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 2019-04-06, 03:16 #1 jrsousa2   Dec 2018 Miami 29 Posts Working on a prime number formula, need a favor! I want to know if the below series converges to 1 even for pseudo-primes, such as 341. It seems to output 1 no matter the value of n. Is anybody here able to run this on a super computer with n=341? (wow, the Tex compiler here renders a very ugly scientific text) $\sum _{k=1}^{\infty } \frac{(-1)^k(2\pi )^{2k}}{2k+1}\sum _{p=1}^{2k} \frac{(-2)^{2k-p}}{p!(2k-p)!}\cdot\frac{2^p-2^{p(n+1)}}{1-2^p}$ Last fiddled with by jrsousa2 on 2019-04-06 at 03:20
 2019-04-06, 03:54 #2 VBCurtis     "Curtis" Feb 2005 Riverside, CA 10001010000102 Posts Why does it need a 'supercomputer'? What values of n have you evaluated, and what is the running time for the largest value?
 2019-04-06, 05:34 #3 jrsousa2   Dec 2018 Miami 358 Posts That was a figure of speech, LOL. As you know, we have to set a number of terms to be summed up (N). On Mathematica in my computer I'm able to run this up to $n=6$, which requires $N=550$ (and takes long to run). If $n=2$, N is just 20, which gives you an idea of how fast that thing is growing. Btw, if confirmed, I will post my formula for $\pi^*(x)$, the prime (and pseudo-primes per Fermat's little theorem) counting function. It's pretty interesting. Omg, the rendering of math symbols here is so ugly that maybe it's preferable to type in plain text, Lol.
2019-04-06, 13:26   #4
Xyzzy

"Mike"
Aug 2002

1E2416 Posts

Quote:
 Originally Posted by jrsousa2 Omg, the rendering of math symbols here is so ugly that maybe it's preferable to type in plain text, Lol.
$\sum _{k=1}^{\infty } \frac{(-1)^k(2\pi )^{2k}}{2k+1}\sum _{p=1}^{2k} \frac{(-2)^{2k-p}}{p!(2k-p)!}\cdot\frac{2^p-2^{p(n+1)}}{1-2^p}$

 2019-04-06, 14:48 #5 VBCurtis     "Curtis" Feb 2005 Riverside, CA 2·472 Posts How does any part of the calculation "require" N = 550? In the first part of your reply you remind us that we set N, but then you say 550 is "required". What are you talking about? Why not sum 100 terms (or 500) for each n first to see what happens? You still haven't told me how long it takes to run; you're on a forum where members routinely run tasks that take multiple weeks for a single calculation to finish (e.g. primality test on a 100-million-digit input, Primo proofs, or NFS factorization of 180+ digit number). If your idea of "a while" is a few hours, let it run overnight, or run your tests until mathematica crashes. Even then, you could break down the first sum into blocks of, say, 500 and then manually add up the outputs.
2019-04-06, 15:17   #6
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

26·7·13 Posts

Quote:
 Originally Posted by jrsousa2 (wow, the Tex compiler here renders a very ugly scientific text)
The tex is still better than the alternative that was posted.

BTW: It appears as though you are trying to prove your formula with empirical evidence.

BTW2: There have already been a few prime generating formulas posted here in the past. My favourite was the twenty-six simultaneous equations in twenty-six variables.
Attached Thumbnails

Last fiddled with by retina on 2019-04-06 at 15:29

2019-04-06, 16:29   #7
jrsousa2

Dec 2018
Miami

29 Posts

Quote:
 Originally Posted by Xyzzy $\sum _{k=1}^{\infty } \frac{(-1)^k(2\pi )^{2k}}{2k+1}\sum _{p=1}^{2k} \frac{(-2)^{2k-p}}{p!(2k-p)!}\cdot\frac{2^p-2^{p(n+1)}}{1-2^p}$
Much better, how did you do it?

2019-04-06, 16:33   #8
jrsousa2

Dec 2018
Miami

29 Posts

Quote:
 Originally Posted by VBCurtis How does any part of the calculation "require" N = 550? In the first part of your reply you remind us that we set N, but then you say 550 is "required". What are you talking about? Why not sum 100 terms (or 500) for each n first to see what happens? You still haven't told me how long it takes to run; you're on a forum where members routinely run tasks that take multiple weeks for a single calculation to finish (e.g. primality test on a 100-million-digit input, Primo proofs, or NFS factorization of 180+ digit number). If your idea of "a while" is a few hours, let it run overnight, or run your tests until mathematica crashes. Even then, you could break down the first sum into blocks of, say, 500 and then manually add up the outputs.
Omg, you're not very good at understanding things, reading between the lines if necessary.
I think I was pretty clear in my post, for n=6 you need N=550 terms of the series summed up, for n=2 you need N=20 terms.
I may be downright wrong, but to me that was a very clear explanation.

Sometimes I have the impression that many people on these math forums are autistic, and though it may sound rude to say this, it may be true.
They need things explained to them in the minimum details.

Last fiddled with by jrsousa2 on 2019-04-06 at 16:47

2019-04-06, 16:35   #9
jrsousa2

Dec 2018
Miami

29 Posts

Quote:
 Originally Posted by retina The tex is still better than the alternative that was posted. BTW: It appears as though you are trying to prove your formula with empirical evidence. BTW2: There have already been a few prime generating formulas posted here in the past. My favourite was the twenty-six simultaneous equations in twenty-six variables.
Lol. Mine should be much simpler. Yes, most of them are just curiosities, they are not very practical or useful. Like the digamma function (ugly) versus my formulae for the harmonic progressions.

Last fiddled with by jrsousa2 on 2019-04-06 at 16:35

 2019-04-06, 16:57 #10 10metreh     Nov 2008 2×33×43 Posts The expression is equal to 1 for all n (including pseudoprimes and composites). First we deal with the finite sum. It turns out it can be written in a much simpler form. $\sum _{p=1}^{2k} \frac{(-2)^{2k-p}}{p!(2k-p)!}\frac{2^p-2^{p(n+1)}}{1-2^p} = \frac{1}{(2k)!}\sum_{p=1}^{2k}{{2k}\choose{p}}(-1)^{2k-p}2^{2k-p}2^p\frac{2^{np}-1}{2^p-1} = \frac{2^{2k}}{(2k)!}\sum_{p=1}^{2k}{{2k}\choose{p}}(-1)^{2k-p}\sum_{r=0}^{n-1}2^{rp}$ Rearrange the order of summation to get $\frac{2^{2k}}{(2k)!}\sum_{r=0}^{n-1}\left(\sum_{p=1}^{2k}{{2k}\choose{p}}(-1)^{2k-p}2^{rp}\right) = \frac{2^{2k}}{(2k)!}\sum_{r=0}^{n-1}\left((2^r-1)^{2k}-1\right)$ Substitute this back into the main expression, and rearrange summation again (we can do this if everything converges nicely, which it does): $\sum _{k=1}^{\infty } \frac{(-1)^k(2\pi )^{2k}}{2k+1}\frac{2^{2k}}{(2k)!}\sum_{r=0}^{n-1}\left((2^r-1)^{2k}-1\right) = \sum_{r=0}^{n-1}\left(\sum_{k=1}^{\infty}\frac{(-1)^k(4\pi)^{2k}}{(2k+1)!}\left((2^r-1)^{2k}-1\right)\right)$ But we have $\sum_{k=1}^{\infty}\frac{(-1)^k x^{2k}}{(2k+1)!} = \frac{\sin x}{x}-1$ as long as x is not zero; when x = 0 it is just 0. So our main expression becomes $\sum_{r=1}^{n-1}\left(\frac{\sin(4\pi(2^r-1))}{4\pi(2^r-1)}-1\right)-\sum_{r=0}^{n-1}\left(\frac{\sin(4\pi)}{4\pi}-1\right) = -(n-1)-(-n) = 1$ (@retina are the  tags in xyzzy's post not working properly for you?)
 2019-04-06, 16:59 #11 VBCurtis     "Curtis" Feb 2005 Riverside, CA 2×472 Posts message deleted; after 10metreh's pretty explanation, further arguing on my part is useless. Last fiddled with by VBCurtis on 2019-04-06 at 17:10

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