Goahead52's Math Posts

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Goahead52
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Goahead52's Math Posts

Postby Goahead52 » Mon Jul 25, 2016 2:27 pm UTC

Hi,

Let me start with examples :

2!=2=1*2 the difference between the factors 2 ans 1 is = 1 and 1 is =1! or 0!
3!=6=2*3 the difference between the factors 3 ans 2 is = 1 and 1 is =1! or 0!
4!=24=4*6 the difference between the factors 6 ans 4 is = 2 and 2 is =2!
5!=120=10*12 the difference between the factors 12 ans 10 is =2 and 2=2!
6!=720=24*30 the difference between the factors 24 ans 30 is = 6 and 6 is =3!
and so on

0! and 1! have no solution the difference of the 2 factors is not equal to some factorial.
The general condition is :
n!=A*B where B-A=k!
Can we always find a solution?

Is the set of factorials satifying the condition above finite or infinite?
Can we prove it?
Thank you for any clue.

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Carlington
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Re: Factorials

Postby Carlington » Mon Jul 25, 2016 2:45 pm UTC

I can prove it's infinite pretty quickly, actually.

n! = n×(n-1)×...×1

For all n > 2, this is 1 × 2 × ... × n, which by definition has factors 1 and 2.
2 - 1 = 1 = 1!

I'm assuming you had a more specific method of choosing the factors A and B. What is your method?
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Re: Factorials

Postby Sabrar » Mon Jul 25, 2016 2:50 pm UTC

Quick work with Excel shows no solution for n=8 (for n=7 we have 70*72=7!)

@Carlington: I think you missed the stipulation that the product of the factors must result in n!.

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Re: Factorials

Postby Carlington » Mon Jul 25, 2016 2:52 pm UTC

You're right, I did. :oops:
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Goahead52
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Finding k such that (n^k) is the biggest number < (n-1)!

Postby Goahead52 » Mon Jul 25, 2016 3:07 pm UTC

Hi,

Solve
Find k such as
(n^k) is the biggest number < (n-1)!

n=5

4!=24
k=1

25=5^2> 24 so k=1

n=7

6!=720
k=3

7^3=343 < 720

as n grows is there any way to find quickly k knowing n ?
When n is not very large it easy to find k
Here are the list of the first few k
0,0,0,1,1,2,3,4,4,5,6,7,7,8,9,10,10,11,12,13,13,14 ,15,16,17,17... (not LISTED in OEIS)
When n is 300 digits or more we have to find another way or algorithm to reach the result quickly.
Some approximation function?

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Re: Factorials

Postby Zohar » Mon Jul 25, 2016 3:12 pm UTC

Yeah I was playing around with 8! and didn't get to a quick solution, but I didn't check all options.
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Re: Factorials

Postby Soupspoon » Mon Jul 25, 2016 3:18 pm UTC

First thoughts:
2!'s A is 1!'s result, B is the latest term
3!'s A is 2!'s result, B is the latest term
4!'s A is the latest term, B is 3!'s result (so why the switch..? Never mind, could be modulo/insigned difference.)
5!'s A is the multiple of (1?)*3*4, B is (1?)*2*5... Where's the pattern gone?
6!'s A is perhaps 1*2*3*4 (result of 4!) and B is 5*6, but could also be 4*6 and 2*3*5 (assign 1 as you wish, but the latter arw consecutive primes without... Not that I think that means anything).
7!'s A is... well, assuming we're looking for B-A of 4!, 60=(3*4*5) and 84=(2*6*7), I think uniquely, but... logic behind that?
8!... 120 difference... From inspection, 140 to 288 is too large a gap, 144 to 280 is also... can't find anything useful (150 partners with 268.8, 118point8 higher, as we advance towards (B-A=0) at root(8!)

Interesting, if confusing, start. But without even getting too far down the line, it starts to look just like loose coincidence. At least what I think you were trying to say.

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Re: Finding k

Postby jaap » Mon Jul 25, 2016 3:29 pm UTC

Use Stirling's approximation.
After a bit of algebra, you get:

k = floor( n - (n-0.9189)/ln(n) - 0.5)

That constant 0.9189 is actually ln(2*pi)/2. Of course it is not valid for n=0 or 1, and with n=2 it gives -1. Otherwise it matches with your list, even though Stirling's approximation is not accurate for small numbers.

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Re: Factorials

Postby Demki » Mon Jul 25, 2016 3:58 pm UTC

Food for thought:

n!=A*B
k!=B-A
So n!=A(A+k!)
Or A^(2)+k!A-n!=0
Solving for A:
A=(-k!±sqrt(k!2+4n!)/2
If you want only positive A then limit it to only +.
This means that k!2+4n! must be a perfect square,
So maybe for given n you could try to find k such that k!2+4n! is a perfect square.

This may or may not help in finding whether there are infinitely many n with thay property.
Last edited by Demki on Mon Jul 25, 2016 6:23 pm UTC, edited 1 time in total.

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Re: Factorials

Postby Tub » Mon Jul 25, 2016 4:46 pm UTC

Lacking some profound insight, I would expect these things to be random chance. It should be easy to write a small program testing at least until 20! (with fast 64 bit integers) or higher (with arbitrary precision integers), but I'm not going to do that.

Instead, let's see if we can gather some insight via stochastics.
If you think about the prime factorization of n!, you will find that it approaches 2^(n/1) * 3^(n/2) * 5^(n/4) * 7^(n/6) ...
Dividing those factors between A and B, the amount of unique factorizations is slightly smaller than half of
sum(p is prime, p < n) ( n/(p-1) + 1 )

We know the partial sums of the reciprocals of the prime, we know the prime counting function, and eventually we can say that the amount of unique factorizations n! = A*B is smaller than n^2 (at least for sufficiently large n).

We also know that |A-B| < n!, and we know that there are only (n-1) factorials k! with k<n in that range. If we assumed a uniform distribution of that difference (which is wrong, but let's do it anyway), we have less than n^2 trials each with less than 1/(n-1)! chance to hit a factorial. The chance for n! to satisfy A-B = k! for any A*B=n! approaches 0 as n increases.

So my guess is: there is an infinite amount of numbers which do not have a solution.


If we stubbornly continue to assume an uniform distribution, the expected amount of numbers satisfying your property is finite. But assuming an uniform distribution gives us too low of a result, so I'm not sure.

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Re: Finding k

Postby Goahead52 » Mon Jul 25, 2016 7:01 pm UTC

jaap wrote:Use Stirling's approximation.
After a bit of algebra, you get:

k = floor( n - (n-0.9189)/ln(n) - 0.5)

That constant 0.9189 is actually ln(2*pi)/2. Of course it is not valid for n=0 or 1, and with n=2 it gives -1. Otherwise it matches with your list, even though Stirling's approximation is not accurate for small numbers.

Thank you mister Jaap for your solution.
As I did not understand clearly your formula I have obtained using logs for :
n=100 k=77
n=200 k=162

Do you have the same results?

Thank you

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Re: Finding k

Postby jaap » Mon Jul 25, 2016 8:42 pm UTC

Goahead52 wrote:
jaap wrote:Use Stirling's approximation.
After a bit of algebra, you get:

k = floor( n - (n-0.9189)/ln(n) - 0.5)

That constant 0.9189 is actually ln(2*pi)/2. Of course it is not valid for n=0 or 1, and with n=2 it gives -1. Otherwise it matches with your list, even though Stirling's approximation is not accurate for small numbers.

Thank you mister Jaap for your solution.
As I did not understand clearly your formula I have obtained using logs for :
n=100 k=77
n=200 k=162

Do you have the same results?

Thank you


77 and 161:
http://www.wolframalpha.com/input/?i=floor(+n+-+(n-ln(2*pi)%2F2)%2Fln(n)+-+1%2F2)+for+n%3D100+and+200

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Re: Finding k such that (n^k) is the biggest number < (n-1)!

Postby Goahead52 » Mon Jul 25, 2016 8:54 pm UTC

Thank you very much.
Now I can use the formula written in Latex.
There is a difference of 1 as n grows to 200. Maybe more when n reach 300 digits.
Anyway I have to use maybe some properties of sequences instead of Stirling approximation.
k=floor (sigma log(i) with i varying from 2 to n-1 divided by log(n))
How to reach a better approximaiion of Sigma(log(i) with i varying from 2 to (n-1)?

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Re: Finding k such that (n^k) is the biggest number < (n-1)!

Postby jaap » Mon Jul 25, 2016 9:25 pm UTC

Goahead52 wrote:Thank you very much.
Now I can use the formula written in Latex.
There is a difference of 1 as n grows to 200. Maybe more when n reach 300 digits.


What do you mean by "a difference of 1"? The approximation gives exactly the right answer, except for n=2.

Actual answers you want:
http://www.wolframalpha.com/input/?i=floor(+ln((n-1)!)+%2F+ln(n)+)+for+n%3D2,+200
Approximation:
http://www.wolframalpha.com/input/?i=floor(+n+-+(n-ln(2*pi)%2F2)%2Fln(n)+-+1%2F2)+for+n%3D2,+200
Difference:
http://www.wolframalpha.com/input/?i=floor(+ln((n-1)!)+%2F+ln(n)+)+-+floor(+n+-+(n-ln(2*pi)%2F2)%2Fln(n)+-+1%2F2)+for+n%3D2,+200

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Re: Finding k such that (n^k) is the biggest number < (n-1)!

Postby Goahead52 » Mon Jul 25, 2016 9:45 pm UTC

I`m very sorry.
It was my fault : instead of reading the line 200 (k=161) on Excel I read 201 (k=162).
Very sorry.
Anyway when n reach 300 digits and more I do not know if your formula gives the right value of k.
Thank you very very much for your help.
I need the exact value of k (that is the first step).
As (n-1)! will be equal n^k+r (where r is some remain) it will be easy to find a primality test 100% accurate.
Finding r is the second step (hardest one). I have yet some ideas on how to compute r

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Re: Factorials

Postby notzeb » Mon Jul 25, 2016 10:13 pm UTC

Demki wrote:This means that k!2+4n! must be a perfect square,
Plugging in k = 2, we get an open problem.

However, if we believe the abc conjecture then every solution has to have n pretty close to 2k:
First, it's pretty clear that n has to be at least twice as large as the largest prime below k. Thus n can't be too much smaller than 2k, so we just have to show that n can't be too much larger than 2k. If n is at least 2k, then (k!)2 divides n!, and we get 1 + (4n!)/(k!)^2 = square, say m2. Then according to the abc conjecture, we should have rad(m2 * (4n!)/(k!)^2) at least as large as something close to m2. That radical is at most as large as m*rad(n!), so we get rad(n!) is at least as large as something close to m. By the prime number theorem, rad(n!) is approximately en, so we get something like en > sqrt(n!)/k!. When n is 2k, the right hand side is the square root of 2k choose k, which is just a little bit less than 2k. When n is a bit more than 2k, say n = 2k + l, the right hand side becomes bigger than sqrt(2k)l, and we get e2k+l > sqrt(2k)l, so l has to be small compared to k, and so n can't be too much larger than 2k.

It seems somehow hard to rule out that there aren't infinitely many solutions with n = 2k. I tried some examples, and n = 2k works for k = 1, 2, 3, 9, and then never seems to work again.

Edit: fixed some weird typos.
Last edited by notzeb on Tue Jul 26, 2016 4:50 pm UTC, edited 1 time in total.
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Re: Factorials

Postby Goahead52 » Tue Jul 26, 2016 11:47 am UTC

Here is an algorithm to test quickly if such k exists.
1. Compute (k!)^2 < n! to find the maximal k
2. Compute from 2 to k the number r=n!/(k!)^2
3. If r is the product of 2 consecutive numbers a*(a+1) then the number k corresponding is solution otherwise compute the next value of k.

Example : n=6
6!=720
int(sqrt(720))=26
k=4 because (4!)^2=576<720
r=720/(3!)^2=720/36=20
20=4*5 hence k=3
720=24*30=(6*4)*(6*5)=(3!*4)*(3!*5)

I hope that anyone can understand the principle of my little algo and why it works.

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Theorem and consequences (powers between x^k and (x+1)^k)

Postby Goahead52 » Thu Jul 28, 2016 12:20 pm UTC

Hi,

Theorem :
k and x are positive integers
k>1
x>1
Between x^k and (x+1)^k non-inclusive there exist either 0 or 1 numbers expressed as y^(k+1).

Example :
Between the 2 squares x^2 and (x+1)^2 non-inclusive there is either 0 or at most 1 cube

What are the consequences of this theorem if proved true?

Thank you

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Re: Theorem and consequences

Postby Cauchy » Thu Jul 28, 2016 6:31 pm UTC

The difference between x^k and (x+1)^k is on the order of kx^{k-1} (by the derivative), and the difference between y^{k+1} and (y+1)^{k+1} is on the order of (k+1)y^k. The smallest k+1-st power number above x^k is larger than (x^{k/(k+1)})^{k+1}, so the difference between it and the next k+1-st number is on the order of (k+1)x^{[k/(k+1)]*k} = (k+1)x^{k-1+1/(k+1)}. which is larger than kx^{k-1}. So I would certainly expect the second k+1-st power number after x^k to be bigger than (x+1)^k, and therefore I would not expect two k+1-st power numbers to be between two kth power numbers.
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Re: Theorem and consequences

Postby Nitrodon » Thu Jul 28, 2016 6:48 pm UTC

Indeed, the theorem can be proven.

The case x=1 is trivial. Assume x>1, and let z be the positive real number satisfying z^(k+1) = x^k. Then z = x^[k/(k+1)] < x, and thus (x+1)/x < (z+1)/z.

Hence [(x+1)/x]^k < [(z+1)/z]^k < [(z+1)/z]^(k+1). Multiplying by x^k = z^(k+1), we obtain (x+1)^k < (z+1)^(k+1)

Now, any y satisfying x^k < y^(k+1) < (x+1)^k must satisfy z^(k+1) < y^(k+1) < (z+1)^(k+1), and hence z < y < z+1. There can only be one integer within this interval, so we are done.

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Re: Theorem and consequences

Postby arbiteroftruth » Thu Jul 28, 2016 7:03 pm UTC

Proof by contradiction:

Assume xk<yk+1<(y+1)k+1<(x+1)k
When all numbers involved are positive, taking the (k+1)th root preserves the inequalities:
xk/(k+1)<y<y+1<(x+1)k/(k+1)
This implies:
(x+1)k/(k+1)-xk/(k+1)>1
Since x is restricted to positive integers and the function xk is differentiable, this further implies:
∃x≥1 d/dx xk/(k+1)>1
∃x≥1 k/(k+1)*x-1/(k+1)>1
But k/(k+1)<1, and for x≥1, x-1/(k+1)≤1, so there is no x satisfying the inequality, and we have our contradiction.

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Re: Theorem and consequences

Postby Goahead52 » Fri Jul 29, 2016 4:01 pm UTC

The idea behind this theorem which is obvious is some sort of personal puzzle.
Now that I know that this theorem is true what can I do with this truth?
I could add another theorem existing or future to make things more complicated.
I have no answer for now.
I`m still thinking on how to use it to reach some non obvious result.
The three proofs are correct.

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Re: Theorem and consequences

Postby Goahead52 » Fri Jul 29, 2016 7:53 pm UTC

Here is my idea.
I expose it as it came to my mind when I posted.
Let us take the sum of cubes :

x^3+y^3=z^3 (1)

We know that the equation (1) does not hold (A.Wiles proved it)

The first x^3 will be between 2 squares a^2 and (a+1)^2
The second y^3 will be between 2 squares b^2 and (b+1)^2
If we sum the lowest values lw=a^2+b^2 and the biggest values bg=(a+1)^2+(b+1)^2
can we have z^3 between lw and bg for any pair of cubes?

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Re: Theorem and consequences

Postby Goahead52 » Fri Jul 29, 2016 8:53 pm UTC

Goahead52 wrote:Here is my idea.
I expose it as it came to my mind when I posted.
Let us take the sum of cubes :

x^3+y^3=z^3 (1)

We know that the equation (1) does not hold (A.Wiles proved it)

The first x^3 will be between 2 squares a^2 and (a+1)^2
The second y^3 will be between 2 squares b^2 and (b+1)^2
If we sum the lowest values lw=a^2+b^2 and the biggest values bg=(a+1)^2+(b+1)^2
can we have z^3 between lw and bg for any pair of cubes?

We could derive from what I wrote above a second theorem not easy to prove.
For any pair of cubes (x^3,y^3) there is no cube z^3 between lw and bg.
If this theorem is true then we have an elementary proof that x^3+y^3 is NOT equal to some cube z^3.

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Re: Theorem and consequences

Postby curiosityspoon » Fri Jul 29, 2016 10:56 pm UTC

The n=3 special case of Fermat was already proven by Euler, long before Wiles came along.

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Re: Theorem and consequences

Postby Goahead52 » Fri Jul 29, 2016 11:06 pm UTC

curiosityspoon wrote:The n=3 special case of Fermat was already proven by Euler, long before Wiles came along.


I gave the example of n=3 but we could extend to n>3.
First : I need to prove that my second theorem is true for n=3.
Second : Any power n>3 could be bounded by 2 squares and so on.
Thank you for your comment.
A new elementary proof is always a plus.
Is my second true or not? That is my question. I need some help at this level.

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Re: Theorem and consequences (powers between x^k and (x+1)^k)

Postby Goahead52 » Sun Jul 31, 2016 3:52 pm UTC

As a<x^3<(a+1)^2 and b<y^3<(b+1)^2 then if (a+b)<z^3<(a+1)^2+(b+1)^3 we could conclude that z^3=x^3+y^3
Am I right?

The same prevail for powers n>3 as long as we use the squares as bounds.

False or true?

Thank you for any comment.

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Re: Theorem and consequences (powers between x^k and (x+1)^k)

Postby Goahead52 » Mon Aug 01, 2016 11:00 pm UTC

Sorry! I have found a reasoning mistake and I need to restate my second theorem by correcting it.

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Primes, sequences and infinite

Postby Goahead52 » Sun Aug 14, 2016 2:51 pm UTC

Hi,

Let
A(p)=(1/2+1/2)+(1/3+1/3+1/3)+(1/5+1/5+1/5+1/5+1/5)+....+(1/p+1/p+...1/p) p times

S(P)=1/2+1/3+1/5+1/7+....1/p

A(p)=pi(p) where pi(p) is the counting function of primes

What is the exact limit of S(p) when p goes to infinite?

Thank you for any comment.

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Re: Primes, sequences and infinite

Postby lightvector » Sun Aug 14, 2016 3:24 pm UTC


Goahead52
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Re: Primes, sequences and infinite

Postby Goahead52 » Sun Aug 14, 2016 3:28 pm UTC


I did it long time ago.
What I mean by limit is asymptotic limit.
I thought it was obvious.
Sorry for not mentioning it.

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Re: Primes, sequences and infinite

Postby Goahead52 » Sun Aug 14, 2016 3:29 pm UTC

In other words how can we express A(P) in S(p)?

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Re: Primes, sequences and infinite

Postby PsiCubed » Sun Aug 14, 2016 4:34 pm UTC

Don't know what the asymptotic limit is, but S grows very very slowly:

S(p59) ~ 2

S(p361139) ~ 3

S(p43922730588128390) ~ 4
Image

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Re: Primes, sequences and infinite

Postby lorb » Sun Aug 14, 2016 4:40 pm UTC

It's ln(ln(n)). You can find that information on the linked wiki page.
Please be gracious in judging my english. (I am not a native speaker/writer.)
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Re: Primes, sequences and infinite

Postby Goahead52 » Sun Aug 14, 2016 4:52 pm UTC

lorb wrote:It's ln(ln(n)). You can find that information on the linked wiki page.

Thank you..
Everything I can read on wiki or on internet I did it before posting.
Here is what I want to expand.
I need to expand A(p) in function of S(p)
here are the first elements :
A(p)=(2*S(P))+(S(p)-(1/2))+(S(p)-(1/2+1/3))+......
=S(p)(2+1+1+......)-(1/2+(1/2+1/3).....)

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Re: Primes, sequences and infinite

Postby Goahead52 » Sun Aug 14, 2016 5:13 pm UTC

I did the computation of A(p) and S(P) up to the 10000th prime :

10000th prime = 104729
When A(p)=10000 S(P)=2.70925824879732
If I could find closed formula for A(p)=g(S(P)) for some given p (g is some function to find) then I could after simplification find more precisely the asymptotic limit of S(P).
If I find more precise than ln(ln(n)) it will be more interesting result.

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Re: Primes, sequences and infinite

Postby cyanyoshi » Sun Aug 14, 2016 7:25 pm UTC

S(p) ≈ ln(ln(p)) + M where M is the Meissel-Mertens constant, so p ≈ exp(exp(S-M)). It's also known that π(p) ≈ p/(ln(p)-1) for large enough p. These together suggest that A(p) ≈ exp(exp(S(p)-M))/(exp(S(p)-M)-1). Is that what you are looking for?

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Re: Primes, sequences and infinite

Postby Goahead52 » Sun Aug 14, 2016 7:38 pm UTC

Thank you anyway.
Results do exist : asymptotic limit of S(p) and A(p). Each one of them was obtained using known results.
That is not my purpose.
I will do it by myself (I mean the expansion of A(p) in function of S(p)) even if it is going to take long time.
I know what I`m looking for.
The result will help me to solve another problem.

Thank you very much to all those who try to help me.

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Re: Primes, sequences and infinite

Postby Goahead52 » Sun Aug 14, 2016 9:09 pm UTC

I start working on the details.
A(p) could be divided in 2 sequences : one positive A`(p) and the second negative A``(p).
A`(p)=p*S(p) (this one is solved
A``(p) depends mainly on the gaps between 2 consecutive p(n+1)-p(n) (it is more complicated to simplify.
I`m not looking for an approximation of A``(p). The exact closed formula.
How to simplify A``(p)?
That is what I have to solve.
I will let you know once finished.

lorb
Posts: 404
Joined: Wed Nov 10, 2010 10:34 am UTC
Location: Austria

Re: Primes, sequences and infinite

Postby lorb » Sun Aug 14, 2016 9:14 pm UTC

It seems you are looking for a closed formula for this?
Please be gracious in judging my english. (I am not a native speaker/writer.)
http://decodedarfur.org/


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