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Old 2013-03-13, 03:43   #1
Batalov
 
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Default Problem E7 of Richard Guy's "Unsolved problems in number theory"

Quote:
Originally Posted by Tomás Oliveira e Silva (NMBRTHRY@LIST)
Richard K. Guy's "Unsolved problems in number theory" problem E7 asks if \sum_n (-1)^n n/p_n, with p_n the n-th prime number, converges or diverges (a question asked by P. Erdos). In its 2004 third edition, no references related to this problem were given.

Has anyone worked on it after 2003?

Preliminary numerical experiments suggest that the sum converges (slowly, of course) to a value close to -0.05216.
Is anyone interested to think about this problem?
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Old 2013-03-13, 04:45   #2
LaurV
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When one approximates n/Pn with 1/log, then it seems that it converges. No idea how to prove, tho... You ask me to prove it if Erdos could not?! :surprised

Last fiddled with by LaurV on 2013-03-13 at 04:46
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Old 2013-03-13, 07:51   #3
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Quote:
Originally Posted by LaurV View Post
You ask me to prove it if Erdos could not?! :surprised
Seems worthwhile to talk about.

Richard K. Guy replied:
Quote:
I haven't received any further information about this, but would be glad to learn if there is any. R.

Last fiddled with by only_human on 2013-03-13 at 08:30
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Old 2013-03-13, 12:20   #4
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Quote:
Originally Posted by Batalov View Post
Is anyone interested to think about this problem?
I'm not really mathematical enough to think about it to a great degree but:

Quote:
Originally Posted by princeps View Post

gives multiple formula for p_n

could these lead to a simplification that might have a easy way to solve ?

Last fiddled with by science_man_88 on 2013-03-13 at 12:23
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Old 2013-03-13, 18:42   #5
only_human
 
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To help frame this question about convergence, I've included the text as stated in Richard Guy's book:
Quote:
If pn is the n th prime, Erdős asks if ∑ (-1)n n/pn converges. He notes that the series ∑ (-1)n (n ln n)/pn diverges.

Last fiddled with by only_human on 2013-03-13 at 18:54 Reason: to use block quote
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Old 2013-03-13, 19:25   #6
Batalov
 
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I reposted the message from NMBRTHRY, because it looked like a fantastic exercise for any skill level (except for a very low one. Think of it as a projectEuler-on-steroids problem, -- only the answer is not known). For a technically minded audience (myself likely included; I have no relevant theoretical knowledge), there's an interesting aspect of summation with error accumulation estimate and control. The zero level thinking would be that "surely, the positive and negative errors will cancel on average"; one needs to depart from that at least one level higher, and get a bound for expected error, or else after a while the summation will become meaningless. The sequence grows very, very slowly.

I've tinkered for half an hour to see where I could get naïvely; ran some simple-minded calculation with a 40-liner and I was barely able to see the value -0.0304 surpassed. (So, that's ways and ways from Oliveira e Silva's value! And it is growing so slow, that I cannot yet imagine what modeling he used...) I run summation in pairs of terms, with the parwise sum rewritten to minimize error. I left it for myslef for the weekend to hijack yafu sieve and refactor with the strength of yafu code behind this. Could be fun for anyone, I thought, ...and that's when I reposted.
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Old 2013-03-13, 19:35   #7
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Quote:
Originally Posted by Batalov View Post
I reposted the message from NMBRTHRY, because it looked like a fantastic exercise for any skill level (except for a very low one. (...//...) Could be fun for anyone, I thought, ...and that's when I reposted.
I agree wholeheartedly.
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Old 2013-03-13, 20:08   #8
R.D. Silverman
 
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Quote:
Originally Posted by Batalov View Post
I reposted the message from NMBRTHRY, because it looked like a fantastic exercise for any skill level
Actually, the problem is much subtler than it seems. The difficulty is
showing that the series does not "diverge by oscillation".

This, in turn, involves showing that there do not exist too many intervals
in which the primes are somewhat denser than average, followed by
intervals in which the primes are somewhat less dense than average.

Quote:
The zero level thinking would be that "surely, the positive and negative errors will cancel on average"; one needs to depart from that at least one level higher, and get a bound for expected error, or else after a while the summation will become meaningless. The sequence grows very, very slowly.
Indeed.

The problem is showing that the errors are essentially a 'random walk';
that they oscillate positive and negative, and do not have long runs
(when primes are too close (or sparse) on average) where the error
remains either positive or negative for long periods.

The answer depends on very subtle issues in the distribution of primes.
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Old 2013-03-13, 21:28   #9
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Questions regarding the distribution of primes are always interesting.
(As an analogy, looking at problems like this with a practiced eye helps. The example of Marion Tinsley shows that you should have a feeling about what approach will lead to a desired outcome.)
I do have some interesting results which may apply to this question but at present I have more questions than answers. A recent prime I posted is a test value for a certain line of inquiry that is associated with this limit process.
I'll have more to post if successful.
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Old 2013-03-13, 22:38   #10
Batalov
 
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I have sieved to p(n)<30e9 (that's up to n<1300005928; it was a toy sieve, and without bit packing I blew up memory already; need to refactor with yafu or A.Kulsha's program?; need to walk in windows after that) and the growth is slower than ln ln n. Of course, it can oscillate on much larger scale. Oliveira e Silva suggests that it converges to -0.05216, which is on my scale (so far) nowhere near in sight...
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Last fiddled with by Batalov on 2013-03-13 at 22:45 Reason: sorry, 30e9; with bit packing can sieve easily to 1e12, of course
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Old 2013-03-14, 13:24   #11
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This is my attempt at the conditional proof:

We will perform the sum two terms at a time so we will always add positive numbers.

On the Riemann hypothesis:

p_{n+1}-p_n = O(\sqrt{p_n}\,\log p_n) (see http://en.wikipedia.org/wiki/Cram%C3%A9r%27s_conjecture )

\frac{1}{p_n} \,- \,\frac{1}{p_n + O(\sqrt{p_n}\,\log p_n)}\, < \,\frac{k(\sqrt{p_n}\,\log p_n)}{p_n^2}\, < \,\frac{k}{p_n^{1.5 - \epsilon}}\, < \,\frac{k}{n^{1.5 - \epsilon}

\sum_{i=1}^{\infty} \frac{k}{{(2i)}^{1.5 - \epsilon} is convergent so the original sum must be convergent as well.

PD: Forget everything I wrote. I understood that the summand was (-1)^n/p_n, not (-1)^n\,n/p_n as posted by the OP.

Last fiddled with by alpertron on 2013-03-14 at 13:38
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