear that one is bigger than another and roughly by how much], but with this number, we can't, so how does it counts as "well defined"?)
It is well defined because f(n) has an actual value for all n. That value may never be known, but it does exist. That is the definition of well defined.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
And if that value is never known, how do people know it's the highest on the thread? Is it based on feelings or estimations?
This part here:
Makes me believe Eliezer's number may not be the biggest number of the thread, it may be "3 boxes" bigger (what WarDaft called "categories") than the second biggest number, or it may be smaller that GoC's number here. We don't know, and the thing here is computability growth expectation:
Assume Eliezer's number is indeed the biggest in the thread, by far.
Make a program that computes its value.
Make a program that computes the value of GoC's number (he gave code and all.)
Have both programs running in parallel.
Now, what we know is that, when Eliezer's number finishes computing, it'll be much bigger, but we don't know when it's done. We know GoC's number will be much smaller, but will be computed much earlier, so much earlier that we can run it again, with its output as its seed, and it'll produce a much bigger number, while Eliezer's number is still computing.
If we do this every time GoC's program is done, how many times does it run before Eliezer's program is done, and, is its output number bigger than Eliezer's number? (if so, maybe there you have the biggest number on the thread right now, even though it would be invalid for me to send it as an entry.)
So, Eliezer's number computes too slowly, if by the time it's done with P = 7, the functions on this thread looping when done would have reached a much bigger number than Eliezer == it's growing too slowly (as measured by time to compute).
This part here:
tomtom2357 wrote: I'm also not sure if his function grows as fast as he says, because it probably takes a lot of symbols to prove halting if it only halts in 10^36534 steps (the current record for 6 state machines).
Makes me believe Eliezer's number may not be the biggest number of the thread, it may be "3 boxes" bigger (what WarDaft called "categories") than the second biggest number, or it may be smaller that GoC's number here. We don't know, and the thing here is computability growth expectation:
Assume Eliezer's number is indeed the biggest in the thread, by far.
Make a program that computes its value.
Make a program that computes the value of GoC's number (he gave code and all.)
Have both programs running in parallel.
Now, what we know is that, when Eliezer's number finishes computing, it'll be much bigger, but we don't know when it's done. We know GoC's number will be much smaller, but will be computed much earlier, so much earlier that we can run it again, with its output as its seed, and it'll produce a much bigger number, while Eliezer's number is still computing.
If we do this every time GoC's program is done, how many times does it run before Eliezer's program is done, and, is its output number bigger than Eliezer's number? (if so, maybe there you have the biggest number on the thread right now, even though it would be invalid for me to send it as an entry.)
So, Eliezer's number computes too slowly, if by the time it's done with P = 7, the functions on this thread looping when done would have reached a much bigger number than Eliezer == it's growing too slowly (as measured by time to compute).

 Posts: 563
 Joined: Tue Jul 27, 2010 8:48 am UTC
Re: My number is bigger!
Vytron wrote:And if that value is never known, how do people know it's the highest on the thread? Is it based on feelings or estimations?
This part here:tomtom2357 wrote: I'm also not sure if his function grows as fast as he says, because it probably takes a lot of symbols to prove halting if it only halts in 10^36534 steps (the current record for 6 state machines).
Makes me believe Eliezer's number may not be the biggest number of the thread, it may be "3 boxes" bigger (what WarDaft called "categories") than the second biggest number, or it may be smaller that GoC's number here. We don't know, and the thing here is computability growth expectation:
Assume Eliezer's number is indeed the biggest in the thread, by far.
Make a program that computes its value.
Make a program that computes the value of GoC's number (he gave code and all.)
Have both programs running in parallel.
Now, what we know is that, when Eliezer's number finishes computing, it'll be much bigger, but we don't know when it's done. We know GoC's number will be much smaller, but will be computed much earlier, so much earlier that we can run it again, with its output as its seed, and it'll produce a much bigger number, while Eliezer's number is still computing.
If we do this every time GoC's program is done, how many times does it run before Eliezer's program is done, and, is its output number bigger than Eliezer's number? (if so, maybe there you have the biggest number on the thread right now, even though it would be invalid for me to send it as an entry.)
So, Eliezer's number computes too slowly, if by the time it's done with P = 7, the functions on this thread looping when done would have reached a much bigger number than Eliezer == it's growing too slowly (as measured by time to compute).
Since we are out so far into huge numbers, the time it takes for a program to compute one of the numbers past page 6 is approximately the number itself. Therefore, the computing angle isn't going to work. Also, Eliezer's number starts slow, so it will probably take until P=100 to get a bigger number than the numbers on this thread. Fortunately for him, his function is applied 10 times, so it is definitely bigger than all the numbers on this thread. I think his function basically kills the thread, because there is (almost) no way any number can pass it.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
 dudiobugtron
 Posts: 1098
 Joined: Mon Jul 30, 2012 9:14 am UTC
 Location: The Outlier
Re: My number is bigger!
This is how I read the above two posts:
I realise there's more to it than that, but it does seem weird from the point of view of someone (me) who doesn't necessarily understand all of it, but is trying to.
tomtom2357 wrote:Vytron wrote:Can't we beat Eliezer's number by applying another one of the functions in this thread multiple times, each time using the result as a new seed?
Elizer's function is applied 10 times, so it beats those functions. 10 is a lot of times.
I realise there's more to it than that, but it does seem weird from the point of view of someone (me) who doesn't necessarily understand all of it, but is trying to.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Hey dudiobugtron
My argument went like this:
"If we created programs to compute these numbers, and we ran them up to the entire life of the universe, these programs would output at that point a number that was bigger than Eliezer's number."
tomtom is basically saying that the numbers from page 6 of this thread would already take more than the entire life of the universe to compute*, so, it makes no sense to bound them by that, so it doesn't matter that Eliezer's number would take much longer to compute, what matters is that while all the other numbers in the thread would require a huge amount of selfrecursions just to reach Eliezer's number, Eliezer can do it in 10.
*The numbers of the first page of the thread, in a plank length of time, are already past the universe's life by a huge magnitude, the point of talking about it is that, if you imagined a universe that lasted {the second biggest number of the thread} years, and had all numbers growing in selfrecursion, at any arbitrary point, you'd see them being bigger than Eliezer's (because it grows so slowly), but there's a point within its recursions (which is reached at least as soon as f(f(f(f(f(f(f()))))))), which it clearly is growing faster than the other numbers of the thread, and it goes up to f(f(f(f(f(f(f(f(f(f()))))))))). If you had a universe with a number of years that selfrecursed Eliezer's number 11 times, that'd be enough for Eliezer's number to have been computed, and it would have been growing much faster than the other numbers on the thread without needing as much selfrecursion.
So the message is: "Abandon hope of beating Eliezer's number, all ye who enter this thread!"
What I don't like is that we don't know exactly at what point it's growing faster than the second number, and exactly by how much it surpasses it, and that we will never know.
But it's kind of cool to see someone managed to basically have a "infinity1" number on the thread, one to which, to beat it you need to use its approach to get bigger numbers but that no different approach can grow as fast (that's tomtom's claim.)
My argument went like this:
"If we created programs to compute these numbers, and we ran them up to the entire life of the universe, these programs would output at that point a number that was bigger than Eliezer's number."
tomtom is basically saying that the numbers from page 6 of this thread would already take more than the entire life of the universe to compute*, so, it makes no sense to bound them by that, so it doesn't matter that Eliezer's number would take much longer to compute, what matters is that while all the other numbers in the thread would require a huge amount of selfrecursions just to reach Eliezer's number, Eliezer can do it in 10.
*The numbers of the first page of the thread, in a plank length of time, are already past the universe's life by a huge magnitude, the point of talking about it is that, if you imagined a universe that lasted {the second biggest number of the thread} years, and had all numbers growing in selfrecursion, at any arbitrary point, you'd see them being bigger than Eliezer's (because it grows so slowly), but there's a point within its recursions (which is reached at least as soon as f(f(f(f(f(f(f()))))))), which it clearly is growing faster than the other numbers of the thread, and it goes up to f(f(f(f(f(f(f(f(f(f()))))))))). If you had a universe with a number of years that selfrecursed Eliezer's number 11 times, that'd be enough for Eliezer's number to have been computed, and it would have been growing much faster than the other numbers on the thread without needing as much selfrecursion.
So the message is: "Abandon hope of beating Eliezer's number, all ye who enter this thread!"
What I don't like is that we don't know exactly at what point it's growing faster than the second number, and exactly by how much it surpasses it, and that we will never know.
But it's kind of cool to see someone managed to basically have a "infinity1" number on the thread, one to which, to beat it you need to use its approach to get bigger numbers but that no different approach can grow as fast (that's tomtom's claim.)
Last edited by Vytron on Mon Jun 24, 2013 7:44 am UTC, edited 1 time in total.

 Posts: 563
 Joined: Tue Jul 27, 2010 8:48 am UTC
Re: My number is bigger!
dudiobugtron wrote:This is how I read the above two posts:tomtom2357 wrote:Vytron wrote:Can't we beat Eliezer's number by applying another one of the functions in this thread multiple times, each time using the result as a new seed?
Elizer's function is applied 10 times, so it beats those functions. 10 is a lot of times.
I realise there's more to it than that, but it does seem weird from the point of view of someone (me) who doesn't necessarily understand all of it, but is trying to.
What I'm saying is that there is a point, call it n, where f(n) (f is Eliezer's function) passes all other numbers on this thread. n is probably less than 10^100. Since f(10)>10^{100} (because BB(10)>1010^{100}, and that did not take 2^(2^10) characters to prove), f(f(10))>f(10^100)>all numbers on this thread. Therefore, f^{10}(10)>f(f(10))>f(10^{100})>all numbers on this thread.
Edit: ninja'd. Anyway, my post said that almost no other method would beat this number. There might be some way to beat the number. For example, define Xi'(n) to be the largest number x such that it is provable in f(n) symbols (using my above definition) (in ZFC or some stronger theory, if you really need that) that Xi(n)>x. The Xi function is found at http://googology.wikia.com/wiki/Xi_function. Now, my number is Xi'^{10}(10). This number is roughly at the same level as Eliezer's number, maybe slightly bigger. Yes, it uses roughly the same method, but still, it proves that Eliezer's number can be beaten.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
tomtom2357 wrote: I think his function basically kills the thread, because there is (almost) no way any number can pass it.
*3 posts later*
tomtom2357 wrote:define Xi'(n) to be the largest number x such that it is provable in f(n) symbols (using my above definition) (in ZFC or some stronger theory, if you really need that) that Xi(n)>x. The Xi function is found at http://googology.wikia.com/wiki/Xi_function. Now, my number is Xi'^{10}(10). This number is roughly at the same level as Eliezer's number, maybe slightly bigger. Yes, it uses roughly the same method, but still, it proves that Eliezer's number can be beaten.
Ha! That didn't take long
So I guess this is just another revolution similar to the collapsing ordinals revolution a few pages back. Behold the age of the Unknowable Numbers (that we can't ever know how big they are.)
My blue number progression can only ever hope to be the third biggest number on the thread, so I guess it's time to give up for now...

 Posts: 563
 Joined: Tue Jul 27, 2010 8:48 am UTC
Re: My number is bigger!
Well (just so no one else does it before me), using Rayo'^{10^100}(10^{100}) is even bigger (I had to use 10^{100}, because Rayo(10)=0), using the same method I used before, and Rayo(n) is defined at http://googology.wikia.com/wiki/Rayo%27s_number.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
Re: My number is bigger!
Vytron wrote:Thanks tomtom. I find it really odd that the biggest number of the thread is agreed on to be much bigger than the second one, and people agree it's computable, yet, nobody has been able to tell exactly how big it is, and not even have some idea about what value of the functions the people have posted it's around? (the creator even accept the structure of his number is unknowable.)
Actually, Yudkowsky, itaibn, meur, and I all came up with numbers based on functions that surpassed all provably recursive functions of strong axiom systems (meur used ZFC, but could have just as easily used ZFC+I0, so we can give it to him I think), so perhaps it can be called a fourway tie.
There's no way to say exactly how big it is, because to do so would mean we have some alternate notation that describes it's size, and there's no alternate notation that grows as fast unless it uses a similar method. Still, we can see that it is big. Perhaps it is easier to explain how we know that BB(googol) will be larger than any legitimate entry in this thread. BB(n) is the largest number of 1's produced by any halting nstate, 2symbol Turing machine. Any entry to this thread will be described by a recursive procedure using a relatively small amount of characters (compared to a googol), and so we can implement the procedure with a Turing machine using far less than a googol states. Ergo it will be less than BB(googol).
It works similarly with Yudkowsky's number. For any recursive notation R that we know terminates, we wil be able to prove that it terminates as ZFC+I0 in some reasonable length, as ZFC+I0 is just about the strongest axiom system out there. Otherwise, we won't have any way to know the notation terminates. (The exception to this rule is to diagonalize out of ZFC+I0, as Yudkowsky et al do. To prove this terminates, you have to go to ZFC+I0 + Con(ZFC + I0). But, if you believe that ZFC+I0 holds, you will also believe that Con(ZFC+I0) holds, so we can believe in ZFC + I0 + Con(ZFC + I0) as well.) So R can be simulated by a Turing machine that can be proven to halt in a reasonable number of steps, say, less than a googolplex. But Yudkowsky's number includes this Turing machine in its construction, so Yudkowsky's number will be bigger than the number outputted by R.
I mean, at least my number has some bounds of reaching F_omega^^n at as soon as xnth symbol (for a blue number without brackets) or as late as 1[1]...n...[1] (for n brackets), but there's no lower bound or upper bound on EliezerYudkowsky's number? And when you say, "almost W1CK", what is the distance's magnitude and, do people just think the number is big because it's reaching the limit of computability, and, how do they know computable numbers can't be bigger in some other way without having to reference theories, axioms, and Turing machines?
I don't think we've established any strong bounds on your notation yet  I would need to understand it better to do it myself. We do know lower bounds on the functional growth rate of Yudkowsky's number  any ordinal less than the proof theoretic ordinal of ZFC+I0 will do. Since all known recursive ordinal notations are less than that, we can make very fast growing functions that are upper bounded by Yudkowsky's function.
Asking for the distance between a recursive ordinal and omega_1^CK is like asking the distance between a finite ordinal and omega  no matter how big the recursive ordinal, the only sensible answer is omega_1^CK.
Because, his, does look like cheating (he went and said "a larger number than the fastgrowing hierarchy for any ordinal for which we could directly write a computable ordering".)
Would like someone to take a closer look at his number and explain what kind of values are passed at every step, because something like "using Turning machines produce mind bogglingly huge numbers" doesn't cut it for me.
I understand that you feel like it is cheating  it's really a different category. I prefer to work in the subcategory of numbers built "from the bottom up" using an explicit recursive process, I feel like those are the most interesting. So, I think we can still talk about such numbers without having to feel like there's no way to win.
Re: My number is bigger!
Vytron wrote:Makes me believe Eliezer's number may not be the biggest number of the thread, it may be "3 boxes" bigger (what WarDaft called "categories") than the second biggest number, or it may be smaller that GoC's number here. We don't know, and the thing here is computability growth expectation:
Assume Eliezer's number is indeed the biggest in the thread, by far.
Make a program that computes its value.
Make a program that computes the value of GoC's number (he gave code and all.)
Have both programs running in parallel.
Now, what we know is that, when Eliezer's number finishes computing, it'll be much bigger, but we don't know when it's done. We know GoC's number will be much smaller, but will be computed much earlier, so much earlier that we can run it again, with its output as its seed, and it'll produce a much bigger number, while Eliezer's number is still computing.
If we do this every time GoC's program is done, how many times does it run before Eliezer's program is done, and, is its output number bigger than Eliezer's number? (if so, maybe there you have the biggest number on the thread right now, even though it would be invalid for me to send it as an entry.)
So, Eliezer's number computes too slowly, if by the time it's done with P = 7, the functions on this thread looping when done would have reached a much bigger number than Eliezer == it's growing too slowly (as measured by time to compute).
Eliezer's program will spend the vast majority of its time running Turing machines, and the number of steps of those machines will be added to Eliezer's number, so the running time and the final number will be approximately the same. If that's not fast enough, you can just compute 2^n, or 2^^n, instead; it makes no real difference to the size of the number.
Rather than use GoC's program, it would be better to use a program that produces large numbers as fast as possible, like a machine language program that keeps setting bits to 1's to create a large binary number. But obviously this is just Eliezer's method, with a minor modification that makes no real difference.
Last edited by Deedlit on Mon Jun 24, 2013 9:32 am UTC, edited 1 time in total.
Re: My number is bigger!
tomtom2357 wrote:What I'm saying is that there is a point, call it n, where f(n) (f is Eliezer's function) passes all other numbers on this thread. n is probably less than 10^100. Since f(10)>10^{100} (because BB(10)>1010^{100}, and that did not take 2^(2^10) characters to prove), f(f(10))>f(10^100)>all numbers on this thread. Therefore, f^{10}(10)>f(f(10))>f(10^{100})>all numbers on this thread.
Edit: ninja'd. Anyway, my post said that almost no other method would beat this number. There might be some way to beat the number. For example, define Xi'(n) to be the largest number x such that it is provable in f(n) symbols (using my above definition) (in ZFC or some stronger theory, if you really need that) that Xi(n)>x. The Xi function is found at http://googology.wikia.com/wiki/Xi_function. Now, my number is Xi'^{10}(10). This number is roughly at the same level as Eliezer's number, maybe slightly bigger. Yes, it uses roughly the same method, but still, it proves that Eliezer's number can be beaten.
How is x being represented? If it is represented in say, decimal, then x will be smaller than 10^f(n). Of course it is still quite large since you are using Eliezer's function, but it doesn't make any real improvement to it.

 Posts: 563
 Joined: Tue Jul 27, 2010 8:48 am UTC
Re: My number is bigger!
Deedlit wrote:Vytron wrote:Thanks tomtom. I find it really odd that the biggest number of the thread is agreed on to be much bigger than the second one, and people agree it's computable, yet, nobody has been able to tell exactly how big it is, and not even have some idea about what value of the functions the people have posted it's around? (the creator even accept the structure of his number is unknowable.)
Actually, Yudkowsky, itaibn, meur, and I all came up with numbers based on functions that surpassed all provably recursive functions of strong axiom systems (meur used ZFC, but could have just as easily used ZFC+I0, so we can give it to him I think), so perhaps it can be called a fourway tie.
Make that a 6 way tie (I defined two numbers above). Also, I count any numbers used in the proof as 1 character.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
Re: My number is bigger!
tomtom2357 wrote:Make that a 6 way tie (I defined two numbers above). Also, I count any numbers used in the proof as 1 character.
But the ZFC language does not have all numbers as symbols. If it did, there would be infinitely many proofs of f(n) symbols or less, so I'm not sure that you can get a recursive procedure to find the largest x such that Rayo(n) > x.

 Posts: 563
 Joined: Tue Jul 27, 2010 8:48 am UTC
Re: My number is bigger!
Deedlit wrote:tomtom2357 wrote:Make that a 6 way tie (I defined two numbers above). Also, I count any numbers used in the proof as 1 character.
But the ZFC language does not have all numbers as symbols. If it did, there would be infinitely many proofs of f(n) symbols or less, so I'm not sure that you can get a recursive procedure to find the largest x such that Rayo(n) > x.
Fine, make it use as many characters that it would be in ZFC.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
Re: My number is bigger!
tomtom2357 wrote:Deedlit wrote:tomtom2357 wrote:Make that a 6 way tie (I defined two numbers above). Also, I count any numbers used in the proof as 1 character.
But the ZFC language does not have all numbers as symbols. If it did, there would be infinitely many proofs of f(n) symbols or less, so I'm not sure that you can get a recursive procedure to find the largest x such that Rayo(n) > x.
Fine, make it use as many characters that it would be in ZFC.
If you mean the shortest statement that uniquely defines a given number, if you are going to do that there's no need to invoke Xi(n) or Rayo(n); just define F(n) as the largest number such that there exists a statement that uniquely defines it, where that is provable in a proof of at most f(n) symbols (where f(n) is your favorite function.) This is easier than proving in addition that x < Rayo(n), so you'll get a larger number.

 Posts: 563
 Joined: Tue Jul 27, 2010 8:48 am UTC
Re: My number is bigger!
Deedlit wrote:tomtom2357 wrote:Deedlit wrote:tomtom2357 wrote:Make that a 6 way tie (I defined two numbers above). Also, I count any numbers used in the proof as 1 character.
But the ZFC language does not have all numbers as symbols. If it did, there would be infinitely many proofs of f(n) symbols or less, so I'm not sure that you can get a recursive procedure to find the largest x such that Rayo(n) > x.
Fine, make it use as many characters that it would be in ZFC.
If you mean the shortest statement that uniquely defines a given number, if you are going to do that there's no need to invoke Xi(n) or Rayo(n); just define F(n) as the largest number such that there exists a statement that uniquely defines it, where that is provable in a proof of at most f(n) symbols (where f(n) is your favorite function.) This is easier than proving in addition that x < Rayo(n), so you'll get a larger number.
Ok then, I'll use that.
Are you the same Deedlit (Deedlit11 on that site) that is on the googology wikia (yes, I am the same tomtom2357)?
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Okay, I won't give up. I mean, I'll give up about getting the biggest number on the thread (though, I may probably could reach it by just "plugging" ZFC and Turning Machines into my number, or not), but not about trying to get my number understood, which may turn out to be even harder.
I also have the hope of managing to break my number somewhere, so, at some bracket, it reaches infinity, and then, I can get arbitrarily close to infinity and maybe, just maybe, that'll work.
I've noticed the bounds of how high my number can grow get bigger whenever I zoom in to the finer detail of it, because, whenever I say "and then go on and on and reach these symbols and n+1s and stuff" people just abstract the progress and think it can't get outside some arbitrary box. I think it can, and I figured the way to imagine imaginary space is with a picture. But first, I'm changing its name, so that it's not confused with imaginary numbers.
Let's call it... Illusory Progression? Yeah, because all it does is defining some kind of progression, continue with it for a long time, and then go back and say "that's step 1 of the real progression". It looks like this:
Here, this line isn't "linear", it's already diagonalizing over hierarchies (otherwise I'd never reach n + ω). Level 3 kicks off the Illusory Progression:
This purple line does what the black line was doing for as long as every point on the black line (so, n such purple lines), and then, takes a look at the whole thing, creates a new kind of purple line of that length, and creates as many lines of that length as there are points in the whole thing (represented by an orange line below). Now, the original black line and purple lines are so small that it's hard to graph, in this line, the purple line is a single point of the orange line, and the black line is a single point in the purple line:
Now that you've done it, a single point beyond the black line is the real progression, and it continues by that angle for as much as the whole thing till now:
This is not to scale. The red line is the limit of the height the orange line can reach, and the green square tells the black line to reach that height in as much as what it took the original black line to grow, in horizontal distance (so, the black line to scale would look vertical, its end wouldn't even surpass the second point of the purple line below). This new black line is the real Level 3 Step 1, and it continues like this...
(n remains a single step of the real progression)
The idea is to travel as few real distance horizontally, travel it within Illusory Progressions, and convert that into verticality, so by the time the number is really where the second purple line would have been placed, the black line has curved to approach a 90 degree line (infinity), faster and faster.
What this means is that the limit of the black line progression is reached at the length of the purple lines, "eats itself" for that many times, and then, this process becomes the real progression.
Since that probably makes no sense (because I have so far failed to depict how Illusory Progressions works, I don't see why those pictures would do it), let's go back to where we left off when showing the progression at the inner levels, the ones that are assumed to "not go very far":
These were left off at the bottom of this post, we were within imaginary space at level 5, let's open those tags [Illusory Progression]
What I'm going to do is suppressing the deepest level because you know it's there (I never put "Level 2 contains Level 1" and it was understood, so here, Level 3 contains Level 2)
Level 5 Step 1 contains: {
Level 4 Step 1 contains: {
Level 3 Step 1: n + ω^{ω} + ω
Level 3 Step 2: n + ω^{ω} + 2ω
Level 3 Step 3: n + ω^{ω} + 3ω
...
Limit of Level 3 = ω^{ω} + ω*ω}
Level 4 Step 2 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{2} + 3ω
...
Limit of Level 3 = ω^{ω} + ω^{3}}
Level 4 Step 3 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{3} + ω
Level 3 Step 2: n + ω^{ω} + ω^{3} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{3} + 3ω
...
Limit of Level 3 = n + ω^{ω} + ω^{4}}
...
Limit of Level 4 = n + ω^{ω} + ω^{ω}}
Level 5 Step 2 contains: {
Level 4 Step 1 contains: {
Level 3 Step 1: n + 2ω^{ω} + ω
Level 3 Step 2: n + 2ω^{ω} + 2ω
Level 3 Step 3: n + 2ω^{ω} + 3ω
...
Limit of Level 3 = n + 2ω^{ω} + ω*ω}
Level 4 Step 2 contains: {
Level 3 Step 1: n + 2ω^{ω} + ω^{2} + ω
Level 3 Step 2: n + 2ω^{ω} + ω^{2} + 2ω
Level 3 Step 3: n + 2ω^{ω} + ω^{2} + 3ω
...
Limit of Level 3 = n + 2ω^{ω} + ω^{3}}
Level 4 Step 3 contains: {
Level 3 Step 1: n + 2ω^{ω} + ω^{3} + ω
Level 3 Step 2: n + 2ω^{ω} + ω^{3} + 2ω
Level 3 Step 3: n + 2ω^{ω} + ω^{3} + 3ω
...
Limit of Level 3 = n + 2ω^{ω} + ω^{4}}
...
Limit of Level 4 = n + 2ω^{ω} + ω^{ω}}
...
Limit of Level 5 = n + ω^{ω + 1}
Level 6
Level 7
Level 8
Level 9
So, the limit of Level 9 is ω^{ω^ω}.
We're still within Illusory Space, and the question is, from the pictures where the black line turned purple, at which point does the purple line turn orange? And that is, at the point the number starts eating itself.
That is, the power of n + ω was reached at some point within the original number progression, so there's a number equivalence with this power, n. If you reach Level n, you'll have:
Level n's limit = ω^{ω^ω^ω... n times ...^ω^ω^ω}
Or in other words, if the power of the fast growing hierachy is reached at x, then, there'll be some level x that outputs a ω tower of length x. This is where the orange line starts.
Note: this isn't really an "x", it's a finite blue number without brackets, it's unknown which number it is, but it's known a finite progression of the levels would eventually reach this power, and as the Illusory Progression is understood, it may be reached really fast, as fast as 7 or something.
Because, that's not the limit of the levels, you can reach Level x  which is equivalent to Level n + ω, as depicted by the power this level had.
Level x (n + ω) Step 1 Contains {
Level x1 (n1 + ω) Step 1 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(ω + 1)}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(ω + 2)}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(ω + 3)}}
Limit of Level x2 (n2 + ω): n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(2ω)}}
Level x1 (n1 + ω) Step 2 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(2ω + 1)}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(2ω + 2)}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(2ω + 3)}}
Limit of Level x2 (n2 + ω): n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(3ω)}}
Level x1 (n1 + ω) Step 3 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(3ω + 1)}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(3ω + 2)}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(3ω + 3)}}
Limit of Level x2 (n2 + ω): n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(4ω)}}
Limit of Level x1 (n1 + ω) = n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(ω^2)}
}
Limit of Level x (n + ω) = n + ω^{ω^ω^ω... n + (ω+1) times ...^ω^ω^ω}
And goes up to the level equivalent of Level n + ω*ω
[/Illusory Progression]
Which is really the value and power of Level 3 Step 1.
I'll stop there to make sure this is clear, because real Level 3 Step 2 does not apply Level 2 at every step like this is doing, it'll apply Level 3 Step 1 at every step.
I also have the hope of managing to break my number somewhere, so, at some bracket, it reaches infinity, and then, I can get arbitrarily close to infinity and maybe, just maybe, that'll work.
I've noticed the bounds of how high my number can grow get bigger whenever I zoom in to the finer detail of it, because, whenever I say "and then go on and on and reach these symbols and n+1s and stuff" people just abstract the progress and think it can't get outside some arbitrary box. I think it can, and I figured the way to imagine imaginary space is with a picture. But first, I'm changing its name, so that it's not confused with imaginary numbers.
Let's call it... Illusory Progression? Yeah, because all it does is defining some kind of progression, continue with it for a long time, and then go back and say "that's step 1 of the real progression". It looks like this:
Here, this line isn't "linear", it's already diagonalizing over hierarchies (otherwise I'd never reach n + ω). Level 3 kicks off the Illusory Progression:
This purple line does what the black line was doing for as long as every point on the black line (so, n such purple lines), and then, takes a look at the whole thing, creates a new kind of purple line of that length, and creates as many lines of that length as there are points in the whole thing (represented by an orange line below). Now, the original black line and purple lines are so small that it's hard to graph, in this line, the purple line is a single point of the orange line, and the black line is a single point in the purple line:
Now that you've done it, a single point beyond the black line is the real progression, and it continues by that angle for as much as the whole thing till now:
This is not to scale. The red line is the limit of the height the orange line can reach, and the green square tells the black line to reach that height in as much as what it took the original black line to grow, in horizontal distance (so, the black line to scale would look vertical, its end wouldn't even surpass the second point of the purple line below). This new black line is the real Level 3 Step 1, and it continues like this...
(n remains a single step of the real progression)
The idea is to travel as few real distance horizontally, travel it within Illusory Progressions, and convert that into verticality, so by the time the number is really where the second purple line would have been placed, the black line has curved to approach a 90 degree line (infinity), faster and faster.
What this means is that the limit of the black line progression is reached at the length of the purple lines, "eats itself" for that many times, and then, this process becomes the real progression.
Since that probably makes no sense (because I have so far failed to depict how Illusory Progressions works, I don't see why those pictures would do it), let's go back to where we left off when showing the progression at the inner levels, the ones that are assumed to "not go very far":
These were left off at the bottom of this post, we were within imaginary space at level 5, let's open those tags [Illusory Progression]
What I'm going to do is suppressing the deepest level because you know it's there (I never put "Level 2 contains Level 1" and it was understood, so here, Level 3 contains Level 2)
Level 5 Step 1 contains: {
Level 4 Step 1 contains: {
Level 3 Step 1: n + ω^{ω} + ω
Level 3 Step 2: n + ω^{ω} + 2ω
Level 3 Step 3: n + ω^{ω} + 3ω
...
Limit of Level 3 = ω^{ω} + ω*ω}
Level 4 Step 2 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{2} + 3ω
...
Limit of Level 3 = ω^{ω} + ω^{3}}
Level 4 Step 3 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{3} + ω
Level 3 Step 2: n + ω^{ω} + ω^{3} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{3} + 3ω
...
Limit of Level 3 = n + ω^{ω} + ω^{4}}
...
Limit of Level 4 = n + ω^{ω} + ω^{ω}}
Level 5 Step 2 contains: {
Level 4 Step 1 contains: {
Level 3 Step 1: n + 2ω^{ω} + ω
Level 3 Step 2: n + 2ω^{ω} + 2ω
Level 3 Step 3: n + 2ω^{ω} + 3ω
...
Limit of Level 3 = n + 2ω^{ω} + ω*ω}
Level 4 Step 2 contains: {
Level 3 Step 1: n + 2ω^{ω} + ω^{2} + ω
Level 3 Step 2: n + 2ω^{ω} + ω^{2} + 2ω
Level 3 Step 3: n + 2ω^{ω} + ω^{2} + 3ω
...
Limit of Level 3 = n + 2ω^{ω} + ω^{3}}
Level 4 Step 3 contains: {
Level 3 Step 1: n + 2ω^{ω} + ω^{3} + ω
Level 3 Step 2: n + 2ω^{ω} + ω^{3} + 2ω
Level 3 Step 3: n + 2ω^{ω} + ω^{3} + 3ω
...
Limit of Level 3 = n + 2ω^{ω} + ω^{4}}
...
Limit of Level 4 = n + 2ω^{ω} + ω^{ω}}
...
Limit of Level 5 = n + ω^{ω + 1}
Level 6
Spoiler:
Level 7
Spoiler:
Level 8
Spoiler:
Level 9
Spoiler:
So, the limit of Level 9 is ω^{ω^ω}.
We're still within Illusory Space, and the question is, from the pictures where the black line turned purple, at which point does the purple line turn orange? And that is, at the point the number starts eating itself.
That is, the power of n + ω was reached at some point within the original number progression, so there's a number equivalence with this power, n. If you reach Level n, you'll have:
Level n's limit = ω^{ω^ω^ω... n times ...^ω^ω^ω}
Or in other words, if the power of the fast growing hierachy is reached at x, then, there'll be some level x that outputs a ω tower of length x. This is where the orange line starts.
Note: this isn't really an "x", it's a finite blue number without brackets, it's unknown which number it is, but it's known a finite progression of the levels would eventually reach this power, and as the Illusory Progression is understood, it may be reached really fast, as fast as 7 or something.
Because, that's not the limit of the levels, you can reach Level x  which is equivalent to Level n + ω, as depicted by the power this level had.
Level x (n + ω) Step 1 Contains {
Level x1 (n1 + ω) Step 1 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(ω + 1)}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(ω + 2)}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(ω + 3)}}
Limit of Level x2 (n2 + ω): n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(2ω)}}
Level x1 (n1 + ω) Step 2 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(2ω + 1)}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(2ω + 2)}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(2ω + 3)}}
Limit of Level x2 (n2 + ω): n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(3ω)}}
Level x1 (n1 + ω) Step 3 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(3ω + 1)}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(3ω + 2)}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(3ω + 3)}}
Limit of Level x2 (n2 + ω): n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(4ω)}}
Limit of Level x1 (n1 + ω) = n + ω^{ω^ω^ω... n + ω times ...^ω^ω^(ω^2)}
}
Limit of Level x (n + ω) = n + ω^{ω^ω^ω... n + (ω+1) times ...^ω^ω^ω}
And goes up to the level equivalent of Level n + ω*ω
[/Illusory Progression]
Which is really the value and power of Level 3 Step 1.
I'll stop there to make sure this is clear, because real Level 3 Step 2 does not apply Level 2 at every step like this is doing, it'll apply Level 3 Step 1 at every step.
Re: My number is bigger!
Yes (assuming your values for the steps are correct), that is the limit of level 3.Vytron wrote:Level 5 Step 1 contains: {
Level 4 Step 1 contains: {
Level 3 Step 1: n + ω^{ω} + ω
Level 3 Step 2: n + ω^{ω} + 2ω
Level 3 Step 3: n + ω^{ω} + 3ω
...
Limit of Level 3 = ω^{ω} + ω*ω}
Unfortunately, no. The limit ω^{ω} + ω^{2} + ω, ω^{ω} + ω^{2} + 2ω, ω^{ω} + ω^{2} + 3ω,... is ω^{ω} + ω^{2} + ω^{2} = ω^{ω} + ω^{2}*2. You have to repeat this process again to get ω^{ω} + ω^{2}*3, then again for ω^{ω} + ω^{2}*4, and continue this chain infinitely many times before you get ω^{ω} + ω^{3}.Level 4 Step 2 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{2} + 3ω
...
Limit of Level 3 = ω^{ω} + ω^{3}}
As for EY's number, we do not truly know if it is well defined. If ZFC+I0 is consistent, then it is well defined.
Also, for the Rayo number and that train of discussion, Zeno machines can produce larger numbers. Much, much larger.
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Okay, thanks, I'm totally fine with that since my number isn't meant to get big that soon anyway.
But I was able to get to ω^{ω} in finite steps, how come now I need to get into an infinite loop to get to ω^{ω} + ω^{3}? It seems I should be able to reach ω^{ω} + ω^{3} within the next ω^{ω} + ω^{3} recursive steps. And a chain of infinitely many times, since the process would never stop, would give me infinity (Like 1+1+1...(infinite times)...+1+1+1 = Infinity, and my steps are increasing the number faster than +1.)
WarDaft wrote:and continue this chain infinitely many times before you get ω^{ω} + ω^{3}.
But I was able to get to ω^{ω} in finite steps, how come now I need to get into an infinite loop to get to ω^{ω} + ω^{3}? It seems I should be able to reach ω^{ω} + ω^{3} within the next ω^{ω} + ω^{3} recursive steps. And a chain of infinitely many times, since the process would never stop, would give me infinity (Like 1+1+1...(infinite times)...+1+1+1 = Infinity, and my steps are increasing the number faster than +1.)
Re: My number is bigger!
That would only be because one particular one of the things you said happens to do that, if it does, I'm still not sure. If you're basing that on me saying 'suppose it does, then...', that would be an ifthen statement, I wasn't offering the antecedent as something I had calculated.
Actually, this might be a more exact definition of what you're aiming for... I think. I haven't proven it to halt, I'm getting rather tired at the moment, so you can do that if you want to use it as a more formal entry (this will also help you decide if it's what you're talking about.)
There are two important functions, the recurseing function, and the rebaseing function. The recurseing function is simple: given a base b, a function f, and a number n, call f on b n times, that is f^{n}(b). The rebaseing function is a little more subtle. Suppose you build a hierarchy of functions with some base function. This is the aim of the rebase function, it is a hierarchy based on some function f, with l layers, as a function of some number n. If you give it (+1) or (*2) or such, then those functions become the base function the rest of the hierarchy is built on. Or you give it (rebase l f) and then that will be the start of the new hierarchy. Note that the recursive step is structured so that you'll end up with things like rebase (l+1) f n = rebase l (rebase l (rebase l (...) n) n) n, n deep. That is, we recursively construct functions that redo everything we've done so far on top of everything we've done so far, including the recursively constructing functions part.
At least, that's what I think it does. Like I said, sleepy. Hopefully this will prove easier to analyse for others, while I take a nap.
Actually, this might be a more exact definition of what you're aiming for... I think. I haven't proven it to halt, I'm getting rather tired at the moment, so you can do that if you want to use it as a more formal entry (this will also help you decide if it's what you're talking about.)
Code: Select all
rebase 0 f n = f (f n)
rebase l f n = rebase (l1) (\x > recurse x (\y > rebase (l1) y) f x) n
recurse 0 f b = b
recurse n f b = f (rec (n1) f b)
There are two important functions, the recurseing function, and the rebaseing function. The recurseing function is simple: given a base b, a function f, and a number n, call f on b n times, that is f^{n}(b). The rebaseing function is a little more subtle. Suppose you build a hierarchy of functions with some base function. This is the aim of the rebase function, it is a hierarchy based on some function f, with l layers, as a function of some number n. If you give it (+1) or (*2) or such, then those functions become the base function the rest of the hierarchy is built on. Or you give it (rebase l f) and then that will be the start of the new hierarchy. Note that the recursive step is structured so that you'll end up with things like rebase (l+1) f n = rebase l (rebase l (rebase l (...) n) n) n, n deep. That is, we recursively construct functions that redo everything we've done so far on top of everything we've done so far, including the recursively constructing functions part.
At least, that's what I think it does. Like I said, sleepy. Hopefully this will prove easier to analyse for others, while I take a nap.
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Ooh, that's very neat WarDaft! Yeah, if I understand it correctly, then, that code outputs the hierarchy growth of Layers within Illusory Progressions, and all Real Level 3 does is applying that code more and more times and not just sticking to its selfrecursions.
By Real Level 4, however, what I call "stories" of layers are applied at every step, with the layers per story being increased as n increases, so an Illusory Progression that goes to "Level n" has to be reached "n" times to know the real value of the first story of Level 4 Step 1, and that goes for n stories to know the real value of Level 4 Step 1, with each story requiring much more layers than the previous one.
But before we get to that I'd really like to know how high Level 3 can get by just increasing number of layers, and it's proving difficult.
I'd like to know:
Given that the power of the number is based on the number of rebases and recursions, does it really matter if the basic, first function is very powerful or starts as "+1"? Given the way blue number progression works, a more powerful function would give access to n + ω sooner but then reaching it at 7 isn't much different from reaching it at g64 given I've equaled 1[1] to 9(xkcd)  I may just equate 1[1] to "the first blue number with a f(n) function that returns n + ω" and then reverseengineer my number to find out at which blue number 1[1] is reached.
Otherwise, it bothers me that I still have disconnected the beyond n + ω growth and the starting growth. So I guess attempting to reach n + ω in a clear way from the start would be a worthwhile effort.
So I'll start with my number, once again.
The seed: 9(xkcd)  where xkcd is A(g64, g64)
And 9(n) is a function, which is the next one:
Take n, then, turn it into uninary. Take that number as if it was decimal and change all its digits for a 9, and add n. Return a number that has that number of digits as 9's, and add n. x=n1. If x is 0, abort. If not, then, take the number you returned and feed it again into the function as the new n, but this time, continue with x=x1 until x=0.
Result:
f(1)=1_{1}, 1 changing all digits for 9=9, 9 digits=999999999+1=10^{9}. 11=0.
9(1)=10^{9}
f(2)=11_{1}, 11 changing all digits for 9=99, 99 digits=999...99 times...999+2=10^{100}+1.
21=1.
f(10^{100}+1)=111...(10^{100}+1 digits)_{1}, 111...(10^{100}+1 digits) changing all digits for 9=999...(10^{100}+1 digits), 999...(10^{100}+1 digits)+2=10^(10^{10^100+1}+1)+1
11=0.
9(2)=10^(10^{10^100+1}+1)+1
f(3)=111_{1}, 111 changing all digits for 9=999, 99 digits=999...999 times...999+3=10^{1000}+2.
31=2.
f(10^{1000}+2)=111...(10^{1000}+2 digits)_{1}, 111...(10^{1000}+2 digits) changing all digits for 9=999...(10^{1000}+2 digits), 999...(10^{1000}+2 digits)+3=10^(10^{10^1000+2}+2)+2
21=1.
f(10^(10^{10^1000+2}+2)+2)=111...(10^(10^{10^1000+2}+2)+2 digits), 111...(10^(10^{10^1000+2}+2)+2 digits) changing all digits for 9=999...(10^(10^{10^1000+2}+2)+2 digits), 999...(10^(10^{10^1000+2}+2)+2 digits) digits+3=10^(10^(10^(10^{10^1000+2}+2)+2)+2)+2
11=0.
9(3)=10^(10^(10^(10^(10^{10^1000+2}+2)+2)+2)+2)+2
On nesting: the 9() function has a peculiar behavior when nesting.
9(9(1)) would solve to 9(10^{9}), so 9(10^{10}) > 9(9(1)), this is undesirable, as the next level of recursion is meant to be more powerful than the previous one, so, it solves instead to:
9(9(1)) = 9(f(f(f(...f(10^{9})+1 times...f(f(f(10^{9})))...)))), with each f() resolving the nests normally. And the very inner nest resolving as above, i.e.:
f(10^{9})=111...(10^{9} digits), to 9s: 999...(10^{9} digits). 999...(10^{9} digits) digits+10^{9}=10^(10^{10^10})+10^9
10^{9}1
f(10^(10^{10^10})+10^9)...
And Step 1 of Level 1 starting at:
9(9(9(...9(xkcd)+1 times...9(xkcd)))...)))=x
Which creates a new function:
g(1)=9(9(9(...9(x)+1 times...9(x)))...)))
Which nests peculiarly to:
g(g(1)) = g(9(9(...9(9(9(...9(x)+1 times...9(x)))...)))+1 times...9(9(9(...9(x)+1 times...9(x)))...)))))...)))
Step 2 of Level 1 continues:
g(g(g(...g(x)+1 times...g(x)))...)))=y
Which creates a new function:
h(1)= g(g(g(...g(g(g(...g(y)+1 times...g(y)))...)))+1 times...g(g(g(...g(y)+1 times...g(y)))...)))))...)))
Which nests peculiarly to:
h(h(1))= h(g(g(...g(g(g(... g(g(g(...g(g(g(...g(y)+1 times...g(y)))...)))+1 times...g(g(g(...g(y)+1 times...g(y)))...)))))...)))+1 times... g(g(g(...g(g(g(...g(y)+1 times...g(y)))...)))+1 times...g(g(g(...g(y)+1 times...g(y)))...)))))...)))))...)))+1 times...g(g(g(...g(g(g(...g(g(g(...g(y)+1 times...g(y)))...)))+1 times...g(g(g(...g(y)+1 times...g(y)))...)))))...)))+1 times...g(g(g(...g(g(g(...g(y)+1 times...g(y)))...)))+1 times...g(g(g(...g(y)+1 times...g(y)))...)))))...)))))...)))))...)))
So besides using higher times and better functions, they also take longer and longer to get to the f() in 9().
This goes to step 9(9(9(...9(xkcd)+1 times...9(xkcd)))...)))=x to produce z, and then goes to Level z, which is the limit, and Level 2 applies Level 1 at every step.
By Real Level 4, however, what I call "stories" of layers are applied at every step, with the layers per story being increased as n increases, so an Illusory Progression that goes to "Level n" has to be reached "n" times to know the real value of the first story of Level 4 Step 1, and that goes for n stories to know the real value of Level 4 Step 1, with each story requiring much more layers than the previous one.
But before we get to that I'd really like to know how high Level 3 can get by just increasing number of layers, and it's proving difficult.
I'd like to know:
Given that the power of the number is based on the number of rebases and recursions, does it really matter if the basic, first function is very powerful or starts as "+1"? Given the way blue number progression works, a more powerful function would give access to n + ω sooner but then reaching it at 7 isn't much different from reaching it at g64 given I've equaled 1[1] to 9(xkcd)  I may just equate 1[1] to "the first blue number with a f(n) function that returns n + ω" and then reverseengineer my number to find out at which blue number 1[1] is reached.
Otherwise, it bothers me that I still have disconnected the beyond n + ω growth and the starting growth. So I guess attempting to reach n + ω in a clear way from the start would be a worthwhile effort.
So I'll start with my number, once again.
The seed: 9(xkcd)  where xkcd is A(g64, g64)
And 9(n) is a function, which is the next one:
Take n, then, turn it into uninary. Take that number as if it was decimal and change all its digits for a 9, and add n. Return a number that has that number of digits as 9's, and add n. x=n1. If x is 0, abort. If not, then, take the number you returned and feed it again into the function as the new n, but this time, continue with x=x1 until x=0.
Result:
f(1)=1_{1}, 1 changing all digits for 9=9, 9 digits=999999999+1=10^{9}. 11=0.
9(1)=10^{9}
f(2)=11_{1}, 11 changing all digits for 9=99, 99 digits=999...99 times...999+2=10^{100}+1.
21=1.
f(10^{100}+1)=111...(10^{100}+1 digits)_{1}, 111...(10^{100}+1 digits) changing all digits for 9=999...(10^{100}+1 digits), 999...(10^{100}+1 digits)+2=10^(10^{10^100+1}+1)+1
11=0.
9(2)=10^(10^{10^100+1}+1)+1
f(3)=111_{1}, 111 changing all digits for 9=999, 99 digits=999...999 times...999+3=10^{1000}+2.
31=2.
f(10^{1000}+2)=111...(10^{1000}+2 digits)_{1}, 111...(10^{1000}+2 digits) changing all digits for 9=999...(10^{1000}+2 digits), 999...(10^{1000}+2 digits)+3=10^(10^{10^1000+2}+2)+2
21=1.
f(10^(10^{10^1000+2}+2)+2)=111...(10^(10^{10^1000+2}+2)+2 digits), 111...(10^(10^{10^1000+2}+2)+2 digits) changing all digits for 9=999...(10^(10^{10^1000+2}+2)+2 digits), 999...(10^(10^{10^1000+2}+2)+2 digits) digits+3=10^(10^(10^(10^{10^1000+2}+2)+2)+2)+2
11=0.
9(3)=10^(10^(10^(10^(10^{10^1000+2}+2)+2)+2)+2)+2
On nesting: the 9() function has a peculiar behavior when nesting.
9(9(1)) would solve to 9(10^{9}), so 9(10^{10}) > 9(9(1)), this is undesirable, as the next level of recursion is meant to be more powerful than the previous one, so, it solves instead to:
9(9(1)) = 9(f(f(f(...f(10^{9})+1 times...f(f(f(10^{9})))...)))), with each f() resolving the nests normally. And the very inner nest resolving as above, i.e.:
f(10^{9})=111...(10^{9} digits), to 9s: 999...(10^{9} digits). 999...(10^{9} digits) digits+10^{9}=10^(10^{10^10})+10^9
10^{9}1
f(10^(10^{10^10})+10^9)...
And Step 1 of Level 1 starting at:
9(9(9(...9(xkcd)+1 times...9(xkcd)))...)))=x
Which creates a new function:
g(1)=9(9(9(...9(x)+1 times...9(x)))...)))
Which nests peculiarly to:
g(g(1)) = g(9(9(...9(9(9(...9(x)+1 times...9(x)))...)))+1 times...9(9(9(...9(x)+1 times...9(x)))...)))))...)))
Step 2 of Level 1 continues:
g(g(g(...g(x)+1 times...g(x)))...)))=y
Which creates a new function:
h(1)= g(g(g(...g(g(g(...g(y)+1 times...g(y)))...)))+1 times...g(g(g(...g(y)+1 times...g(y)))...)))))...)))
Which nests peculiarly to:
h(h(1))= h(g(g(...g(g(g(... g(g(g(...g(g(g(...g(y)+1 times...g(y)))...)))+1 times...g(g(g(...g(y)+1 times...g(y)))...)))))...)))+1 times... g(g(g(...g(g(g(...g(y)+1 times...g(y)))...)))+1 times...g(g(g(...g(y)+1 times...g(y)))...)))))...)))))...)))+1 times...g(g(g(...g(g(g(...g(g(g(...g(y)+1 times...g(y)))...)))+1 times...g(g(g(...g(y)+1 times...g(y)))...)))))...)))+1 times...g(g(g(...g(g(g(...g(y)+1 times...g(y)))...)))+1 times...g(g(g(...g(y)+1 times...g(y)))...)))))...)))))...)))))...)))
So besides using higher times and better functions, they also take longer and longer to get to the f() in 9().
This goes to step 9(9(9(...9(xkcd)+1 times...9(xkcd)))...)))=x to produce z, and then goes to Level z, which is the limit, and Level 2 applies Level 1 at every step.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Anyway, this is Level 5 within Illusory Progression Redux with WarDaft's fix. As long as the base remains unknown, the difference this makes is unknown as well, though.
[Illusory Progression]
Level 5 Step 1 contains: {
Level 4 Step 1 contains: {
Level 3 Step 1: n + ω^{ω} + ω
Level 3 Step 2: n + ω^{ω} + 2ω
Level 3 Step 3: n + ω^{ω} + 3ω
...
Limit of Level 3 = ω^{ω} + ω*ω}
Level 4 Step 2 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{2} + 3ω
...
Limit of Level 3 = ω^{ω} + 2ω^{2}}
Level 4 Step 3 contains: {
Level 3 Step 1: n + ω^{ω} + 2ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + 2ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + 2ω^{2} + 3ω
...
Limit of Level 3 = n + ω^{ω} + 3ω^{2}}
...
Limit of Level 4 = n + ω^{ω} + ω^{3}}
Level 5 Step 2 contains: {
Level 4 Step 1 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{3} + ω
Level 3 Step 2: n + ω^{ω} + ω^{3} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{3} + 3ω
...
Limit of Level 3 = n + ω^{ω} + ω^{3} + ω*ω}
Level 4 Step 2 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{3} + ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + ω^{3} + ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{3} + ω^{2} + 3ω
...
Limit of Level 3 = n + ω^{ω} + ω^{3} + 2ω^{2}}
Level 4 Step 3 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{3} + 2ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + ω^{3} + 2ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{3} + 2ω^{2} + 3ω
...
Limit of Level 3 = n + ω^{ω} + ω^{3} + 3ω^{2}}
...
Limit of Level 4 = n + ω^{ω} + 2ω^{3}}
Level 5 Step 3 contains: {
Level 4 Step 1 contains: {
Level 3 Step 1: n + ω^{ω} + 2ω^{3} + ω
Level 3 Step 2: n + ω^{ω} + 2ω^{3} + 2ω
Level 3 Step 3: n + ω^{ω} + 2ω^{3} + 3ω
...
Limit of Level 3 = n + ω^{ω} + 2ω^{3} + ω*ω}
Level 4 Step 2 contains: {
Level 3 Step 1: n + ω^{ω} + 2ω^{3} + ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + 2ω^{3} + ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + 2ω^{3} + ω^{2} + 3ω
...
Limit of Level 3 = n + ω^{ω} + ω^{ω} + 2ω^{3}] + 2ω^{2}}
Level 4 Step 3 contains: {
Level 3 Step 1: n + ω^{ω} + 2ω^{3} + 2ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + 2ω^{3} + 2ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + 2ω^{3} + 2ω^{2} + 3ω
...
Limit of Level 3 = n + ω^{ω} + 2ω^{3} + 3ω^{2}}
...
Limit of Level 4 = n + ω^{ω} + 3ω^{3}}
...
Limit of Level 5 = n + ω^{ω} + ω^{4}
Level 6
Level 7
Is that right? Man! The number is surely growing a lot slower
But it wasn't meant to grow so fast this soon, it was meant to grow clearly. This has a correspondence on the Levels of Level k = n + ω^{ω} + ω^{k1}
I'll repost what I did in the other post, this time with clearer results:
We're still within Illusory Space, and the question is, from the pictures where the black line turned purple, at which point does the purple line turn orange? And that is, at the point the number starts eating itself.
That is, the power of n + ω was reached at some point within the original number progression, so there's a number equivalence with this power, n. If you reach Level n, you'll have:
Level n's limit = ω^{ω} + ω^{n}
Or in other words, if the power of the fast growing hierachy is reached at x, then, there'll be some level x that outputs ω^{ω} + ω^{x}. This is where the orange line starts.
Note: this isn't really an "x", it's a finite blue number without brackets, it's unknown which number it is, but it's known a finite progression of the levels would eventually reach this power, and as the Illusory Progression is understood, it may be reached really fast, as fast as g7 or something.
Because, that's not the limit of the levels, you can reach Level x  which is equivalent to Level n + ω, as depicted by the power this level had.
Level x (n + ω) Step 1 Contains {
Level x1 (n1 + ω) Step 1 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω} + ω^{x2}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω} + 2ω^{x2}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω} + 3ω^{x2}}
Limit of Level x2 (n2 + ω): n + ω^{ω} + ω^{x1}}
Level x1 (n1 + ω) Step 2 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω} + ω^{x1} + ω^{x2}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω} + ω^{x1} + 2ω^{x2}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω} + ω^{x1} + 3ω^{x2}}
Limit of Level x2 (n2 + ω): n + ω^{ω} + 2ω^{x1}}
Level x1 (n1 + ω) Step 3 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω} + 2ω^{x1} + ω^{x2}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω} + 2ω^{x1} + 2ω^{x2}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω} + 2ω^{x1} + 3ω^{x2}}
Limit of Level x2 (n2 + ω): n + ω^{ω} + 4ω^{x1}}
Limit of Level x1 (n1 + ω) = n + ω^{ω} + ω^{x}
}
Limit of Level x (n + ω) = n + 2ω^{ω}
And then, for every step of the original progression, there's a level.
Level x + 1 (n + ω + 1) Step 1 Contains{
Level x Step 1 = 3ω^{ω}
Level x Step 2 = 4ω^{ω}
Level x Step 3 = 5ω^{ω}
...
Limit of Level x (n + ω) = n + ω^{ω + 1}}
Limit of Level x + 1 (n + ω + 1) = n + ω^{2ω}
Level x+2 (n + ω + 2) Step 1 Contains{
Level x+1 Step 1 = ω^{2ω + 1}
Level x+1 Step 2 = ω^{2ω + 2}
Level x+1 Step 3 = ω^{2ω + 3}
...
Limit of Level x+1 (n + ω + 1) = n + ω^{3ω}}
Limit of Level x+2 (n + ω + 2) = n + ω^{ω^ω}
Level x+3 (n + ω + 3) Step 1 Contains{
Level x+2 Step 1 = 2ω^{ω^ω}
Level x+2 Step 2 = 3ω^{ω^ω}
Level x+2 Step 3 = 4ω^{ω^ω}
...
Limit of Level x+2 (n + ω + 2) = n + ω^{ω^ω + 2ω}}
Limit of Level x+3 (n + ω + 3) = n + ω^{2ω^ω}
Level x+4 (n + ω + 4) Step 1 Contains{
Level x+3 Step 1 = 2ω^{2ω^ω}
Level x+3 Step 2 = 3ω^{2ω^ω}
Level x+3 Step 3 = 4ω^{2ω^ω}
...
Limit of Level x+3 (n + ω + 3) = n + ω^{(2ω+1)^ω}}
Limit of Level x+4 (n + ω + 4) = n + ω^{3ω^ω}
Level x+5 (n + ω + 5) Step 1 Contains{
Level x+4 Step 1 = 2ω^{3ω^ω}
Level x+4 Step 2 = 3ω^{3ω^ω}
Level x+4 Step 3 = 4ω^{3ω^ω}
...
Limit of Level x+4 (n + ω + 4) = n + ω^{(3ω+1)^ω}}
Limit of Level x+5 (n + ω + 5) = n + ω^{4ω^ω}
...
Level x*2 (n + 2ω) Step 1 Contains{
Level (x*2)1 Step 1 = 2ω^{xω^ω}
Level (x*2)1 Step 2 = 3ω^{xω^ω}
Level (x*2)1 Step 3 = 4ω^{xω^ω}
...
Limit of Level (x*2)1 (n + ω + x) = n + ω^{ω^ω + 1}}
Limit of Level x*2 (n + 2ω) = n + ω^{ω^2ω}
Limit of Level x*3 (n + 3ω) = n + ω^{ω^3ω}
Limit of Level x*4 (n + 3ω) = n + ω^{ω^4ω}
...
Limit of Level x^{2} (n + ω^{2}) = n + ω^{ω^ω^2}
Limit of Level x^{3} (n + ω^{3}) = n + ω^{ω^ω^3}
Limit of Level x^{4} (n + ω^{4}) = n + ω^{ω^ω^4}
...
Limit of Level x^{x} (n + ω^{ω}) = n + ω^{ω^ω^ω}
[/Illusory Progression]
Which is really the value and power of Level 3 Step 1.
I'll stop there to make sure this is clear, because real Level 3 Step 2 does not apply Level 2 at every step like this is doing, it'll apply Level 3 Step 1 at every step with the limit of the steps not being x anymore, but x^{x}, so, instead of stopping at Step ω, they go all the way to step ω^{ω^ω^ω}.
[Illusory Progression]
Level 5 Step 1 contains: {
Level 4 Step 1 contains: {
Level 3 Step 1: n + ω^{ω} + ω
Level 3 Step 2: n + ω^{ω} + 2ω
Level 3 Step 3: n + ω^{ω} + 3ω
...
Limit of Level 3 = ω^{ω} + ω*ω}
Level 4 Step 2 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{2} + 3ω
...
Limit of Level 3 = ω^{ω} + 2ω^{2}}
Level 4 Step 3 contains: {
Level 3 Step 1: n + ω^{ω} + 2ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + 2ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + 2ω^{2} + 3ω
...
Limit of Level 3 = n + ω^{ω} + 3ω^{2}}
...
Limit of Level 4 = n + ω^{ω} + ω^{3}}
Level 5 Step 2 contains: {
Level 4 Step 1 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{3} + ω
Level 3 Step 2: n + ω^{ω} + ω^{3} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{3} + 3ω
...
Limit of Level 3 = n + ω^{ω} + ω^{3} + ω*ω}
Level 4 Step 2 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{3} + ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + ω^{3} + ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{3} + ω^{2} + 3ω
...
Limit of Level 3 = n + ω^{ω} + ω^{3} + 2ω^{2}}
Level 4 Step 3 contains: {
Level 3 Step 1: n + ω^{ω} + ω^{3} + 2ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + ω^{3} + 2ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + ω^{3} + 2ω^{2} + 3ω
...
Limit of Level 3 = n + ω^{ω} + ω^{3} + 3ω^{2}}
...
Limit of Level 4 = n + ω^{ω} + 2ω^{3}}
Level 5 Step 3 contains: {
Level 4 Step 1 contains: {
Level 3 Step 1: n + ω^{ω} + 2ω^{3} + ω
Level 3 Step 2: n + ω^{ω} + 2ω^{3} + 2ω
Level 3 Step 3: n + ω^{ω} + 2ω^{3} + 3ω
...
Limit of Level 3 = n + ω^{ω} + 2ω^{3} + ω*ω}
Level 4 Step 2 contains: {
Level 3 Step 1: n + ω^{ω} + 2ω^{3} + ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + 2ω^{3} + ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + 2ω^{3} + ω^{2} + 3ω
...
Limit of Level 3 = n + ω^{ω} + ω^{ω} + 2ω^{3}] + 2ω^{2}}
Level 4 Step 3 contains: {
Level 3 Step 1: n + ω^{ω} + 2ω^{3} + 2ω^{2} + ω
Level 3 Step 2: n + ω^{ω} + 2ω^{3} + 2ω^{2} + 2ω
Level 3 Step 3: n + ω^{ω} + 2ω^{3} + 2ω^{2} + 3ω
...
Limit of Level 3 = n + ω^{ω} + 2ω^{3} + 3ω^{2}}
...
Limit of Level 4 = n + ω^{ω} + 3ω^{3}}
...
Limit of Level 5 = n + ω^{ω} + ω^{4}
Level 6
Spoiler:
Level 7
Spoiler:
Is that right? Man! The number is surely growing a lot slower
But it wasn't meant to grow so fast this soon, it was meant to grow clearly. This has a correspondence on the Levels of Level k = n + ω^{ω} + ω^{k1}
I'll repost what I did in the other post, this time with clearer results:
We're still within Illusory Space, and the question is, from the pictures where the black line turned purple, at which point does the purple line turn orange? And that is, at the point the number starts eating itself.
That is, the power of n + ω was reached at some point within the original number progression, so there's a number equivalence with this power, n. If you reach Level n, you'll have:
Level n's limit = ω^{ω} + ω^{n}
Or in other words, if the power of the fast growing hierachy is reached at x, then, there'll be some level x that outputs ω^{ω} + ω^{x}. This is where the orange line starts.
Note: this isn't really an "x", it's a finite blue number without brackets, it's unknown which number it is, but it's known a finite progression of the levels would eventually reach this power, and as the Illusory Progression is understood, it may be reached really fast, as fast as g7 or something.
Because, that's not the limit of the levels, you can reach Level x  which is equivalent to Level n + ω, as depicted by the power this level had.
Level x (n + ω) Step 1 Contains {
Level x1 (n1 + ω) Step 1 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω} + ω^{x2}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω} + 2ω^{x2}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω} + 3ω^{x2}}
Limit of Level x2 (n2 + ω): n + ω^{ω} + ω^{x1}}
Level x1 (n1 + ω) Step 2 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω} + ω^{x1} + ω^{x2}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω} + ω^{x1} + 2ω^{x2}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω} + ω^{x1} + 3ω^{x2}}
Limit of Level x2 (n2 + ω): n + ω^{ω} + 2ω^{x1}}
Level x1 (n1 + ω) Step 3 Contains {
Level x2 (n2 + ω) Step 1 Contains {n + ω^{ω} + 2ω^{x1} + ω^{x2}}
Level x2 (n2 + ω) Step 2 Contains {n + ω^{ω} + 2ω^{x1} + 2ω^{x2}}
Level x2 (n2 + ω) Step 3 Contains {n + ω^{ω} + 2ω^{x1} + 3ω^{x2}}
Limit of Level x2 (n2 + ω): n + ω^{ω} + 4ω^{x1}}
Limit of Level x1 (n1 + ω) = n + ω^{ω} + ω^{x}
}
Limit of Level x (n + ω) = n + 2ω^{ω}
And then, for every step of the original progression, there's a level.
Level x + 1 (n + ω + 1) Step 1 Contains{
Level x Step 1 = 3ω^{ω}
Level x Step 2 = 4ω^{ω}
Level x Step 3 = 5ω^{ω}
...
Limit of Level x (n + ω) = n + ω^{ω + 1}}
Limit of Level x + 1 (n + ω + 1) = n + ω^{2ω}
Level x+2 (n + ω + 2) Step 1 Contains{
Level x+1 Step 1 = ω^{2ω + 1}
Level x+1 Step 2 = ω^{2ω + 2}
Level x+1 Step 3 = ω^{2ω + 3}
...
Limit of Level x+1 (n + ω + 1) = n + ω^{3ω}}
Limit of Level x+2 (n + ω + 2) = n + ω^{ω^ω}
Level x+3 (n + ω + 3) Step 1 Contains{
Level x+2 Step 1 = 2ω^{ω^ω}
Level x+2 Step 2 = 3ω^{ω^ω}
Level x+2 Step 3 = 4ω^{ω^ω}
...
Limit of Level x+2 (n + ω + 2) = n + ω^{ω^ω + 2ω}}
Limit of Level x+3 (n + ω + 3) = n + ω^{2ω^ω}
Level x+4 (n + ω + 4) Step 1 Contains{
Level x+3 Step 1 = 2ω^{2ω^ω}
Level x+3 Step 2 = 3ω^{2ω^ω}
Level x+3 Step 3 = 4ω^{2ω^ω}
...
Limit of Level x+3 (n + ω + 3) = n + ω^{(2ω+1)^ω}}
Limit of Level x+4 (n + ω + 4) = n + ω^{3ω^ω}
Level x+5 (n + ω + 5) Step 1 Contains{
Level x+4 Step 1 = 2ω^{3ω^ω}
Level x+4 Step 2 = 3ω^{3ω^ω}
Level x+4 Step 3 = 4ω^{3ω^ω}
...
Limit of Level x+4 (n + ω + 4) = n + ω^{(3ω+1)^ω}}
Limit of Level x+5 (n + ω + 5) = n + ω^{4ω^ω}
...
Level x*2 (n + 2ω) Step 1 Contains{
Level (x*2)1 Step 1 = 2ω^{xω^ω}
Level (x*2)1 Step 2 = 3ω^{xω^ω}
Level (x*2)1 Step 3 = 4ω^{xω^ω}
...
Limit of Level (x*2)1 (n + ω + x) = n + ω^{ω^ω + 1}}
Limit of Level x*2 (n + 2ω) = n + ω^{ω^2ω}
Limit of Level x*3 (n + 3ω) = n + ω^{ω^3ω}
Limit of Level x*4 (n + 3ω) = n + ω^{ω^4ω}
...
Limit of Level x^{2} (n + ω^{2}) = n + ω^{ω^ω^2}
Limit of Level x^{3} (n + ω^{3}) = n + ω^{ω^ω^3}
Limit of Level x^{4} (n + ω^{4}) = n + ω^{ω^ω^4}
...
Limit of Level x^{x} (n + ω^{ω}) = n + ω^{ω^ω^ω}
[/Illusory Progression]
Which is really the value and power of Level 3 Step 1.
I'll stop there to make sure this is clear, because real Level 3 Step 2 does not apply Level 2 at every step like this is doing, it'll apply Level 3 Step 1 at every step with the limit of the steps not being x anymore, but x^{x}, so, instead of stopping at Step ω, they go all the way to step ω^{ω^ω^ω}.
Re: My number is bigger!
I'm still not sure that I'm certain how illusory space works, but it looks like you're getting a good grasp of ordinal notation so I'll take your word for the above matching, it looks reasonable at first glance, I'll take a closer look later.
And to answer how it might matter whether you start with +1 as your function or something else, the best answer is that IF it actually matters, that's a bad sign about the rest of your notation. You should end up in basically the same place no matter what you start with, otherwise you're not really adding much to whatever you started with.
And to answer how it might matter whether you start with +1 as your function or something else, the best answer is that IF it actually matters, that's a bad sign about the rest of your notation. You should end up in basically the same place no matter what you start with, otherwise you're not really adding much to whatever you started with.
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
WarDaft wrote:I'm still not sure that I'm certain how illusory space works
I'm just basically showing some progression and then making it so that next step of that progression is really what you'd expect to find by going to the limits of that progression many times. For instance, without Illusory space, the limit of Level 3 Step 1 was n + 3ω, with it I can reach ω^{(ω^(ω^ω))} instead.
The reason I had to come up with it is that, previously I was showing a progression like:
Level 1's limit: n + ω
Level 2's limit: n + 2ω
Level 3's limit: n + ω^{2}
Level 4's limit: n + ω^{ω}
I was getting complaints that I couldn't do it like that, because I can't jump from 2ω to ω^{2} and then to ω^{ω}. That was true due to the implicit limits of the level's steps, so I thought that "branching off" progressions that went over those limits was the solution, and since those levels and steps don't actually exist, I thought about saying they happened in "imaginary space", which later was changed to illusory space to avoid confusion.
Illusory space ensures that I can jump from ω^{2} to ω^{ω} in one level, and not only that, it turns out I can jump from ω^{2} to ω^{(ω^(ω^ω))} in one step, because all intermediate values between those happened within illusory space.
Yes, I think I can reach the power of ε_{0} within the levels, I don't even have computed what's the real limit of Level 3, but limit 4 step 1 is going to eat it for lunch just like ω^{(ω^(ω^ω))} is eating 2ω here. It's still meaningless because my number's base hasn't been defined yet, but I hope it will, eventually.
WarDaft wrote:And to answer how it might matter whether you start with +1 as your function or something else, the best answer is that IF it actually matters, that's a bad sign about the rest of your notation. You should end up in basically the same place no matter what you start with, otherwise you're not really adding much to whatever you started with.
Thanks. Welp! So it turns out I wasted my time improving the 9() function since the power of my number should be in the number of recursions. So I should start as weak as possible, I mean, within reason (I don't think starting at f(n)=n+(n/xkcd) should be necessary...)
So I'll start from scratch with the +1 function and a 1 seed. I guess it'll start hilariously slow now since the number of steps and levels depends on numbers reached. Back when I started with f(n)=n*2 I could reach f(n)=n^(16^(n^(16^{65536}))) fairly soon, but now that'll be much harder...
Seed: 1
Level 1 Step 1:
f(1)=2
And that's the only step
Okay, so now, Level 2 will create a new function based on the first. What happened was 1+1, so the function is rebased to n+n. Or, n*2.
Level 2 Step 1:
f(2)=4
Level 2 Step 2:
f(4)=8
That's the maximum number of steps... Level 3 will create a new function, what happened was (1+1)*2*2, or (n+n)*(n+n)*(n+n), or (n*2)^{3}, which is the new base.
Level 3 introduces [Illusory Space]
First, start a counter, this counter is 8 (highest number reached).
Level 3 Step 1:
f(8)=8*2=16. 16^{3} = 2^{12}
Level 3 Step 2:
f(2^{12}) = 2^{12}*2= 2^{13}. (2^{13})^3 = 2^{39}
Level 3 Step 3:
f(2^{39})= 2^{39}*2 = 2^{40}. (2^{13})^3 = 2^{120}
Level 3 Step 4:
f(2^{120})=2^{363}
Level 3 Step 5:
f(2^{363})=2^{1092}
Level 3 Step 6:
f(2^{1092})=2^{3279}
Level 3 Step 7:
f(2^{3279})=2^{9840}
Level 3 Step 8:
f(2^{9840})=2^{29523}
Step (highest level reached in Level 2) was reached. Decrease counter by 1 (counter: 7). Rebase: (((((((((((((((((n2^3)*2)^3)*2)^3)*2)^3)*2)^3)*2)^3)*2)^3)*2)^3)*2)^3) = ((((((((((((((((((n*(n+n))^(n3))*(n+n))^(n3))*(n+n))^(n3))*(n+n))^(n3))*(n+n))^(n3))*(n+n))^(n3))*(n+n))^(n3))*(n+n))^(n3))*(n+n))^(n3))
Go to step 2^{29523} before decreasing counter.
Level 3 Step 9:
f(2^{29523}) = ((((((((((((((((((2^{59047}))^(3*(2^{29523}) ))*(2^{29524} ))^(3*(2^{29523}) ))*(2^{29524} ))^(3*(2^{29523}) ))*(2^{29524} ))^(3*(2^{29523}) ))*(2^{29524} ))^(3*(2^{29523}) ))*(2^{29524} ))^(3*(2^{29523}) ))*(2^{29524} ))^(3*(2^{29523}) ))*(2^{29524} ))^(3*(2^{29523}) ))*(2^{29524} ))^(3*(2^{29523}) ))
Level 3 Step 10:
Spoiler:
Anyway, reach Step 2^{29523}, get a "big number", rebase, counter is 6. Reach Step "big number", get "bigger number", rebase, counter is 5. Reach Step "bigger number", and so on.
When counter reaches 0, you leave [/Illusory Space], with the real value and function base of Level 3 Step 1, and for Level 3 Step 2, you open a big [Illusory Space] for the entire step, and one small [Illusory Space] for every Illusory level and step in the progression, and a layer of counters that keep growing every time you meet them. When you've closed all [/Illusory Space]s and all counters have reached 0, you have the true value of Level 3 Step 2.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
*Sigh* I don't like how my number works, it becomes too complex too quick, but I have to blame math itself for it. I mean, there's no way to simplify (2^9)*3? It says one has to calculate them and then write the number down? Yeah, that's easy, 1536. But when you are working with (12345^67890)*3, then it's not pretty anymore.
So I'll go back to an idea I had and ditched previously in this thread: Inventing my own math signs are their rules. Of course, they'll be the blue signs.
They allow to deal with problems as the above in a tidy manner and are more powerful than normal signs, but that's not the reason I'm switching to them, the reason is because I couldn't even write out the number of Level 3 Step 11 on my previous attempt. So much for clarity...
So here it goes, the + sign works like this:
n^{k} + x^{y} = (n+x)^{k+y}
Simple enough. Now I can show numbers clearly:
Seed: 1
Function: f(n)=n+1
Level 1 Step 1:
f(1)=2
Rebase: 1+1=2 = n+n=n*2. Turn the sign blue.
f(n)=n*2
Level 2 applies Level 1 at every step with new base.
Level 2 Step 1:
{
Level 1 Step 1:
f(2)=2*2. 2^{1}+2^{1}=4^{2}=16 (2^{4})
Level 1 Step 2:
f(2^{4})=2^{4}*2. 2^{4}+2^{4}=4^{8} (2^{16})}
Rebase: (1+1)^{16} = (n*2)^{16}. Turn it blue.
f(n)=(n*2)^^{16}
Go to step 2^{16}.
Level 2 Step 2:
{Level 1 Step 1:
f(2^{16})=(2^{16}*2)^16. (2^{16}+2^{16})=4^{32}^16 (2^{64})^^{16}.
(2^{64}*2^{64})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
(2^{64}+2^{64}...[2^{64} times])*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{64*128^64})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{128^64+128^64 [64 times]})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{8192^4096})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((2^{2^2^2^15})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((2^{2^2^2^15}+2^{2^2^2^15} [2^{64} times])*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{64*128^128^128^61440})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{128^128^128^61440+128^128^128^61440 [64 times]})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{8192^8192^8192^3932160})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
...}
And, wow! I keep getting stuck in big number calculation anyway, I need a way to simplify this...
So I'll go back to an idea I had and ditched previously in this thread: Inventing my own math signs are their rules. Of course, they'll be the blue signs.
They allow to deal with problems as the above in a tidy manner and are more powerful than normal signs, but that's not the reason I'm switching to them, the reason is because I couldn't even write out the number of Level 3 Step 11 on my previous attempt. So much for clarity...
So here it goes, the + sign works like this:
n^{k} + x^{y} = (n+x)^{k+y}
Simple enough. Now I can show numbers clearly:
Seed: 1
Function: f(n)=n+1
Level 1 Step 1:
f(1)=2
Rebase: 1+1=2 = n+n=n*2. Turn the sign blue.
f(n)=n*2
Level 2 applies Level 1 at every step with new base.
Level 2 Step 1:
{
Level 1 Step 1:
f(2)=2*2. 2^{1}+2^{1}=4^{2}=16 (2^{4})
Level 1 Step 2:
f(2^{4})=2^{4}*2. 2^{4}+2^{4}=4^{8} (2^{16})}
Rebase: (1+1)^{16} = (n*2)^{16}. Turn it blue.
f(n)=(n*2)^^{16}
Go to step 2^{16}.
Level 2 Step 2:
{Level 1 Step 1:
f(2^{16})=(2^{16}*2)^16. (2^{16}+2^{16})=4^{32}^16 (2^{64})^^{16}.
(2^{64}*2^{64})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
(2^{64}+2^{64}...[2^{64} times])*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{64*128^64})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{128^64+128^64 [64 times]})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{8192^4096})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((2^{2^2^2^15})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((2^{2^2^2^15}+2^{2^2^2^15} [2^{64} times])*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{64*128^128^128^61440})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{128^128^128^61440+128^128^128^61440 [64 times]})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
((4^{8192^8192^8192^3932160})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
...}
And, wow! I keep getting stuck in big number calculation anyway, I need a way to simplify this...
Re: My number is bigger!
There's simply no point in saying (2^9)*3. Just say x^y. Eventually you'll get to the point where saying x^y^....^z is also pointless, and you just simplify to x^^n because the individual numbers are just so unimportant compared to the height of the stack. 100000000000000000000^^100 is less than 2^^105, so the 100000000000000000000 becomes pointless to type out. You especially don't need to know each and every digit if they aren't all zero. Once you get to the next arrow, 3^^^(n+2) will be larger than any k^^^n where it actually makes sense to specify k^^^n rather than just k, because 3^^^(n+2) = 3^^^n^^3^^3 ~ 3^^^n^^7500000000000 = 3^^^(n1)^^3^^7500000000000 = 3^^^(n1)^^3^3^3^... 7500000000000 later ... ^3^3, which is to say, you start with 3, then you jump to a number with approximately 3 digits and call that x_1, then you jump to a number with approximately x_1 digits, call that x_2, then you jump to a number with approximately x_3 digits and call that x_4, then where you repeat that sentence about seven and a half trillion times. Do not get caught up in fiddly bits, they are not important.
I really should be doing homework, but I can't get an idea I had for multidimensional hydras out of my head. But if I don't get this compiler done... it will be very bad =/
I really should be doing homework, but I can't get an idea I had for multidimensional hydras out of my head. But if I don't get this compiler done... it will be very bad =/
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Oh, thanks WarDaft. So I've been wasting valuable time on precision when those numbers aren't going to matter in the future. So, if I have something like 2^{3932160}, it only matters that is bigger than 2^2^21 and I don't really care about what is cut off due to its insignificance.
I think what I can do is, instead of using = signs as I've been doing, I'll switch to > signs, so it's known I pass some value at some point, and it doesn't matter by how much.
Define +: n^{k} + x^{y} = (n+x)^{k+y}
Seed: 1
Function: f(n)=n+1
Level 1 Step 1:
f(1)=2
Rebase: 1+1=2 = n+n=n*2. Turn the sign blue.
f(n)=n*2
Level 2 applies Level 1 at every step with new base.
Level 2 Step 1:
{
Level 1 Step 1:
f(2)=2*2. 2^{1}+2^{1}=4^{2}=16 (2^{4})
Level 1 Step 2:
f(2^{4})=2^{4}*2. 2^{4}+2^{4}=4^{8} (2^{16})}
Rebase: (1+1)^{16} = (n*2)^{16}. Turn it blue.
f(n)=(n*2)^^{16}}
Level 2 Step 2:
{Level 1 Step 1:
f(2^{16})=(2^{16}*2)^16. (2^{16}+2^{16})=4^{32}^16 (2^{64})^^{16}.
(2^{64}*2^{64})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
Each multiplication adds one ^ more than the last one, so: 2+3+4+...+62+63+64, or 2^2^2...2^10 + 2^9 times...^2^2^64.
f(2^{16})>2↑↑10
Level 1 Step 2:
f(2↑↑10)=(2↑↑10*2)^16)=(2↑↑10+2↑↑10)^16)=4↑↑20^16 (2↑↑21^16)
(2↑↑21*2↑↑21)*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21
(2↑↑21+2↑↑21 [2↑↑21 times])*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21
(2↑↑↑(2↑↑21))*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21
(2↑↑↑↑(2↑↑↑21(2↑↑21)))*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21
f(2↑↑10)>2↑^{18}2↑^{17}...
Level 1 Step 3:
f(2↑^{18}2↑^{17}...)=(2↑^{18}2↑^{17}...*2)^16)=(2↑^{18}2↑^{17}...+2↑^{18}2↑^{17}...)^16)=4↑^{36}4↑^{34}...^16
(4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...)*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...
(4↑^{36}4↑^{34}...+4↑^{36}4↑^{34}... [4↑^{36}4↑^{34}... times])*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...
(4↑^{36↑^36↑34...})*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...
f(2↑^{18}2↑^{17}...)>4→36→34→2
Level 1 Step 4:
f(4→36→34→2)=(4→36→34→2*2)^16
(8→72→68→4*8→72→68→4)*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4
(8→72→68→4+8→72→68→4 [8→72→68→4 times])*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4
(8→72→68→32→288→560→16)*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4
f(4→36→34→2)>3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3
Level 1 Step 5:
f(3→^{52}3)=(3→^{52}3*2)^16
(6→^{104}6*6→^{104}6)*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6
(6→^{104}6+6→^{104}6 [6→^{104}6 times])*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6
(6→^{624→10816→624}6)*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6
...
Go to step 2^16}
Is this growing clearly? With every step being this much powerful than the last, I find it hard to see what's at Level 1 Step 2^16 (which is the limit of Level 2):
I think what I can do is, instead of using = signs as I've been doing, I'll switch to > signs, so it's known I pass some value at some point, and it doesn't matter by how much.
Define +: n^{k} + x^{y} = (n+x)^{k+y}
Seed: 1
Function: f(n)=n+1
Level 1 Step 1:
f(1)=2
Rebase: 1+1=2 = n+n=n*2. Turn the sign blue.
f(n)=n*2
Level 2 applies Level 1 at every step with new base.
Level 2 Step 1:
{
Level 1 Step 1:
f(2)=2*2. 2^{1}+2^{1}=4^{2}=16 (2^{4})
Level 1 Step 2:
f(2^{4})=2^{4}*2. 2^{4}+2^{4}=4^{8} (2^{16})}
Rebase: (1+1)^{16} = (n*2)^{16}. Turn it blue.
f(n)=(n*2)^^{16}}
Level 2 Step 2:
{Level 1 Step 1:
f(2^{16})=(2^{16}*2)^16. (2^{16}+2^{16})=4^{32}^16 (2^{64})^^{16}.
(2^{64}*2^{64})*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}*2^{64}
Each multiplication adds one ^ more than the last one, so: 2+3+4+...+62+63+64, or 2^2^2...2^10 + 2^9 times...^2^2^64.
f(2^{16})>2↑↑10
Level 1 Step 2:
f(2↑↑10)=(2↑↑10*2)^16)=(2↑↑10+2↑↑10)^16)=4↑↑20^16 (2↑↑21^16)
(2↑↑21*2↑↑21)*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21
(2↑↑21+2↑↑21 [2↑↑21 times])*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21
(2↑↑↑(2↑↑21))*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21
(2↑↑↑↑(2↑↑↑21(2↑↑21)))*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21*2↑↑21
f(2↑↑10)>2↑^{18}2↑^{17}...
Level 1 Step 3:
f(2↑^{18}2↑^{17}...)=(2↑^{18}2↑^{17}...*2)^16)=(2↑^{18}2↑^{17}...+2↑^{18}2↑^{17}...)^16)=4↑^{36}4↑^{34}...^16
(4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...)*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...
(4↑^{36}4↑^{34}...+4↑^{36}4↑^{34}... [4↑^{36}4↑^{34}... times])*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...
(4↑^{36↑^36↑34...})*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...*4↑^{36}4↑^{34}...
f(2↑^{18}2↑^{17}...)>4→36→34→2
Level 1 Step 4:
f(4→36→34→2)=(4→36→34→2*2)^16
(8→72→68→4*8→72→68→4)*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4
(8→72→68→4+8→72→68→4 [8→72→68→4 times])*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4
(8→72→68→32→288→560→16)*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4*8→72→68→4
f(4→36→34→2)>3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3→3
Level 1 Step 5:
f(3→^{52}3)=(3→^{52}3*2)^16
(6→^{104}6*6→^{104}6)*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6
(6→^{104}6+6→^{104}6 [6→^{104}6 times])*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6
(6→^{624→10816→624}6)*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6*6→^{104}6
...
Go to step 2^16}
Is this growing clearly? With every step being this much powerful than the last, I find it hard to see what's at Level 1 Step 2^16 (which is the limit of Level 2):
Re: My number is bigger!
How do you differentiate between:Vytron wrote:Define +: n^{k} + x^{y} = (n+x)^{k+y}
16^{1} + 16^{1} = 32^{2}
4^{2} + 4^{2} = 8^{4}
Because clearly, 32^{2} =/= 8^{4}
Maybe require that n^{k} and x^{y} be written with the smallest n and x possible before the operation is allowed to take place? Not sure if this would mess up other parts of your logic, cause most of it goes straight over my head.
Dashboard Confessional wrote:I want to give you whatever you need. What is it you need? Is it within me?
Avatar by Matt
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Snark wrote:Because clearly, 32^{2} =/= 8^{4}
Maybe require that n^{k} and x^{y} be written with the smallest n and x possible before the operation is allowed to take place?
Erm, yeah, they're meant to take the highest k equivalence before the operation is carried out. I stopped doing it in the calculations because at some point, as WarDaft said, it doesn't matter to use precise values. The number appears smaller than it actually is, but that doesn't matter in the long run.
For instance, at the end of Level 1 Step 5, you end with a number much bigger than:
_{3→}3→^{3→3→624→3}→3_{→3}
(because that 624 sould actually be a number with ↑'s followed by other huge numbers with ↑'s, and we're just cropping the whole thing)
But much smaller than:
_{3→}3→^{3→3→624→3}→3_{→4}
So it suffices to say that my number is on there somewhere, and it doesn't matter that the number shown is much smaller than the actual number because by that point we don't know exactly how big is it anyway (knowing exactly how big is very time consuming, so bounding sames a lot of time).
E.g. by the difference of magnitude the number has at this point, 32^{2} ≈ 8^{4}, because:
_{3→}3→^{3→3→624→3}→3_{→3} + 32^{2}
And
_{3→}3→^{3→3→624→3}→3_{→3} + 8^{4}
or even:
_{3→}3→^{3→3→624→3}→3_{→3} + 3→^{3→3→624→3}→3
Will result in a number that is:
<_{3→}3→^{3→3→624→3}→3_{→4}
So, 32^{2} ≈ 3→^{3→3→624→3}→3 at this magnitude difference, like when ω2=ω.
Snark wrote:Not sure if this would mess up other parts of your logic
Even if the number was reduced to the lowest possible n and x before carrying out operations, the number at the end of Level 1 Step 5 would be much smaller than _{3→}3→^{3→3→624→3}→3_{→4} (which is much smaller than _{3→}3→^{3→3→624→3}→4_{→3}, which is much smaller than _{3→}3→^{3→3→625→3}→3_{→3} ), so reduction makes no difference in the long run.
Snark wrote:cause most of it goes straight over my head.
I had to resort to narrow operators stacked on conway notation to show my number.
3→^{52}3 is 3→3...52 times...→3→3
_{3→}3→^{52}3_{3} is 3→3→...3→^{52}3 times...
_{3→}3→^{3→3→624→3}→3_{→3} actually has a ^{3→3→624→3} for every arrow on the floor below, cropped for clarity, but that doesn't matter.
It gives me an idea, though!
We can define a tower like that as →^{*n} so that:
_{3→}3→^{3→3→624→3}→3_{→3} > 3→^{*3}624→^{*3}→3 (and < 3→^{*3}624→^{*3}→4)
Now, the value of Step 1 Level 6 should exceed:
3→^{*6}1872→^{*6}→3
And the number finally plateaus on its growth!
Step 1 Level 7:
3→^{*18}5616→^{*18}→3
Step 1 Level 8:
3→^{*54}(2^12)→^{*54}→3
Which bounds the value of Step 1 Level 2^16 to:
3→^{*(2^40)}(2^480)→^{*(2^40)}→3
(A tower of _{→}→^{→} with 2^40 elements).
And that's the limit.
So now, the next function will be:
f(n)=n3→^{*(n2^n40)}(n2^n480)→^{*(n2^n40)}→n3, turned blue.
[Illusory Space]
Level 3 Step 1 {
Level 2 Step 1 {
Level 1 Step 1 {
f(3→^{*(2^40)}(2^480)→^{*(2^40)}→3)=
n=3→^{*(2^40)}(2^480)→^{*(2^40)}→3
(3*n)→^{*((2*n)^(40*n))}((2*n)^(480*n))→^{*((2*n)^(40*n))}→(3*n)
...}
Um, perhaps this needs a different kind of notation...

 Posts: 6
 Joined: Sun Jan 20, 2013 5:20 am UTC
Re: My number is bigger!
Just a brief note: Using ZFC+10 will not get you close to my number, because "ZFC + a large cardinal axiom" is vastly stronger with vastly more reflectivity than ZFC + Con(ZFC) + Con(Con(ZFC)) for 10 iterations, or [imath]ZFC + \omega[/imath], etc. That is, ZFC plus a large cardinal axiom would prove all of those systems consistent (and [imath]\omega[/imath]consistent which is more to the point). Entries which only use base ZFC without the largest known large cardinal axiom are blown entirely out of the water thereby. Despite this, checking whether something can be proven in ZFC+(a large cardinal axiom) in n steps is a fully computable operation, so my number is still quite computable.
Rayo's number is not computable so is not eligible for this contest. Rayo's number talks about what is true in ZF, not just what is provable in it.
Rayo's number is not computable so is not eligible for this contest. Rayo's number talks about what is true in ZF, not just what is provable in it.
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Anyway, I have no idea how big is
(3*n)→^{*((2*n)^(40*n))}((2*n)^(480*n))→^{*((2*n)^(40*n))}→(3*n)
But I can bound it.
It's much higher than
n=3→^{*(2^40)}(2^480)→^{*(2^40)}→3
3→^{*(n)}(n)→^{*(n)}→3
But if a new sign was introduced that said how many towers of →^{*}'s were stacked on each other so that
_{3→}*3^{*3→...}..._{...} = 3→^^{2}...
Then, the number would be much lower than:
3→^{^(n)}(n)→^{^(n)}→3
So, x, the number at [Illusory Space] Level 3 Step 1 { Level 2 Step 1 {
Level 1 Step 1 { }
Will be:
3→^{*(n)}(n)→^{*(n)}→3 < x < 3→^{^(n)}(n)→^{^(n)}→3
For [Illusory Space] Level 3 Step 1 { Level 2 Step 1 {
Level 1 Step 1 { }, the process is repeated for:
(3*x)→^{*((2*x)^(40*x))}((2*x)^(480*x))→^{*((2*x)^(40*x))}→(3*x)
And the process is repeated:
It's much higher than
n=3→^{*(x)}(x)→^{*(x)}→3
3→^{^(n)}(n)→^{^(n)}→3
But if a new sign was introduced that said how many towers of →^{^}'s were stacked on each other so that
_{3→}^3^{^3→...}..._{...} = 3→↑^{2}...
Then, the number would be much lower than:
3→^{↑(n)}(n)→^{↑(n)}→3
So, x, the number at [Illusory Space] Level 3 Step 1 { Level 2 Step 1 {
Level 1 Step 2 { }
Will be:
3→^{^(n)}(n)→^{^(n)}→3 < x < 3→^{↑(n)}(n)→^{↑(n)}→3
And so on.
Every step introduces a new sign, and it goes to Level 1 Step (3*n)→^{*((2*n)^(40*n))}((2*n)^(480*n))→^{*((2*n)^(40*n))}→(3*n), or, the 3→^{*(n)}(n)→^{*(n)}→3th sign. Let's call that x.
I can specify the kth number of the sign like this:
3→^{(x)}x→^{(x)}3, and rebase to:
3→^{(x)}x→^{(x)}3
So, x, the number at [Illusory Space] Level 3 Step 1 { Level 2 Step 2 {
Level 1 Step 1 { } will be:
f(n)=_{3→}(3→^{(n)}n→^{(n)}3)_{3→}^{(n)}n→^{(n)}_{3→}(3→^{(n)}n→^{(n)}3)_{3}
Which can be simplified to:
3→^{((x))}x→^{((x))}3
So that every step adds a new parenthesis nest. This goes up to Step 3→^{(x)}x→^{(x)}3
So it will have:
_{3→}(((... 3→^{((x))}x→^{((x))}3 times...(((x)))... 3→^{((x))}x→^{((x))}3 times...)))_{x→}(((... 3→^{((x))}x→^{((x))}3 times...(((x)))... 3→^{((x))}x→^{((x))}3 times...)))_{3}
I can specify the kth number of nested parenthesis like this:
3→^{[n]}3 = 3→^{(((...n times...(((x)))... n times...)))}x→^{(((... n times...(((x)))... n times ...)))}3
So the current highest number is:
n=3→^{((x))}x→^{((x))}3
3→^{[n]}3
So I rebase to:
f(n)=3→^{[n]}3
So, x, the number at [Illusory Space] Level 3 Step 1 { Level 2 Step 3 {
Level 1 Step 1 { } will be:
n=3→^{((x))}x→^{((x))}3
x=3→^{[n]}3
f(x)=3→^{[x]}3
Which can be simplified to:
3→^{[[x]]}x→^{[[x]]}3
Now, every Step does that, and goes up to Step 3→^{[[x]]}x→^{[[x]]}3 so in the (), [], <> progression of nests, there'll be y symbols, and it goes to the 3→^{[[x]]}x→^{[[x]]}3th symbol.
Now, I can use a new operator, the down pointing arrow:
3↓^{n}3 = 3→^{nth nesting symbol}3
So that the biggest number reached is:
n=3→^{[[x]]}x→^{[[x]]}3
3↓^{n}3
So I can rebase to:
3↓^{n}3
So, x, the number at [Illusory Space] Level 3 Step 1 { Level 2 Step 4 {
Level 1 Step 1 { } will be:
n=3→^{[[x]]}x→^{[[x]]}3
k=3↓^{n}3
f(a)=3↓^{k*a}3
...
Now, I believe I have reached a landmark. Here, the power of this number by the nth nesting symbol should match up roughly with Actaeus's number of this post on page 3 of this thread!
His entry is ↻(A(G,G)), so I can pass his number with _{3↓}3↓^{3}3_{3}.
This is important because I've been babbling for months but had yet to enter a valid entry for the thread. I think this is it and I don't mind it being punningly small because I'm just at [Illusory Space] Level 3 Step 1 { Level 2 Step 4 {
Level 1 Step 1 { }, which is fine. And I started from +1 so didn't have to use or depend from the xkcd number for this.
Now, things get interesting.
GoC made comparisons to the numbers posted in ordinal notation here, and Actaeus's number had an equivalence of ↻(n) = ω^{ω^ω} + n, which means _{3↓}n↓^{3}3_{3} has a higher value than ω^{ω^ω} + n, and 3↓^{k*a}3 has a much higher value than any value in _{3↓}n↓^{3}3_{3}.
So the value at
[Illusory Space]
Level 3 Step 1
{ Level 2 Step 4 {
Level 1 Step 1 { x }
Has surpassed:
x = n + ω^{ω^ω}
Now I can check back at what point I was meant to reach that value before. It was...
Limit of Level x+2 (n + ω + 2) = n + ω^{ω^ω}
Which is known to be:
[Illusory Space]
Level 3 Step 1
{n + ω^{ω^ω} < Limit of Level 2 < 2ω^{ω^ω}
Which continues:
Level 3 Step 2:
{n + 2ω^{ω^ω} < Limit of Level 2 < 3ω^{ω^ω}}
Level 3 Step 3:
{n + 3ω^{ω^ω} < Limit of Level 2 < 4ω^{ω^ω}}
Correspondence: Level 3 Step n = n + (n+1)ω^{ω^ω}
Limit of Level 3: n + ω^{ω^ω + ω}
Level 4 Step 1:
{ω^{ω^ω + ω2} < Limit of Level 3 < ω^{ω^ω + ω3}}
Level 4 Step 2:
{ω^{ω^ω + ω3} < Limit of Level 3 < ω^{ω^ω + ω4}
Level 4 Step 3:
{ω^{ω^ω + ω4} < Limit of Level 3 < ω^{ω^ω + ω5}
Correspondence: Level 4 Step n = n + ω^{ω^ω + ωn}
Limit of Level 4: n + ω^{2ω^ω}
Level 5 Step 1:
{2ω^{2ω^ω} < Limit of Level 4 < 3ω^{2ω^ω}}
Level 5 Step 2:
{3ω^{2ω^ω} < Limit of Level 4 < 4ω^{2ω^ω}}
Level 5 Step 3:
{4ω^{2ω^ω} < Limit of Level 4 < 5ω^{2ω^ω}}
Correspondence: Level 5 Step n = n + (n+1)ω^{2ω^ω}
Limit of Level 5: n + ω^{(2ω+1)^ω}
Level 6 Step 1:
{ω^{3ω^ω} < Limit of Level 5 < ω^{4ω^ω}}
Level 6 Step 2:
{ω^{4ω^ω} < Limit of Level 5 < ω^{5ω^ω}}
Level 6 Step 3:
{ω^{5ω^ω} < Limit of Level 5 < ω^{6ω^ω}}
Correspondence: Level 6 Step n = n + ω^{(n+3)ω^ω}
Limit of Level 6: n + ω^{ω^(ω + 1)}
Level 7 Step 1:
{ω^{ω^(ω + 2)} < Limit of Level 6 < ω^{ω^(ω + 3)}}
Level 7 Step 2:
{ω^{ω^(ω + 3)} < Limit of Level 6 < ω^{ω^(ω + 4)}}
Level 7 Step 3:
{ω^{ω^(ω + 4)} < Limit of Level 6 < ω^{ω^(ω + 5)}}
Correspondence: Level 7 Step n = n + ω^{ω^(ω + (n+2))}
Limit of Level 7: n + ω^{ω^(2ω)}
Limit of Level 8: n + 2ω^{ω^(2ω)}
Limit of Level 9: n + ω^{2ω^(2ω)}
Limit of Level 10: n + ω^{ω^(3ω)}
Limit of Level 11: n + ω^{ω^(ω^2)}
Limit of Level 12: n + 2ω^{ω^(ω^2)}
Limit of Level 13: n + ω^{2ω^(ω^2)}
Limit of Level 14: n + ω^{ω^(2ω^2)}
Limit of Level 15: n + ω^{ω^(ω^3)}
Limit of Level 19: n + ω^{ω^(ω^4)}
Limit of Level 23: n + ω^{ω^(ω^5)}
Correspondence: Level n = n + ω^{ω^(ω^n)}
Limit 1 of the levels (Level n+ω): n + ω^{ω^(ω^ω)}
(Level n+ω+1): n + 2ω^{ω^(ω^ω)}
(Level n+ω+2): n + ω^{2ω^(ω^ω)}
(Level n+ω+3): n + ω^{ω^(2ω^ω)}
(Level n+ω+4): n + ω^{ω^(ω^2ω)}
(Level n+ω+5): n + ω^{ω^ω^(ω^2)}
(Level n+ω+6): n + 2ω^{ω^ω^(ω^2)}
(Level n+ω+7): n + ω^{2ω^ω^(ω^2)}
(Level n+ω+8): n + ω^{ω^2ω^(ω^2)}
(Level n+ω+9): n + ω^{ω^ω^(2ω^2)}
(Level n+ω+10): n + ω^{ω^ω^(ω^3)}
(Level n+ω+15): n + ω^{ω^ω^(ω^4)}
(Level n+ω+20): n + ω^{ω^ω^(ω^5)}
Limit 2 of the levels (Level n+2ω): n + ω^{ω^ω^(ω^ω)}
(Level n+2ω+1): n + 2ω^{ω^ω^(ω^ω)}
(Level n+2ω+2): n + ω^{2ω^ω^(ω^ω)}
(Level n+2ω+3): n + ω^{ω^2ω^(ω^ω)}
(Level n+2ω+4): n + ω^{ω^ω^(2ω^ω)}
(Level n+2ω+5): n + ω^{ω^ω^(ω^2ω)}
(Level n+2ω+6): n + ω^{ω^ω^ω^(ω^2)}
(Level n+2ω+7): n + 2ω^{ω^ω^ω^(ω^2)}
(Level n+2ω+8): n + ω^{2ω^ω^ω^(ω^2)}
(Level n+2ω+9): n + ω^{ω^2ω^ω^(ω^2)}
(Level n+2ω+10): n + ω^{ω^ω^2ω^(ω^2)}
(Level n+2ω+11): n + ω^{ω^ω^ω^(2ω^2)}
(Level n+2ω+12): n + ω^{ω^ω^ω^(ω^3)}
(Level n+2ω+18): n + ω^{ω^ω^ω^(ω^4)}
(Level n+2ω+24): n + ω^{ω^ω^ω^(ω^5)}
Limit 3 of the levels (Level n+3ω): ω^{ω^ω^ω^(ω^ω)}
...
Go to Limit n + ω^{ω^(ω^ω)} of the levels. That's, really, the true value of Level 3 Step 1.
[/Illusory Space]
(3*n)→^{*((2*n)^(40*n))}((2*n)^(480*n))→^{*((2*n)^(40*n))}→(3*n)
But I can bound it.
It's much higher than
n=3→^{*(2^40)}(2^480)→^{*(2^40)}→3
3→^{*(n)}(n)→^{*(n)}→3
But if a new sign was introduced that said how many towers of →^{*}'s were stacked on each other so that
_{3→}*3^{*3→...}..._{...} = 3→^^{2}...
Then, the number would be much lower than:
3→^{^(n)}(n)→^{^(n)}→3
So, x, the number at [Illusory Space] Level 3 Step 1 { Level 2 Step 1 {
Level 1 Step 1 { }
Will be:
3→^{*(n)}(n)→^{*(n)}→3 < x < 3→^{^(n)}(n)→^{^(n)}→3
For [Illusory Space] Level 3 Step 1 { Level 2 Step 1 {
Level 1 Step 1 { }, the process is repeated for:
(3*x)→^{*((2*x)^(40*x))}((2*x)^(480*x))→^{*((2*x)^(40*x))}→(3*x)
And the process is repeated:
It's much higher than
n=3→^{*(x)}(x)→^{*(x)}→3
3→^{^(n)}(n)→^{^(n)}→3
But if a new sign was introduced that said how many towers of →^{^}'s were stacked on each other so that
_{3→}^3^{^3→...}..._{...} = 3→↑^{2}...
Then, the number would be much lower than:
3→^{↑(n)}(n)→^{↑(n)}→3
So, x, the number at [Illusory Space] Level 3 Step 1 { Level 2 Step 1 {
Level 1 Step 2 { }
Will be:
3→^{^(n)}(n)→^{^(n)}→3 < x < 3→^{↑(n)}(n)→^{↑(n)}→3
And so on.
Every step introduces a new sign, and it goes to Level 1 Step (3*n)→^{*((2*n)^(40*n))}((2*n)^(480*n))→^{*((2*n)^(40*n))}→(3*n), or, the 3→^{*(n)}(n)→^{*(n)}→3th sign. Let's call that x.
I can specify the kth number of the sign like this:
3→^{(x)}x→^{(x)}3, and rebase to:
3→^{(x)}x→^{(x)}3
So, x, the number at [Illusory Space] Level 3 Step 1 { Level 2 Step 2 {
Level 1 Step 1 { } will be:
f(n)=_{3→}(3→^{(n)}n→^{(n)}3)_{3→}^{(n)}n→^{(n)}_{3→}(3→^{(n)}n→^{(n)}3)_{3}
Which can be simplified to:
3→^{((x))}x→^{((x))}3
So that every step adds a new parenthesis nest. This goes up to Step 3→^{(x)}x→^{(x)}3
So it will have:
_{3→}(((... 3→^{((x))}x→^{((x))}3 times...(((x)))... 3→^{((x))}x→^{((x))}3 times...)))_{x→}(((... 3→^{((x))}x→^{((x))}3 times...(((x)))... 3→^{((x))}x→^{((x))}3 times...)))_{3}
I can specify the kth number of nested parenthesis like this:
3→^{[n]}3 = 3→^{(((...n times...(((x)))... n times...)))}x→^{(((... n times...(((x)))... n times ...)))}3
So the current highest number is:
n=3→^{((x))}x→^{((x))}3
3→^{[n]}3
So I rebase to:
f(n)=3→^{[n]}3
So, x, the number at [Illusory Space] Level 3 Step 1 { Level 2 Step 3 {
Level 1 Step 1 { } will be:
n=3→^{((x))}x→^{((x))}3
x=3→^{[n]}3
f(x)=3→^{[x]}3
Which can be simplified to:
3→^{[[x]]}x→^{[[x]]}3
Now, every Step does that, and goes up to Step 3→^{[[x]]}x→^{[[x]]}3 so in the (), [], <> progression of nests, there'll be y symbols, and it goes to the 3→^{[[x]]}x→^{[[x]]}3th symbol.
Now, I can use a new operator, the down pointing arrow:
3↓^{n}3 = 3→^{nth nesting symbol}3
So that the biggest number reached is:
n=3→^{[[x]]}x→^{[[x]]}3
3↓^{n}3
So I can rebase to:
3↓^{n}3
So, x, the number at [Illusory Space] Level 3 Step 1 { Level 2 Step 4 {
Level 1 Step 1 { } will be:
n=3→^{[[x]]}x→^{[[x]]}3
k=3↓^{n}3
f(a)=3↓^{k*a}3
...
Now, I believe I have reached a landmark. Here, the power of this number by the nth nesting symbol should match up roughly with Actaeus's number of this post on page 3 of this thread!
His entry is ↻(A(G,G)), so I can pass his number with _{3↓}3↓^{3}3_{3}.
This is important because I've been babbling for months but had yet to enter a valid entry for the thread. I think this is it and I don't mind it being punningly small because I'm just at [Illusory Space] Level 3 Step 1 { Level 2 Step 4 {
Level 1 Step 1 { }, which is fine. And I started from +1 so didn't have to use or depend from the xkcd number for this.
Now, things get interesting.
GoC made comparisons to the numbers posted in ordinal notation here, and Actaeus's number had an equivalence of ↻(n) = ω^{ω^ω} + n, which means _{3↓}n↓^{3}3_{3} has a higher value than ω^{ω^ω} + n, and 3↓^{k*a}3 has a much higher value than any value in _{3↓}n↓^{3}3_{3}.
So the value at
[Illusory Space]
Level 3 Step 1
{ Level 2 Step 4 {
Level 1 Step 1 { x }
Has surpassed:
x = n + ω^{ω^ω}
Now I can check back at what point I was meant to reach that value before. It was...
Limit of Level x+2 (n + ω + 2) = n + ω^{ω^ω}
Which is known to be:
[Illusory Space]
Level 3 Step 1
{n + ω^{ω^ω} < Limit of Level 2 < 2ω^{ω^ω}
Which continues:
Level 3 Step 2:
{n + 2ω^{ω^ω} < Limit of Level 2 < 3ω^{ω^ω}}
Level 3 Step 3:
{n + 3ω^{ω^ω} < Limit of Level 2 < 4ω^{ω^ω}}
Correspondence: Level 3 Step n = n + (n+1)ω^{ω^ω}
Limit of Level 3: n + ω^{ω^ω + ω}
Level 4 Step 1:
{ω^{ω^ω + ω2} < Limit of Level 3 < ω^{ω^ω + ω3}}
Level 4 Step 2:
{ω^{ω^ω + ω3} < Limit of Level 3 < ω^{ω^ω + ω4}
Level 4 Step 3:
{ω^{ω^ω + ω4} < Limit of Level 3 < ω^{ω^ω + ω5}
Correspondence: Level 4 Step n = n + ω^{ω^ω + ωn}
Limit of Level 4: n + ω^{2ω^ω}
Level 5 Step 1:
{2ω^{2ω^ω} < Limit of Level 4 < 3ω^{2ω^ω}}
Level 5 Step 2:
{3ω^{2ω^ω} < Limit of Level 4 < 4ω^{2ω^ω}}
Level 5 Step 3:
{4ω^{2ω^ω} < Limit of Level 4 < 5ω^{2ω^ω}}
Correspondence: Level 5 Step n = n + (n+1)ω^{2ω^ω}
Limit of Level 5: n + ω^{(2ω+1)^ω}
Level 6 Step 1:
{ω^{3ω^ω} < Limit of Level 5 < ω^{4ω^ω}}
Level 6 Step 2:
{ω^{4ω^ω} < Limit of Level 5 < ω^{5ω^ω}}
Level 6 Step 3:
{ω^{5ω^ω} < Limit of Level 5 < ω^{6ω^ω}}
Correspondence: Level 6 Step n = n + ω^{(n+3)ω^ω}
Limit of Level 6: n + ω^{ω^(ω + 1)}
Level 7 Step 1:
{ω^{ω^(ω + 2)} < Limit of Level 6 < ω^{ω^(ω + 3)}}
Level 7 Step 2:
{ω^{ω^(ω + 3)} < Limit of Level 6 < ω^{ω^(ω + 4)}}
Level 7 Step 3:
{ω^{ω^(ω + 4)} < Limit of Level 6 < ω^{ω^(ω + 5)}}
Correspondence: Level 7 Step n = n + ω^{ω^(ω + (n+2))}
Limit of Level 7: n + ω^{ω^(2ω)}
Limit of Level 8: n + 2ω^{ω^(2ω)}
Limit of Level 9: n + ω^{2ω^(2ω)}
Limit of Level 10: n + ω^{ω^(3ω)}
Limit of Level 11: n + ω^{ω^(ω^2)}
Limit of Level 12: n + 2ω^{ω^(ω^2)}
Limit of Level 13: n + ω^{2ω^(ω^2)}
Limit of Level 14: n + ω^{ω^(2ω^2)}
Limit of Level 15: n + ω^{ω^(ω^3)}
Limit of Level 19: n + ω^{ω^(ω^4)}
Limit of Level 23: n + ω^{ω^(ω^5)}
Correspondence: Level n = n + ω^{ω^(ω^n)}
Limit 1 of the levels (Level n+ω): n + ω^{ω^(ω^ω)}
(Level n+ω+1): n + 2ω^{ω^(ω^ω)}
(Level n+ω+2): n + ω^{2ω^(ω^ω)}
(Level n+ω+3): n + ω^{ω^(2ω^ω)}
(Level n+ω+4): n + ω^{ω^(ω^2ω)}
(Level n+ω+5): n + ω^{ω^ω^(ω^2)}
(Level n+ω+6): n + 2ω^{ω^ω^(ω^2)}
(Level n+ω+7): n + ω^{2ω^ω^(ω^2)}
(Level n+ω+8): n + ω^{ω^2ω^(ω^2)}
(Level n+ω+9): n + ω^{ω^ω^(2ω^2)}
(Level n+ω+10): n + ω^{ω^ω^(ω^3)}
(Level n+ω+15): n + ω^{ω^ω^(ω^4)}
(Level n+ω+20): n + ω^{ω^ω^(ω^5)}
Limit 2 of the levels (Level n+2ω): n + ω^{ω^ω^(ω^ω)}
(Level n+2ω+1): n + 2ω^{ω^ω^(ω^ω)}
(Level n+2ω+2): n + ω^{2ω^ω^(ω^ω)}
(Level n+2ω+3): n + ω^{ω^2ω^(ω^ω)}
(Level n+2ω+4): n + ω^{ω^ω^(2ω^ω)}
(Level n+2ω+5): n + ω^{ω^ω^(ω^2ω)}
(Level n+2ω+6): n + ω^{ω^ω^ω^(ω^2)}
(Level n+2ω+7): n + 2ω^{ω^ω^ω^(ω^2)}
(Level n+2ω+8): n + ω^{2ω^ω^ω^(ω^2)}
(Level n+2ω+9): n + ω^{ω^2ω^ω^(ω^2)}
(Level n+2ω+10): n + ω^{ω^ω^2ω^(ω^2)}
(Level n+2ω+11): n + ω^{ω^ω^ω^(2ω^2)}
(Level n+2ω+12): n + ω^{ω^ω^ω^(ω^3)}
(Level n+2ω+18): n + ω^{ω^ω^ω^(ω^4)}
(Level n+2ω+24): n + ω^{ω^ω^ω^(ω^5)}
Limit 3 of the levels (Level n+3ω): ω^{ω^ω^ω^(ω^ω)}
...
Go to Limit n + ω^{ω^(ω^ω)} of the levels. That's, really, the true value of Level 3 Step 1.
[/Illusory Space]
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Continuing where I left off, the true value of Level 3 Step 1 should exceed:
n + ω↑↑ω^(ω^ω)
Now the rebase is that, and it's done at every step. Level 3 Step 2 will now create an [Illusory Space] for the entire Step, but will also create one for every step after its imaginary level 3s. This is two layers of Illusory space at once, and it's very strong.
Level 3 Step 2 Contains [Illusory Space]
Level 1 Step 1: {
n + (ω+1)↑↑ω^(ω^ω) }
Level 1 Step 2: {
n + (ω+2)↑↑ω^(ω^ω)}
Level 1 Step 3: {
n + (ω+3)↑↑ω^(ω^ω)}
...
Limit of Level 1: n + (2ω)↑↑ω^(ω^ω)
Rebase, now, this happens at every Step.
Level 2 Step 1: {
Level 1 Step 1: {
n + (3ω)↑↑ω^(ω^ω)
}
Level 1 Step 2: {
n + (4ω)↑↑ω^(ω^ω)
}
Level 1 Step 3: {
n + (5ω)↑↑ω^(ω^ω)}
Limit of Level 1: n + (ω^{2})↑↑ω^(ω^ω)}
Rebase
Level 2 Step 2: {
Level 2 Step 3:
Level 2 Step 4:
Level 2 Step 5:
Level 2 Step 6:
Level 2 Step 7:
Level 2 Step 8:
Level 2 Step 9:
Level 2 Step 10:
...
Limit of Level 2: ω↑↑(ω^{(ω^ω)}+ω^{ω}+ω^{3})
Rebase
[Illusory Space Layer 2]
Level 3 Step 1:
Level 3 Step 2:
Level 3 Step 3:
Limit of Level 3: ω↑↑(2ω^{(ω^ω)})
And after rebase, Level 1 Step 1 will have gained this growth, so will contain ω↑↑(3ω^{(ω^ω)}), ω↑↑(4ω^{(ω^ω)}), ω↑↑(5ω^{(ω^ω)})... Is this growing right?
n + ω↑↑ω^(ω^ω)
Now the rebase is that, and it's done at every step. Level 3 Step 2 will now create an [Illusory Space] for the entire Step, but will also create one for every step after its imaginary level 3s. This is two layers of Illusory space at once, and it's very strong.
Level 3 Step 2 Contains [Illusory Space]
Level 1 Step 1: {
n + (ω+1)↑↑ω^(ω^ω) }
Level 1 Step 2: {
n + (ω+2)↑↑ω^(ω^ω)}
Level 1 Step 3: {
n + (ω+3)↑↑ω^(ω^ω)}
...
Limit of Level 1: n + (2ω)↑↑ω^(ω^ω)
Rebase, now, this happens at every Step.
Level 2 Step 1: {
Level 1 Step 1: {
n + (3ω)↑↑ω^(ω^ω)
}
Level 1 Step 2: {
n + (4ω)↑↑ω^(ω^ω)
}
Level 1 Step 3: {
n + (5ω)↑↑ω^(ω^ω)}
Limit of Level 1: n + (ω^{2})↑↑ω^(ω^ω)}
Rebase
Level 2 Step 2: {
Spoiler:
Level 2 Step 3:
Spoiler:
Level 2 Step 4:
Spoiler:
Level 2 Step 5:
Spoiler:
Level 2 Step 6:
Spoiler:
Level 2 Step 7:
Spoiler:
Level 2 Step 8:
Spoiler:
Level 2 Step 9:
Spoiler:
Level 2 Step 10:
Spoiler:
...
Limit of Level 2: ω↑↑(ω^{(ω^ω)}+ω^{ω}+ω^{3})
Rebase
[Illusory Space Layer 2]
Level 3 Step 1:
Spoiler:
Level 3 Step 2:
Spoiler:
Level 3 Step 3:
Spoiler:
Limit of Level 3: ω↑↑(2ω^{(ω^ω)})
And after rebase, Level 1 Step 1 will have gained this growth, so will contain ω↑↑(3ω^{(ω^ω)}), ω↑↑(4ω^{(ω^ω)}), ω↑↑(5ω^{(ω^ω)})... Is this growing right?
 Vytron
 Posts: 429
 Joined: Mon Oct 19, 2009 10:11 am UTC
 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Infinity is not a number.
Who is online
Users browsing this forum: SecondTalon and 37 guests