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Chukchi
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Joined: December 15th, 2001, 3:43 am

Binomial sum

November 11th, 2009, 8:49 pm

Prove that p^2 divides binomial sum a(n) = 16^n * Sum( Binomial(2k,k) / 16^k, {k, 0, n} ) for n = (p - 3)/2 and prime p of the form p = 6k + 1, such as {7,13,19,31,37,43,61,...}.Binomial(2k,k) = (2k)!/(k!)! are the central binomial coefficients.For example, for p = 7 and n = (p - 3)/2 = 2a(2) == 16^2 * ( C(4,2)/16^2 + C(2,1)/16^1 + C(0,0)/16^0 ) = = C(4,2) + 16 * C(2,1) + 16^2 * C(0,0) == 4!/(2!)^2 + 16 * 2!/(1!)^2 + 16^2 * 0!/(0!)^2 == 6 + 16 * 2 + 16^2 * 1 == 6 + 32 + 256 == 294 == 2 * 3 * 7^2 Thus p^2 = 7^2 divides a(2).
Last edited by Chukchi on November 10th, 2009, 11:00 pm, edited 1 time in total.
 
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beaker
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Joined: June 25th, 2009, 5:59 pm

Binomial sum

November 11th, 2009, 10:02 pm

Just to clarify, prove p^2 divides a(n) for only those p where p = 6k+1 is prime for some natural number k? The only issue here may be, who knows how many of those p are prime?
 
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Chukchi
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Binomial sum

November 11th, 2009, 11:43 pm

beaker, there are infinitely many primes of the form p = 6k + 1for all such primes p^2 divides a(n) = 16^n * Sum( Binomial(2k,k) / 16^k, {k, 0, n} ) where n = (p - 3)/2it is a conjecture to prove
 
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beaker
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Joined: June 25th, 2009, 5:59 pm

Binomial sum

November 12th, 2009, 1:15 am

Infinitely many for sure. What I meant was that there exists some natural number X, such that for all k > X no one knows (there is no way to tell) if p = 6k + 1 is prime.In other words, as far as I know (which is limited certainly) there is no way to mathematically describe "all prime numbers of form .....". Or stated a bit differently, if you lay out, for all natural numbers k, the sequence 6k+1, and then put a 0 for all numbers in the sequence which are composite and a 1 for all numbers which are prime, then there is no way to predict when a 1 will occur. So I dont know how to prove anything for each of those 1s. Cheers.
Last edited by beaker on November 11th, 2009, 11:00 pm, edited 1 time in total.
 
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wileysw
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Binomial sum

November 12th, 2009, 3:49 pm

Chukchi, not sure if you have already known the following reference.couple of recent papers by Zhi-Wei Sun especially this paper, seems to me, is addressing exactly the problem you posed (and this one).
Last edited by wileysw on November 11th, 2009, 11:00 pm, edited 1 time in total.
 
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Chukchi
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Binomial sum

November 13th, 2009, 5:30 am

Thank you, wileysw.I didn't know about this paper. It looks very interesting but a little bit too technical and too general for an amateur like me.Funny thing is that an author Zhi-wei Sun has wrote already half a dozen papers on this subject. In the very first paper dated back to 2007 he and his co-author named Roberto Tauraso kindly refered to my own early observations as a starting point of their research.Three years ago Roberto wrote me an e-mail asking if I have a proof to some of my statements and conjectures.Unfortunatelly in later papers they refer only to their own publications and preprints. It is very common practice.I'm glad that some of my conjectures and observations became such an interesting topic for professionals. But for me it is just a hobby. I enjoy to play with numbers.Anyway, the problem in this topic is based on my most recent observations. I believe it should have an elegant proof. It could be a nice topic for somebody to write a research paper as well.It seems that the Zhi-Wei Sun's paper doesn't have an exact solution but it has almost all necessary tools.