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Let's have the following zeta binomial $\sum\limits_{n=1}^\infty (1/n-1/(n+1))^k$, where $k$ a natural number and $k>1$. From the expansion of these binomials we obtain polynomials of $\pi$ where one of the terms is always an integer. Does anyone know how to calculate this integer, for all values of $k$, without expanding the zeta binomial?

Vassilis Parassidis
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  • Why don't you tell us the values for the first few $k$. Maybe there's a pattern to be found. – Gerry Myerson Feb 19 '12 at 01:02
  • @Gerry Myerson,You can tell us this pattern since you are suggesting that one exists. – Vassilis Parassidis Feb 19 '12 at 01:09
  • Even with the employment of the binomial expansions the recursion is particularly hard to wade through, or at least so I predict. Figuring this out without such expansion I have even less of an idea of how to go about. – anon Feb 19 '12 at 01:20
  • The first six terms are 1, -3, 10, -35, 126, -462. – anon Feb 19 '12 at 01:32
  • OEIS lists several sequences starting that way, e.g., http://oeis.org/A110556. – Gerry Myerson Feb 19 '12 at 02:18
  • From the responses to my two previous questions it seems to me that there must be an algorithm on the web which calculates the pi polynomials. I regret to say that I do not know who first calculated the zeta binomials in closed form. Apart from that I don't know whether anyone has created a formula to calculate the constant. If not then such a formula would be original mathematical work. I would greatly appreciate it if you could inform me of existing knowledge on this mathematical subject. – Vassilis Parassidis Feb 19 '12 at 02:19
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    Vassilis, write $(1/n-1/(n+1))^k$ as $1/(n(n+1))^k$ and use partial fraction decomposition, then summing everything up, you'll see that terms with odd powers will form telescoping series (thus easy to deal with) and the terms with even powers with contribute powers of $\pi$ via zeta values. Actually calculating the individual numbers in the partial fraction decomposition for general $k$ seems like a chore, though it is easy for any particular $k$. – B R Feb 19 '12 at 02:35
  • @BR.I looked at the site Gerry suggested and the information I got does not say any such formula exists. Also, someone has calculated all k up to k=24. I should mention here the exponents of the pi polynomials are the same as the exponents of the binomial cyclotomic equations. – Vassilis Parassidis Feb 19 '12 at 03:56
  • I want to Inform you I found the formula to calculate the constant.(2m_1-1)(2m_2-1)....(2m_k-1)2^(k-1)/k! m takes all values from 1 to k.If k=11 then we have. – Vassilis Parassidis Feb 19 '12 at 04:29
  • (21-1)(22-1)..(211-1)2^10/11!=352716 – Vassilis Parassidis Feb 19 '12 at 04:38
  • Good. Can you prove it? – Gerry Myerson Feb 19 '12 at 05:08
  • @Gerry Myerson.From the formula above we now obtain a new mathematical series which converges to a decimal. I will try to express this decimal in closed form. – Vassilis Parassidis Feb 19 '12 at 07:33
  • That wasn't the question. You have written down a formula for a constant. Do you have a proof of that formula? If so, will you present it here? – Gerry Myerson Feb 19 '12 at 07:45
  • @BR: The partial fraction decomposition would just be the binomial expansion of the original $k$th power of a difference. Vassilis: There is a fairly straightforward recursive procedure for computing these polynomials, but it requires the binomial expansion, and in order to find a closed form (if it exists) would require some combinatorics I can't think of. What I'm curious about is why you're asking for this to be done "without expanding the zeta binomial" which pretty much ruins all hope you have of getting an answer, in my opinion. – anon Feb 24 '12 at 06:46

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