My number is bigger!
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Re: My number is bigger!
Sorry, tvtropes refrence.
Re: My number is bigger!
EliezerYudkowsky wrote: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.
Theoretically. If your large cardinal axiom assumption is not actually consistent with ZFC, then you risk returning a ZFC statement which represents an infinite value.
Worse, how do we compare your entry to a similar entry based on say, ZF + the least Reinhardt Cardinal, should such a cardinal actually exist? Or some other proof system not even based on set theory and where it is not known which is stronger. Proof system strength based entries are messy at best.
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Re: My number is bigger!
The Mnib function, M(n) = the largest number appearing on or before page n of this thread that is not generated by using the Mnib function.
My number is M(xkcd)1
*Hey, the rules only banned adding to previously posted numbers
My number is M(xkcd)1
*Hey, the rules only banned adding to previously posted numbers
Last edited by ericmn on Tue Oct 08, 2013 9:47 pm UTC, edited 1 time in total.
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Re: My number is bigger!
Extrapolation of what? Because the only way that could work is by referencing previously posted numbers.
Re: My number is bigger!
Vytron wrote:Extrapolation of what? Because the only way that could work is by referencing previously posted numbers.
Just extrapolate from the numbers posted on the next few pages.
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Re: My number is bigger!
0
Okay, now that your number uses this to extrapolate, I don't think it's very big...
Okay, now that your number uses this to extrapolate, I don't think it's very big...
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Re: My number is bigger!
When the number was first posted, it did refer to numbers posted prior. So it is invalid.
On top of that, when it was first posted, it was not well defined.
Finally, while you might think your answer is clever, it isn't: this is a thread where people use ridiculous infinite ordinals, obtuse computation theory, and (I think in at least one case?) large cardinals, to win the "my number is bigger" game.
*That* is clever. "+1" is not.
On top of that, when it was first posted, it was not well defined.
Finally, while you might think your answer is clever, it isn't: this is a thread where people use ridiculous infinite ordinals, obtuse computation theory, and (I think in at least one case?) large cardinals, to win the "my number is bigger" game.
*That* is clever. "+1" is not.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
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Re: My number is bigger!
...
Oh.
What Yakk said.
Oh.
What Yakk said.
Re: My number is bigger!
Yakk wrote:Finally, while you might think your answer is clever, it isn't: this is a thread where people use ridiculous infinite ordinals, obtuse computation theory, and (I think in at least one case?) large cardinals, to win the "my number is bigger" game.
I was just speculating on the magnitude of the numbers if this thread's idea progression continued to the (xkcd)th page. No need to be a dick about it.
Re: My number is bigger!
We can't really speculate about that. It's not going to reach an XKCD'th page, and as the thread entries do not form a closed system, it does not have to actually be a computable function.
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Re: My number is bigger!
I am going to be quite incredibly dumb and say:
My number is lim x > 0 (xkcd/x/x/x/x/x/x/x/......../x), where /x repeats xkcd/x times.
My number is lim x > 0 (xkcd/x/x/x/x/x/x/x/......../x), where /x repeats xkcd/x times.
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Re: My number is bigger!
^Beaten by first page of the thread.
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Re: My number is bigger!
Actually, no. The limit is infinity. Which means that it isn't allowed according to the rules of the game.
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Re: My number is bigger!
just cause a topic like this wouldn't be complete without the Fibinnochi sequence.
let F(0) = 1 and F(1) = 1
let F(n) = F(n1) +F(n2)
let G(n) = F(F(F(...(n)) n times.
let H(n) = G(n) > G(n^2) >G(n^3)... where > is chained arrow notation and continues until the power is n.
my number is H(100).
let F(0) = 1 and F(1) = 1
let F(n) = F(n1) +F(n2)
let G(n) = F(F(F(...(n)) n times.
let H(n) = G(n) > G(n^2) >G(n^3)... where > is chained arrow notation and continues until the power is n.
my number is H(100).
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 Vytron
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Re: My number is bigger!
At some point, the only thing that matters is the number of recursions that your number has, so yours has 3 levels of recursive recursions.
Mouffles beats your number here with g_64 levels of recursive recursions.
Mouffles beats your number here with g_64 levels of recursive recursions.

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Re: My number is bigger!
Infinity is not a number.
Re: My number is bigger!
I don't know what the current record is, but I'll try to make a very large number starting with using only standard notation of up arrows and the number googolplex.
Imagine the number googolplex^^^^^^....^^^^^^googolplex , with 1 googolplex up arrows. Call this A_{1}. Define A_{n} to be A_{n1}^^^^^.....^^^^^A_{n1} with A_{n1} arrows.
Define B_{0} to be the number ^{A}A_{A}...... with ^{A}A_{A}......A's with ^{A}A_{A}......A's with (repeat that phrase ^{A}A_{A}...... with ^{A}A_{A} A's times)... with ^{A}A_{A} A's
...crap, my brain may have just asploded. Will continue with the series of huge numbers when I'm not so tired... (I expect B_{0} is nothing compared to some of the numbers here)
Imagine the number googolplex^^^^^^....^^^^^^googolplex , with 1 googolplex up arrows. Call this A_{1}. Define A_{n} to be A_{n1}^^^^^.....^^^^^A_{n1} with A_{n1} arrows.
Define B_{0} to be the number ^{A}A_{A}...... with ^{A}A_{A}......A's with ^{A}A_{A}......A's with (repeat that phrase ^{A}A_{A}...... with ^{A}A_{A} A's times)... with ^{A}A_{A} A's
...crap, my brain may have just asploded. Will continue with the series of huge numbers when I'm not so tired... (I expect B_{0} is nothing compared to some of the numbers here)
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Re: My number is bigger!
patzer wrote:I'll try to make a very large number starting with using only standard notation of up arrows and the number googolplex.
Imagine the number googolplex^^^^^^....^^^^^^googolplex , with 1 googolplex up arrows. Call this A_{1}. Define A_{n} to be A_{n1}^^^^^.....^^^^^A_{n1} with A_{n1} arrows.
Define B_{0} to be the number ^{A}A_{A}...... with ^{A}A_{A}......A's with ^{A}A_{A}......A's with (repeat that phrase ^{A}A_{A}...... with ^{A}A_{A} A's times)... with ^{A}A_{A} A's
...crap, my brain may have just asploded. Will continue with the series of huge numbers when I'm not so tired... (I expect B_{0} is nothing compared to some of the numbers here)
I'm not sure, but I think that your B_{0} would be comparable, in the fastgrowinghierarchy, to f_{ω·2}(googleplex). And if that's the case, this number was already surpassed in the beginning of page 2.
patzer wrote:I don't know what the current record is...
Niether do I. The current record on this thread is so completely beyond my grasp, that I don't even bother trying to figure it out. I pretty much lost count somewhere before page 10.
Re: My number is bigger!
Okay, let's continue.
Let us name the process that was just described to transform 1 googolplex into B_{0}, as ^ .
Thus 1 googleplex^googleplex = B_{0}.
Note that this redefines the meaning of ^.
Now I use a similar process to the first post:
Imagine the number B_{0}^^^^^^....^^^^^^B_{0} , with B_{0} up arrows. Call this A_{1}. Define A_{n} to be A_{n1}^^^^^.....^^^^^A_{n1} with A_{n1} arrows.
Define B_{0} to be the number ^{A}A_{A}...... with ^{A}A_{A}......A's with ^{A}A_{A}......A's with (repeat that phrase ^{A}A_{A}...... with ^{A}A_{A1} A's times)... with ^{A}A_{A1} A's
(minor nitpick:)
Yes, B_{0} is redefined now.
Repeat, redefining ^ and B_{0} again.
Repeat this process B_{0} times (that's the original value of B_{0}), in fact. Set B_{1} to be the value of B_{0} after its final iteration, and reset B_{0} to its original value.

B_{1} is already enormous. I'll continue this later once I have thought of more ways to continue the series.
Let us name the process that was just described to transform 1 googolplex into B_{0}, as ^ .
Thus 1 googleplex^googleplex = B_{0}.
Note that this redefines the meaning of ^.
Now I use a similar process to the first post:
Imagine the number B_{0}^^^^^^....^^^^^^B_{0} , with B_{0} up arrows. Call this A_{1}. Define A_{n} to be A_{n1}^^^^^.....^^^^^A_{n1} with A_{n1} arrows.
Define B_{0} to be the number ^{A}A_{A}...... with ^{A}A_{A}......A's with ^{A}A_{A}......A's with (repeat that phrase ^{A}A_{A}...... with ^{A}A_{A1} A's times)... with ^{A}A_{A1} A's
(minor nitpick:
Spoiler:
Yes, B_{0} is redefined now.
Repeat, redefining ^ and B_{0} again.
Repeat this process B_{0} times (that's the original value of B_{0}), in fact. Set B_{1} to be the value of B_{0} after its final iteration, and reset B_{0} to its original value.

B_{1} is already enormous. I'll continue this later once I have thought of more ways to continue the series.
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Re: My number is bigger!
Oh, yay, competition! Let's compare our numbers!
Note that this is nothing compared to, say:
googolplex → googolplex ... → googolplex → googolplex with 1 googolplex up arrows. So I'm beating this number at Level 1 Step 4.
Here you pass Level 1 Step 4 at A_{23}
At this point B_{n} looks somewhat equivalent to my 3→^{n}3, but my Level 1 Step 5 starts at (3→^{52}3*2)^16, so I think you'd need to repeat this process for A, B, C, D... for the (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}6th symbol to pass Level 1 Step 5.
Nice, here your number should take a level in badass.
Here, I don't see why you stay at ^ and B_{0} instead of moving on to C or B_{2}, though I see you want to move there for "much more powerful things", I think you're just having here B_{1} being equivalent to ^{[{originalB}0^{]}th symbol]}B_{0}th redefinition of ^, but the problem is it's still being powered by original B_{0} which is tiny at this point, so I don't think B_{1} passes Level 1 Step 5 as the B_{0}th symbol (what you'd call the B_{0}th redefinition) is still weaker than recursive redefinitions. You'd probably need to repeat the whole process of redefining B_{1} and going to the B_{1}th recursion 6→^{104}6*6→^{104}6 times to pass Level 1 Step 5.
patzer wrote:I don't know what the current record is, but I'll try to make a very large number starting with using only standard notation of up arrows and the number googolplex.
Imagine the number googolplex^^^^^^....^^^^^^googolplex , with 1 googolplex up arrows.
Note that this is nothing compared to, say:
googolplex → googolplex ... → googolplex → googolplex with 1 googolplex up arrows. So I'm beating this number at Level 1 Step 4.
patzer wrote:Call this A1. Define An to be An1^^^^^.....^^^^^An1 with An1 arrows.
Here you pass Level 1 Step 4 at A_{23}
patzer wrote:Define B_{0} to be the number ^{A}A_{A}...... with ^{A}A_{A}......A's with ^{A}A_{A}......A's with (repeat that phrase ^{A}A_{A}...... with ^{A}A_{A} A's times)... with ^{A}A_{A} A's
At this point B_{n} looks somewhat equivalent to my 3→^{n}3, but my Level 1 Step 5 starts at (3→^{52}3*2)^16, so I think you'd need to repeat this process for A, B, C, D... for the (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}6th symbol to pass Level 1 Step 5.
patzer wrote:Let us name the process that was just described to transform 1 googolplex into B0, as ^ .
Thus 1 googleplex^googleplex = B0.
Note that this redefines the meaning of ^.
Nice, here your number should take a level in badass.
patzer wrote:Okay, let's continue.
Let us name the process that was just described to transform 1 googolplex into B_{0}, as ^ .
Thus 1 googleplex^googleplex = B_{0}.
Note that this redefines the meaning of ^.
Now I use a similar process to the first post:
Imagine the number B_{0}^^^^^^....^^^^^^B_{0} , with B_{0} up arrows. Call this A_{1}. Define A_{n} to be A_{n1}^^^^^.....^^^^^A_{n1} with A_{n1} arrows.
Define B_{0} to be the number ^{A}A_{A}...... with ^{A}A_{A}......A's with ^{A}A_{A}......A's with (repeat that phrase ^{A}A_{A}...... with ^{A}A_{A1} A's times)... with ^{A}A_{A1} A's
(minor nitpick:)Spoiler:
Yes, B_{0} is redefined now.
Repeat, redefining ^ and B_{0} again.
Repeat this process B_{0} times (that's the original value of B_{0}), in fact. Set B_{1} to be the value of B_{0} after its final iteration, and reset B_{0} to its original value.

B_{1} is already enormous. I'll continue this later once I have thought of more ways to continue the series.
Here, I don't see why you stay at ^ and B_{0} instead of moving on to C or B_{2}, though I see you want to move there for "much more powerful things", I think you're just having here B_{1} being equivalent to ^{[{originalB}0^{]}th symbol]}B_{0}th redefinition of ^, but the problem is it's still being powered by original B_{0} which is tiny at this point, so I don't think B_{1} passes Level 1 Step 5 as the B_{0}th symbol (what you'd call the B_{0}th redefinition) is still weaker than recursive redefinitions. You'd probably need to repeat the whole process of redefining B_{1} and going to the B_{1}th recursion 6→^{104}6*6→^{104}6 times to pass Level 1 Step 5.
Re: My number is bigger!
Yeah I guess you're right, I didn't realize how small B_{0} actually was. I did want to reserve the higher letters and numbers for more powerful functions, but I will use redefinition instead as it's easier.
Okay...
So I described a process to get to B_{1} from B_{0}. Using the same process, but replacing B_{0} with B_{1} in all cases, will result in B_{2}. Keep defining the rest of the Bnumbers in the same pattern.
Define C_{0} to be the number ^{B}B_{B}...... with ^{B}B_{B}......B's with ^{B}B_{B}......B's with (repeat that phrase ^{B}B_{B}...... with ^{B}B_{B} B's times)... with ^{B}B_{B} B's
Then define the rest of the Cnumbers in relation to the Bnumbers as the Bnumbers had been defined in relation to the Anumbers.
Define the rest of the Dnumbers in relation to the Cnumbers as the Cnumbers had been defined in relation to the Bnumbers.
Define the rest of the letters and numbers similarly.
Now we reach the second iteration of my plan.
Redefine all the numbers so that New A_{0} = Old A_{0}; New A_{1} = Old B_{0}; New A_{2} = Old C_{0}; New A_{3} = Old D_{0}; etc.
These new Anumbers have now been defined. Define the new Bnumbers from the new Anumbers using exactly the same formula I used to derive the old Bnumbers from the old Anumbers.
Continue the pattern with the rest of the letters.
Repeat the letter redefinition.
Repeat again.
I'll use the notation B_{1; 24} (for example), where the second subscript refers to the iteration of the plan (how many redefinitions have taken place), if disambiguation is needed.
Now I was going to redefine again, but it would get too confusing.
So, α_{0} = Z_{tower of Zs Z9; 9 high followed by a 9;tower of Zs Z9; 9 high followed by a 9}
α_{0} will be the basis of a future system.
Okay...
So I described a process to get to B_{1} from B_{0}. Using the same process, but replacing B_{0} with B_{1} in all cases, will result in B_{2}. Keep defining the rest of the Bnumbers in the same pattern.
Define C_{0} to be the number ^{B}B_{B}...... with ^{B}B_{B}......B's with ^{B}B_{B}......B's with (repeat that phrase ^{B}B_{B}...... with ^{B}B_{B} B's times)... with ^{B}B_{B} B's
Then define the rest of the Cnumbers in relation to the Bnumbers as the Bnumbers had been defined in relation to the Anumbers.
Define the rest of the Dnumbers in relation to the Cnumbers as the Cnumbers had been defined in relation to the Bnumbers.
Define the rest of the letters and numbers similarly.
Now we reach the second iteration of my plan.
Redefine all the numbers so that New A_{0} = Old A_{0}; New A_{1} = Old B_{0}; New A_{2} = Old C_{0}; New A_{3} = Old D_{0}; etc.
These new Anumbers have now been defined. Define the new Bnumbers from the new Anumbers using exactly the same formula I used to derive the old Bnumbers from the old Anumbers.
Continue the pattern with the rest of the letters.
Repeat the letter redefinition.
Repeat again.
I'll use the notation B_{1; 24} (for example), where the second subscript refers to the iteration of the plan (how many redefinitions have taken place), if disambiguation is needed.
Now I was going to redefine again, but it would get too confusing.
So, α_{0} = Z_{tower of Zs Z9; 9 high followed by a 9;tower of Zs Z9; 9 high followed by a 9}
α_{0} will be the basis of a future system.
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Re: My number is bigger!
Actually, I can make an α_{0}bigger than that! Ignore my previous definition of it.
Here, where I use the construction Z_{2} for brevity (for example), I mean Z_{2; 2}
Let A_{0; 0; 1} = the number ^{Z}Z_{Z}...... with ^{Z}Z_{Z}......Z's with ^{Z}Z_{Z}......Z's with (repeat that phrase ^{Z}Z_{Z}...... with ^{Z}Z_{Z} Z's times)... with ^{Z}Z_{Z9} Z's.

And α_{0} = Z_{[tower of Zs Z9; 9 high followed by a 9];[tower of Zs Z9; 9 high followed by a 9];[tower of Zs Z9; 9 high followed by a 9];[tower of Zs Z9; 9 high followed by a 9]}..... , with the number of subscripts separated by semicolons present being Z_{[tower of Zs Z9; 9 high followed by a 9];[tower of Zs Z9; 9 high followed by a 9];[tower of Zs Z9; 9 high followed by a 9]}.
Obviously I have not yet defined how there can be more than three subscripts for a letter! I will do so later. This section is a work in progress. In the meantime A_{0; 0; 1} is currently my highest defined number.
Here, where I use the construction Z_{2} for brevity (for example), I mean Z_{2; 2}
Let A_{0; 0; 1} = the number ^{Z}Z_{Z}...... with ^{Z}Z_{Z}......Z's with ^{Z}Z_{Z}......Z's with (repeat that phrase ^{Z}Z_{Z}...... with ^{Z}Z_{Z} Z's times)... with ^{Z}Z_{Z9} Z's.

And α_{0} = Z_{[tower of Zs Z9; 9 high followed by a 9];[tower of Zs Z9; 9 high followed by a 9];[tower of Zs Z9; 9 high followed by a 9];[tower of Zs Z9; 9 high followed by a 9]}..... , with the number of subscripts separated by semicolons present being Z_{[tower of Zs Z9; 9 high followed by a 9];[tower of Zs Z9; 9 high followed by a 9];[tower of Zs Z9; 9 high followed by a 9]}.
Obviously I have not yet defined how there can be more than three subscripts for a letter! I will do so later. This section is a work in progress. In the meantime A_{0; 0; 1} is currently my highest defined number.
If it looks like a duck, and quacks like a duck, we have at least to consider the possibility that we have a small aquatic bird of the family Anatidae on our hands. –Douglas Adams
 Vytron
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Re: My number is bigger!
Um, let's take a break, I think your number could be significantly bigger if...
So this is 26 iterations, right? One for each letter of the alphabet? I think this is a mistake, here, instead of using 26 symbols, you should be using [highest defined number] symbols, say A_{0} symbols, then B_{0} (biggest one defined) symbols, then C_{0} (biggest one defined) symbols... If you have a clear, strong definition right here of how many times you make your symbols recursive, you can surpass Level 1 Step 5 with ease before reaching α_{0}, because the way you have it, A_{0; 0; 1} is being powered by those 26 recursions (and every time you repeat all recursions you just go from A to Z), and at this point, there's not much difference from using 26 symbols or using one googolplex symbols, so that, say, ^{A}A_{0; 0; 1}; A_{0; 0; 1}; A_{0; 0; 1} is still inside Level 1 Step 5, because every time you do a recursion again you stop at Z. In other words, you keep adding new, and new levels over the old levels, but those new levels aren't much more powerful than previous ones because you're not increasing the number of recursions of them ("do it another 26 times, and then do something slightly stronger another 26 times...").
patzer wrote:Define the rest of the letters and numbers similarly.
Now we reach the second iteration of my plan.
Redefine all the numbers so that New A_{0} = Old A_{0}; New A_{1} = Old B_{0}; New A_{2} = Old C_{0}; New A_{3} = Old D_{0}; etc.
These new Anumbers have now been defined. Define the new Bnumbers from the new Anumbers using exactly the same formula I used to derive the old Bnumbers from the old Anumbers.
Continue the pattern with the rest of the letters.
So this is 26 iterations, right? One for each letter of the alphabet? I think this is a mistake, here, instead of using 26 symbols, you should be using [highest defined number] symbols, say A_{0} symbols, then B_{0} (biggest one defined) symbols, then C_{0} (biggest one defined) symbols... If you have a clear, strong definition right here of how many times you make your symbols recursive, you can surpass Level 1 Step 5 with ease before reaching α_{0}, because the way you have it, A_{0; 0; 1} is being powered by those 26 recursions (and every time you repeat all recursions you just go from A to Z), and at this point, there's not much difference from using 26 symbols or using one googolplex symbols, so that, say, ^{A}A_{0; 0; 1}; A_{0; 0; 1}; A_{0; 0; 1} is still inside Level 1 Step 5, because every time you do a recursion again you stop at Z. In other words, you keep adding new, and new levels over the old levels, but those new levels aren't much more powerful than previous ones because you're not increasing the number of recursions of them ("do it another 26 times, and then do something slightly stronger another 26 times...").
Re: My number is bigger!
Ah.
I'll admit, I have completely forgotten my pattern of logic for this now
I may take another look later, but for now, this thread is just too hard for me.
I'll admit, I have completely forgotten my pattern of logic for this now
I may take another look later, but for now, this thread is just too hard for me.
If it looks like a duck, and quacks like a duck, we have at least to consider the possibility that we have a small aquatic bird of the family Anatidae on our hands. –Douglas Adams
 orangedragonfire
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 Vytron
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Re: My number is bigger!
Also, an addendum on big numbers: Once you reach some big function to make some number big applying this function over and over isn't going to make your number much bigger, and applying the next level of that function over and over just means you're going to be stuck adding stronger functions without going nowhere fast.
For comparison, let's say you have some function:
f(n)=n+1+1+1... f(n) times ...+1+1+1
At this point, feeding it into itself will just cause it to go:
f(n)=n+1+1+1... f(n)*2 times ...+1+1+1
Which is nothing compared to:
f(n)=n*2*2*2... f(n) times ...*2*2*2
Which is nothing compared to:
f(n)=((...((n^2)^2)*2... f(n) times ...^2)^2)^2)
Which is nothing compared to:
f(n)=n^2^2^2... f(n) times ...^2^2^2
...
So you get the idea.
Well, once you get to the recursive power of:
f(n) = f(f(f(... f(n) times ...f(f(f(n)))...)))
Or some supra level that includes those by some
f_{n}= ^{f(n)}f(n)_{f(n)} (for a tower of f(n) length)
You have reached that power, but doing it over and over isn't going to advance you to the levels very fast.
I.e. if you redefine the + sign to mean "adding the next tower of powers, so that two +'s means a towers of towers", you are back to:
f(n)=n+1+1+1... f(n) times ...+1+1+1
Which in this case is much, much larger, and powerful than the one above, because this one has recursive towers of towers. The problem is, being recursive over it again is going to get you to:
f(n)=n+1+1+1... f(n)*2 times ...+1+1+1
You need to iterate the recursion and make the number self recursive, or you're not going to reach
f(n)=n*2*2*2... f(n) times ...*2*2*2
Any time soon (because in f(n)=n+1+1+1... f(n)*k times ...+1+1+1, k is not growing fast enough, it's not going to reach f(n) any time soon, so you hit a block that can't be passed by just feeding the number to itself, you have to increase the time its feeds into itself while it's growing so that they increase by the same magnitude that the powers increase.)
For comparison, let's say you have some function:
f(n)=n+1+1+1... f(n) times ...+1+1+1
At this point, feeding it into itself will just cause it to go:
f(n)=n+1+1+1... f(n)*2 times ...+1+1+1
Which is nothing compared to:
f(n)=n*2*2*2... f(n) times ...*2*2*2
Which is nothing compared to:
f(n)=((...((n^2)^2)*2... f(n) times ...^2)^2)^2)
Which is nothing compared to:
f(n)=n^2^2^2... f(n) times ...^2^2^2
...
So you get the idea.
Well, once you get to the recursive power of:
f(n) = f(f(f(... f(n) times ...f(f(f(n)))...)))
Or some supra level that includes those by some
f_{n}= ^{f(n)}f(n)_{f(n)} (for a tower of f(n) length)
You have reached that power, but doing it over and over isn't going to advance you to the levels very fast.
I.e. if you redefine the + sign to mean "adding the next tower of powers, so that two +'s means a towers of towers", you are back to:
f(n)=n+1+1+1... f(n) times ...+1+1+1
Which in this case is much, much larger, and powerful than the one above, because this one has recursive towers of towers. The problem is, being recursive over it again is going to get you to:
f(n)=n+1+1+1... f(n)*2 times ...+1+1+1
You need to iterate the recursion and make the number self recursive, or you're not going to reach
f(n)=n*2*2*2... f(n) times ...*2*2*2
Any time soon (because in f(n)=n+1+1+1... f(n)*k times ...+1+1+1, k is not growing fast enough, it's not going to reach f(n) any time soon, so you hit a block that can't be passed by just feeding the number to itself, you have to increase the time its feeds into itself while it's growing so that they increase by the same magnitude that the powers increase.)
Re: My number is bigger!
Vytron, I welcome you to join the Googology Wiki. There are a bunch of people there interested in large numbers, and you will find a good audience for your large number notation.
Come join the fun!
Come join the fun!
 Vytron
 Posts: 429
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 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Thanks Deedlit, I recently reviewed my number, and... I don't think I understand how it works correctly anymore, the Actaeus's power of recursion could have been passed much earlier than I thought, and Illusory space seems so strong that I got stuck defining how it works at earlier levels (defining 3 layers was very hard, and the next part would have "biggest number defined so far" layers...), but I'll take a look!
Re: My number is bigger!
Complaint about the transfinite which i withdraw later due to having been uninformed
Last edited by Daggoth on Thu Apr 20, 2017 7:16 pm UTC, edited 1 time in total.
 Yakk
 Poster with most posts but no title.
 Posts: 11073
 Joined: Sat Jan 27, 2007 7:27 pm UTC
 Location: E pur si muove
Re: My number is bigger!
Daggoth, using cardinals (etc) to describe the system that produces your finite integer, with the process itself being (theoretically) fi ite and checkable, is what most folk referring to cardinals are doing. And they are using cardinals to label (finite) numbers, as in "this finite number can be named via this process by using this large cardinal as an input to the process".
You can do finite deteminsitic math where some symbols are not finite, and have the entire process be computable and provably halting.
You can do finite deteminsitic math where some symbols are not finite, and have the entire process be computable and provably halting.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: My number is bigger!
Second complaint, also withdrawn
Last edited by Daggoth on Thu Apr 20, 2017 7:16 pm UTC, edited 1 time in total.
Re: My number is bigger!
Daggoth wrote:Ive thought of another reason Ordinals shouldnt be allowed
remember the rule saying its not allowed to say "that guys number+1"
Well omega, the starter point of ordinals is the limit
Of successor{0,1,2.....} so if, say any finite number has a successor
Omega is like saying "that guys number +alot"
And building up from that.
If we are not allowed to win by succesorship then omega is banned by definition
That's not quite what we're doing. When we use infinite ordinals, we're using them to label functions. You can most definitely have an enumerable infinite list of functions just fine. We're not actually saying ω is the submission.
A simple example: f1(n) = n+1
f2(n) = f1^{n[/sub](2)fx(n) = f(x1)[sup]n}(2)
fω(n) = fn(n)
So f3(4) = f2(f2(f2(f2(2)))) = f2(f2(f2(f1(f1(2))))) = f2(f2(f2(4))) = f2(f2(6)) = f2(8) = 10
f4(4) = f3(f3(f3(f3(2)))) = f3(f3(f3(6))) = f3(f3(14)) = f3(30) = 62
fω(6) = f6(6) ~~ 2^^^6
When we use ordinal collapsing functions, we define something somewhat (but not exactly) like:
f(o,0) = {0,1,ω}
f(o,n+1) = {a+b,ab,a^{b} a < f(o,n), b <f(o,n} union {f(a,m)  a < f(o,n), m < f(o,n), a < o, m < ω}
So we have f(0,0) = {0,1,ω}, f(0,1) = {0,1,2,ω,ω^{2},ω^{ω}}, f(0,2) = {0,1,2,3,4,ω...}, and then
f(1,0) = {0,1,ω}, f(1,1) = {0,1,2,ω,ω^{2},ω^{ω}}, f(1,2) = {0,1,2,3,4,ω..}, f(1,3) = {0,1,2,3,4..16,ω...}, f(1,4) = {0,1,2,3..2^{16},ω...}
If we adopt the convention that f(o,n) is the largest finite number in the (always finite, for finite n!) set f(o,n), then we'll have things like:
f(2,2) = {0,1,2...f(1,f(2,1)),ω...}
In general f(o,n) will be approximately Ackerman(o,n), for finite o.
But f(ω,0) is still well defined, as {0,1,ω}.
And f(ω,1) = {0,1,2,ω..}
f(ω,2) = {0,1,2...f(1,f(2,1)),ω...}
f(ω,3) = {0,1,2...f(f(ω,2),f(ω,2)),ω...}  we have just magically gained the function a > Ackermann(a,a), in a fully computable and guaranteed to terminate way.
There's nothing at all wrong with it.
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.
Re: My number is bigger!
Yea i can see how it is useful and constructible, but to me that "feels" sort of like cheating, at least in the scope of this competition.
Its like saying, you enter a special kind of race and the guy who can run the most distance wins. ω is like saying, "I've ran to farthest possible point. If i could run one more meter, then i am not at the place where i begin". Then i run another meter for ω+1, another ω meters for ω2, and i do that ω times for ω^2 etc.."
I'd be like "HOW did you run that long?"
Famous big numbers don't come from ordinals or constructions with the help of ordinals, graham's number a. came from ramsey theory and b. can be step by step "reached" without any notion of ordinals. Same as with goodstein sequences, tree(n), hydras, etc.. Wether you could use ω and other ordinals to measure them is one thing, but it's not how they were spawned in the first place
Also, how can you show that ω+1 > ω and not =
Its like saying, you enter a special kind of race and the guy who can run the most distance wins. ω is like saying, "I've ran to farthest possible point. If i could run one more meter, then i am not at the place where i begin". Then i run another meter for ω+1, another ω meters for ω2, and i do that ω times for ω^2 etc.."
I'd be like "HOW did you run that long?"
Famous big numbers don't come from ordinals or constructions with the help of ordinals, graham's number a. came from ramsey theory and b. can be step by step "reached" without any notion of ordinals. Same as with goodstein sequences, tree(n), hydras, etc.. Wether you could use ω and other ordinals to measure them is one thing, but it's not how they were spawned in the first place
Also, how can you show that ω+1 > ω and not =
 orangedragonfire
 Posts: 13
 Joined: Wed Sep 28, 2011 3:45 am UTC
 Location: It exists. Probably.
Re: My number is bigger!
Well, basically, ordinals aren't really used as infinite numbers in the context of the Fast Growing Hirachy, but as lists. Lists of lists of lists.
f1(n) = n+1
fx(n) = f(x1)^{n}(n)
Gives us functions for all natural numbers x. Now we extend this to include a list as an argument instead of x; let A be a list with elements {a_{1}, a_{2}, a_{3}, ...} These elements may be lists themselves, since we can just use the same procedure to deal with that.
fA(n) = fa_{n}^{n}(n)
Ordinals in this context are just easy references to various lists.
ω = {1, 2, 3, ...}
ω2 = {ω+1, ω+2, ω+3, ...} (yes, this is a list of lists)
ωn = {ω(n1)+1, ω(n1)+2, ω(n1)+3, ...}
ω^{2} = {ω, ω2, ω3, ...}
ω^{n} = {ω^{n1}, ω^{n1}2, ω^{n1}3, ...}
...
I'm not quite sure about ordinal collapsing functions, but I think what they are actually trying to get is the size of a set. It doesn't really matter what sort of elements you put into the set, since each element will only count as 1.
f1(n) = n+1
fx(n) = f(x1)^{n}(n)
Gives us functions for all natural numbers x. Now we extend this to include a list as an argument instead of x; let A be a list with elements {a_{1}, a_{2}, a_{3}, ...} These elements may be lists themselves, since we can just use the same procedure to deal with that.
fA(n) = fa_{n}^{n}(n)
Ordinals in this context are just easy references to various lists.
ω = {1, 2, 3, ...}
ω2 = {ω+1, ω+2, ω+3, ...} (yes, this is a list of lists)
ωn = {ω(n1)+1, ω(n1)+2, ω(n1)+3, ...}
ω^{2} = {ω, ω2, ω3, ...}
ω^{n} = {ω^{n1}, ω^{n1}2, ω^{n1}3, ...}
...
I'm not quite sure about ordinal collapsing functions, but I think what they are actually trying to get is the size of a set. It doesn't really matter what sort of elements you put into the set, since each element will only count as 1.
 Vytron
 Posts: 429
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 Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P
Re: My number is bigger!
Yeah, for instance, in my number, which if well defined should defeat the first 24 pages on the thread, or something, I define a ladder of functions that power up new functions, each more powerful than the last, and if, say, I manage to prove that some function has reached this:
f(x)=3→^{[x]}3=y
f(a)=ω + a=b
y>b
Then I can go and say "hey, look, this function here has surpassed ω + n, so instead of having to say 'repeat some new process 3→^{[<highest number defined >]}3 times' I can go and say 'repeat some new process for ω + a times'". Collapsible ω will have less power than my actual number, but what matters is my function is between f(n)=ω+n and f(n)=nω.
This way, my number gains meaning.
Because, if it comes to the point where I start using different colored brackets and show some stepbystep process and end with some
3→^{[[[x]]]}3
And claim it's some really humongous function, what does that tell you? Nothing.
But if I show that that function is like this:
f(x)=3→^{[[[x]]]}3=y
f(a)=ε_{a}=b
y>b
Then I have reached somewhere, and next time I can do something like 3→^{[[[x]]]}3 where x is 3→^{[[[y]]]}3 where y is 3→^{[[[z]]]}3 ... repeating the process collapsible ε_{<biggest number defined>} times because I have reached that power.
How is that cheating?
Using the ZermeloFraenkel set theory plus the Axiom of Choice plus the axiom that there exists an I0 rankintorank cardinal and turing machines and not bothering to explain how your number grows looks more like cheating, and the current Champion of the thread does that...
f(x)=3→^{[x]}3=y
f(a)=ω + a=b
y>b
Then I can go and say "hey, look, this function here has surpassed ω + n, so instead of having to say 'repeat some new process 3→^{[<highest number defined >]}3 times' I can go and say 'repeat some new process for ω + a times'". Collapsible ω will have less power than my actual number, but what matters is my function is between f(n)=ω+n and f(n)=nω.
This way, my number gains meaning.
Because, if it comes to the point where I start using different colored brackets and show some stepbystep process and end with some
3→^{[[[x]]]}3
And claim it's some really humongous function, what does that tell you? Nothing.
But if I show that that function is like this:
f(x)=3→^{[[[x]]]}3=y
f(a)=ε_{a}=b
y>b
Then I have reached somewhere, and next time I can do something like 3→^{[[[x]]]}3 where x is 3→^{[[[y]]]}3 where y is 3→^{[[[z]]]}3 ... repeating the process collapsible ε_{<biggest number defined>} times because I have reached that power.
How is that cheating?
Using the ZermeloFraenkel set theory plus the Axiom of Choice plus the axiom that there exists an I0 rankintorank cardinal and turing machines and not bothering to explain how your number grows looks more like cheating, and the current Champion of the thread does that...
Re: My number is bigger!
Daggoth wrote:Yea i can see how it is useful and constructible, but to me that "feels" sort of like cheating, at least in the scope of this competition.
Its like saying, you enter a special kind of race and the guy who can run the most distance wins. ω is like saying, "I've ran to farthest possible point. If i could run one more meter, then i am not at the place where i begin". Then i run another meter for ω+1, another ω meters for ω2, and i do that ω times for ω^2 etc.."
I'd be like "HOW did you run that long?"
Famous big numbers don't come from ordinals or constructions with the help of ordinals, graham's number a. came from ramsey theory and b. can be step by step "reached" without any notion of ordinals. Same as with goodstein sequences, tree(n), hydras, etc.. Wether you could use ω and other ordinals to measure them is one thing, but it's not how they were spawned in the first place
Also, how can you show that ω+1 > ω and not =
Goodstein and Hydras are quite literally ordinal collapse under another name, that is exactly how they work. It's just an ordering property, we don't need infinite numbers. We could replace ω with the real number one, and the naturals with the sequence 0.5, 0.75, 0.875... 11/2^{n}... and construct an analogous set of functions  even though it's a lot harder to name things clearly, we can rename a the fundamental sequence for any ordinal into the reals just fine.
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.
Re: My number is bigger!
I'll give it a shot then.
the seed (n) function outputs the Stem function with n arguments
The value for the first argument is n, the second argument is successor n, the third value is the sum of all the previous arguments, the fourth value is the product of all the previous arguments, the fifth value is a power tower with all the previous arguments in ascending order, the sixth value is a tetration tower, the seventh a ^^^ sequence etc..
Seeding: The recursion step
Seed(1) = Stem (1)
Seed(2) = Stem (2,2+1) =(2,3)
Seed(3) = Stem (3,4,7) = (3,3+1,3+4)
Seed(4) = Stem (4,5,9,180) = (4, 4+1, 4+5,4*5*9)
Seed(5) = Stem (5,6,11,330, 5^6^11^330)(Please remember that power towers evaluate from the top down, so first you do 11^330 which should yield a 340 digit number, and this is the exponent of 5^6^) At this point, scientific notation breaks
Seed(6) = Stem (6,7,13,546,6^7^13^546,6^^7^^13^^546^^(6^7^13^546)) (a tower of 546^'s, with 6^7^13^546 as the height)
Seed(7) = Stem (7,8,15,840,7^8^15^840,7^^8^^15^^840^^(7^8^15^840), 7^^^..you get the point)
Seed(8) = Stem (8,9,17,1224,8^9^17^1224,8^^9..1224^^(8^9^17^1224), 8^^^9...1224^^^(8^^9..1224^^(8^9^17^1224)), 8^^^^(((...))))
Stem is a variablearguments function
It goes as follows
Stemming, the diagonalization step
With Only one argument stem is basically leaf (10^n)
Stem (1) = Leaf (10)
Stem (2) = Leaf (100)
Stem (3) = Leaf (1000) (n is basically the amount of zeroes)
Stem (x,y) = Stem (X) when y = 1)
Stem (x,2) = Leaf(Leaf(X^10))
Stem (x,3) = Leaf(Leaf(Leaf(X^10)))
Stem (x,y) = You end up with nested leaf functions with (x) at the core, Y deep.
So For example: Stem (4,3) = Leaf (Leaf (Leaf(10000)))
With three arguments
The final two arguments, become x and y as in the two argument stem function, the next argument to the left becomes
the power of the leaf function. The function recurses that many times.
Stem (x, y, z) = LeafX(LeafX...(z deep... leaf(y)))
Stem (2, 3, 4) = Leaf2(Leaf2(Leaf2(Leaf2(3)))), Where Leaf2(n)=Leaf(leaf(n))
With four arguments
Stem (w,x,y,z) = Stem(x,y,z^w)
With six arguments
Stem (u,v,w,x,y,z) = Stem(...x^^^u ,y^^v, z^w)
Lets say 20 arguments
Stem (a b ... u) = Stem (a^^^^^^^^^^d... r^^^o s^^p t^q) This is a new 10 argument stem and so again it reduces to a 5 argument stem and finally to a 3 argument stem
For stems beyond 27 arguments, you may use doubleletters for variables, beyond 729 arguments, triple letters
Isnt the alphabet 26 letters?...whatever
These become new symbols and do not indicate multiplication in the traditional algebraic sense
Now onto Leaf, the matrixation step
Leaf (1) = is a one dimensional matrix with 1 in all places, you resolve it by applying successor to it, so Leaf (1) = 2

Leaf (2) = 2 dimensional matrix with rows(left right), columns(up down) of size 2 and 2 in all places.
2,2
2,2
This is an additive matrix you resolve it by doing addition so that each element sums every other element once, for example
a b
d c
a+b,a+c,a+d,b+a,b+c,b+d...
So Leaf(2) should be 48 (12 pairs of twos)

Leaf (3) = (3,3,3)(3,3,3)(3,3,3)
This is a 3d matrix of 3's its resolved by multiplication of all pairs (again, non commutatively so a*b does not overwrite b*a) and then taking the product of the results, or
abc jkl rst
def mnñ uvw
ghi opq xyz
 indicates a break into 3d space.
So, (a*b)*(a*c)*.....(y*x)(y*z). This should be 9^702 according to combinatorics. Around a 700 digit figure only using Leaf(3), nice.
Note: ai can be visualized as the bottom part of an oreo, jq the filling and rz the top part, i figured, if you're is still here up to this part
you deserve a cookie
(i'm mexican so i am allowed to use ñ, so if you look back you'll see why my alphabet is 27 in size)

Leaf (4) = (4,4,4,4)(4,4,4,4)(4,4,4,4)(4,4,4,4)
This is a 4d Matrix of 4's. It is resolved by Exponentiating all pairs then making an exponentiation tower with the results
abcd pqrs fghi uvwx \
efgh tuvwjklm yzabb \
ijkl xyzaa*nñop cdef \ Copy everything another 3 times
mnño bcdeqrst ghij \
So, (a^b)^(a^c).... up to ...^(jn^jm)
* indicates that the letters continue from ab to az, again, avoiding the symbol limit of the alphabet
This has 256 unique elements so 65280 pairs of (4^4) so a tower of 256^256^...^256 of height 65280 in other words 256 ^^ 65280
So Leaf(4) beats mosers number but isn't too impressive yet
 indicates a break into 3d space. up from the computer screen, i guess
\ indicates a break into 4d space, that special spacey space where tesseracts rotate

Leaf (5) is a 5d tetration matrix
Not even gonna try to write it down, nor visualize it, but there should be 3125 unique elements leading to 9762500 pairs. In knuths it should look like (5^^5)^^(5^^5)^^...9762497 times...^^(5^^5) or a tower of tetration with 9762500 height. Remember 5^^5 is 5^5^5^5^5 so around 450 digits at every step of that staircase. Or 5^^^9762500
Leaf (6) is a 6d ^^^ matrix
46656 unique elements, 2176735680 pairs. Each (6^^^6). Thats, (6000 digit number)^^^(itself)...2176735680 times. Now to remind the power of ^^^, 3^^^3 breaks single up arrow and power tower notations, as you would need a 3^3 tower 7 trillion high. Then to work your way down. Leaf (6) is 6^^^6 pentated to itself 2176735678 times. Or 6^^^^2176735678
So my number is seed(99) because you can't store more than 99 of them in your item slot.
And so, we shall go to war
Sri Sumbhajee
let me know what you think guys, i trimmed it (hehehe) from my original idea which had flowers, fruits and branches to make it more concise.
the seed (n) function outputs the Stem function with n arguments
The value for the first argument is n, the second argument is successor n, the third value is the sum of all the previous arguments, the fourth value is the product of all the previous arguments, the fifth value is a power tower with all the previous arguments in ascending order, the sixth value is a tetration tower, the seventh a ^^^ sequence etc..
Seeding: The recursion step
Seed(1) = Stem (1)
Seed(2) = Stem (2,2+1) =(2,3)
Seed(3) = Stem (3,4,7) = (3,3+1,3+4)
Seed(4) = Stem (4,5,9,180) = (4, 4+1, 4+5,4*5*9)
Seed(5) = Stem (5,6,11,330, 5^6^11^330)(Please remember that power towers evaluate from the top down, so first you do 11^330 which should yield a 340 digit number, and this is the exponent of 5^6^) At this point, scientific notation breaks
Seed(6) = Stem (6,7,13,546,6^7^13^546,6^^7^^13^^546^^(6^7^13^546)) (a tower of 546^'s, with 6^7^13^546 as the height)
Seed(7) = Stem (7,8,15,840,7^8^15^840,7^^8^^15^^840^^(7^8^15^840), 7^^^..you get the point)
Seed(8) = Stem (8,9,17,1224,8^9^17^1224,8^^9..1224^^(8^9^17^1224), 8^^^9...1224^^^(8^^9..1224^^(8^9^17^1224)), 8^^^^(((...))))
Stem is a variablearguments function
It goes as follows
Stemming, the diagonalization step
With Only one argument stem is basically leaf (10^n)
Stem (1) = Leaf (10)
Stem (2) = Leaf (100)
Stem (3) = Leaf (1000) (n is basically the amount of zeroes)
Stem (x,y) = Stem (X) when y = 1)
Stem (x,2) = Leaf(Leaf(X^10))
Stem (x,3) = Leaf(Leaf(Leaf(X^10)))
Stem (x,y) = You end up with nested leaf functions with (x) at the core, Y deep.
So For example: Stem (4,3) = Leaf (Leaf (Leaf(10000)))
With three arguments
The final two arguments, become x and y as in the two argument stem function, the next argument to the left becomes
the power of the leaf function. The function recurses that many times.
Stem (x, y, z) = LeafX(LeafX...(z deep... leaf(y)))
Stem (2, 3, 4) = Leaf2(Leaf2(Leaf2(Leaf2(3)))), Where Leaf2(n)=Leaf(leaf(n))
With four arguments
Stem (w,x,y,z) = Stem(x,y,z^w)
With six arguments
Stem (u,v,w,x,y,z) = Stem(...x^^^u ,y^^v, z^w)
Lets say 20 arguments
Stem (a b ... u) = Stem (a^^^^^^^^^^d... r^^^o s^^p t^q) This is a new 10 argument stem and so again it reduces to a 5 argument stem and finally to a 3 argument stem
For stems beyond 27 arguments, you may use doubleletters for variables, beyond 729 arguments, triple letters
Isnt the alphabet 26 letters?...whatever
These become new symbols and do not indicate multiplication in the traditional algebraic sense
Now onto Leaf, the matrixation step
Leaf (1) = is a one dimensional matrix with 1 in all places, you resolve it by applying successor to it, so Leaf (1) = 2

Leaf (2) = 2 dimensional matrix with rows(left right), columns(up down) of size 2 and 2 in all places.
2,2
2,2
This is an additive matrix you resolve it by doing addition so that each element sums every other element once, for example
a b
d c
a+b,a+c,a+d,b+a,b+c,b+d...
So Leaf(2) should be 48 (12 pairs of twos)

Leaf (3) = (3,3,3)(3,3,3)(3,3,3)
This is a 3d matrix of 3's its resolved by multiplication of all pairs (again, non commutatively so a*b does not overwrite b*a) and then taking the product of the results, or
abc jkl rst
def mnñ uvw
ghi opq xyz
 indicates a break into 3d space.
So, (a*b)*(a*c)*.....(y*x)(y*z). This should be 9^702 according to combinatorics. Around a 700 digit figure only using Leaf(3), nice.
Note: ai can be visualized as the bottom part of an oreo, jq the filling and rz the top part, i figured, if you're is still here up to this part
you deserve a cookie
(i'm mexican so i am allowed to use ñ, so if you look back you'll see why my alphabet is 27 in size)

Leaf (4) = (4,4,4,4)(4,4,4,4)(4,4,4,4)(4,4,4,4)
This is a 4d Matrix of 4's. It is resolved by Exponentiating all pairs then making an exponentiation tower with the results
abcd pqrs fghi uvwx \
efgh tuvwjklm yzabb \
ijkl xyzaa*nñop cdef \ Copy everything another 3 times
mnño bcdeqrst ghij \
So, (a^b)^(a^c).... up to ...^(jn^jm)
* indicates that the letters continue from ab to az, again, avoiding the symbol limit of the alphabet
This has 256 unique elements so 65280 pairs of (4^4) so a tower of 256^256^...^256 of height 65280 in other words 256 ^^ 65280
So Leaf(4) beats mosers number but isn't too impressive yet
 indicates a break into 3d space. up from the computer screen, i guess
\ indicates a break into 4d space, that special spacey space where tesseracts rotate

Leaf (5) is a 5d tetration matrix
Not even gonna try to write it down, nor visualize it, but there should be 3125 unique elements leading to 9762500 pairs. In knuths it should look like (5^^5)^^(5^^5)^^...9762497 times...^^(5^^5) or a tower of tetration with 9762500 height. Remember 5^^5 is 5^5^5^5^5 so around 450 digits at every step of that staircase. Or 5^^^9762500
Leaf (6) is a 6d ^^^ matrix
46656 unique elements, 2176735680 pairs. Each (6^^^6). Thats, (6000 digit number)^^^(itself)...2176735680 times. Now to remind the power of ^^^, 3^^^3 breaks single up arrow and power tower notations, as you would need a 3^3 tower 7 trillion high. Then to work your way down. Leaf (6) is 6^^^6 pentated to itself 2176735678 times. Or 6^^^^2176735678
So my number is seed(99) because you can't store more than 99 of them in your item slot.
And so, we shall go to war
Sri Sumbhajee
let me know what you think guys, i trimmed it (hehehe) from my original idea which had flowers, fruits and branches to make it more concise.
 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, let's see, I'm going to compare your number to Actaeus's number here, because I think it's the landmark on the thread for a clean, fast growing function hard to surpass.
The base power of your number runs on Leaves, and, at some point, the number of matrices and dimensions that they have aren't going to be as important as the operations you run on them. I.e. applying the next successor to the elements you already have is going to be a lot more powerful than adding n (say, Leaf(n1)) dimensions of elements to the same successor.
So, Leave (n) doesn't increase as fast as a→^{n} b increases.
Next, you have Stem with arguments. To simplify, they make Leaves recursive, but their power can't exceed the power of recursive arrows (because they are more powerful than leaves), so:
a→_{n} b > Stem (x,n)
Every argument added adds a new meta recursive layer. This is because, at this point your number plateaus in its growth. That is, the Leaves are going to be using more powerful successors than what you're using here, applying a recursive lead x^^^u times isn't going to make much of a difference than applying it x^^^^u times because the successors inside the Leaves have surpassed the ^^^^ power long time ago, and you could as well be using !!! and !!!! factorials instead.
This means the power of your number will now depend on the number of arguments. This can be simplified as:
Stem (x,... n number of argumens ...,u) < a→_{b},_{n} a.
Finally, we reach the Seed step. Once again the powers in the Leaves are going to dominate the seeds, so the power of the last seed will not be more powerful than ↑^{99}, which was surpassed by the Leaves long time ago.
We know that Seed(99) can't have more than 99 arguments, so we know that, using Acteaus's notation posted on page 3 of the thread, 99→_{99,100}99 will have surpassed your number.
The base power of your number runs on Leaves, and, at some point, the number of matrices and dimensions that they have aren't going to be as important as the operations you run on them. I.e. applying the next successor to the elements you already have is going to be a lot more powerful than adding n (say, Leaf(n1)) dimensions of elements to the same successor.
So, Leave (n) doesn't increase as fast as a→^{n} b increases.
Next, you have Stem with arguments. To simplify, they make Leaves recursive, but their power can't exceed the power of recursive arrows (because they are more powerful than leaves), so:
a→_{n} b > Stem (x,n)
Every argument added adds a new meta recursive layer. This is because, at this point your number plateaus in its growth. That is, the Leaves are going to be using more powerful successors than what you're using here, applying a recursive lead x^^^u times isn't going to make much of a difference than applying it x^^^^u times because the successors inside the Leaves have surpassed the ^^^^ power long time ago, and you could as well be using !!! and !!!! factorials instead.
This means the power of your number will now depend on the number of arguments. This can be simplified as:
Stem (x,... n number of argumens ...,u) < a→_{b},_{n} a.
Finally, we reach the Seed step. Once again the powers in the Leaves are going to dominate the seeds, so the power of the last seed will not be more powerful than ↑^{99}, which was surpassed by the Leaves long time ago.
We know that Seed(99) can't have more than 99 arguments, so we know that, using Acteaus's notation posted on page 3 of the thread, 99→_{99,100}99 will have surpassed your number.
Re: My number is bigger!
Well, i wont be making anyone feel like virgin's today, but it was nice to see it's well explained enough to be given a size comparison, i like your ideas of illusion space, and i do believe it's one of the nicest sumbissions here since you use ordinals only for measuring.
Do you think seed(n) is far from ↻(n)?
Do you think seed(n) is far from ↻(n)?
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