## Binomial sums and the zeta function

Recall the three sequences,

$C_n = \sum_{k=0}^n {\binom n k} {\binom {n+k}k} = 1, 3, 13, 63, 321, 1683\dots$

$B_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k} = 1, 3, 19, 147, 1251, 11253\dots$

$A_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k}^2 = 1, 5, 73, 1445, 33001, 819005\dots$

Equivalently,

$C_n = \,_2F_1(-n,n+1;\,1;\,-1)$

$B_n = \,_3F_2(-n,-n,n+1;\,1,1;\,1)$

$A_n = \,_4F_3(-n,-n,n+1,n+1;\,1,1,1;\,1)$

where $_pF_q$ is the generalized hypergeometric function.  Then it is known that,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\, C_n C_{n+1}} = \frac{1}{2}\log(2)\\ &\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2 B_n B_{n+1}} = \frac{1}{5}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3 A_n A_{n+1}} = \frac{1}{6}\, \zeta(3)\end{aligned}

Beautiful, aren’t they? Since the numbers increase at a near-geometric rate (for example, $\frac{A_{n+1}}{A_n} \to (1+\sqrt{2}\,)^4 \approx 33.97$  as n  goes to infinity), then the convergence is very fast.

We also have the nice evaluations,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\ &\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = \frac{1}{3}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} = -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right) \\ &\sum_{n=1}^\infty \frac{1}{n^4\,\binom {2n}n} = \frac{17}{36}\,\zeta(4)\\ &\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n} = -\frac{19}{3}\,\zeta(5) +\frac{2}{3}\,\zeta(2)\zeta(3)+\frac{\pi\sqrt{3}}{2^3\cdot 3^2}\left(\zeta(4, \tfrac{1}{3})-\zeta(4,\tfrac{2}{3}) \right)\\ &\sum_{n=1}^\infty \frac{1}{n^6\,\binom{2n}n} = \;\;?\\ &\sum_{n=1}^\infty \frac{1}{n^7\,\binom{2n}n} = -\frac{493}{24}\zeta(7)+2\zeta(2)\zeta(5)+\frac{17}{18}\zeta(3)\zeta(4)+\frac{11\pi\sqrt{3}}{2^5\cdot 3^4}\left(\zeta(6,\tfrac{1}{3})-\zeta(6,\tfrac{2}{3})\right)\\ \end{aligned}

with the Riemann zeta function $\zeta(s)$ and the more general Hurwitz zeta function $\zeta(s,a)$,

\begin{aligned} &\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}\\&\zeta(s,a) = \sum_{n=0}^\infty \frac{1}{(n+a)^s} \end{aligned}

respectively.  (Note that for $a = 1$, the Hurwitz reduces into the Riemann.)  The expression for p = 5 in the paper here used Dirichlet L-functions, but a poster from mathstackexchange gave it in terms of the Hurwitz zeta.  The one for p = 7 is from Mathworld’s article on central binomial coefficients.

However, none are known for p > 7 (as well as p = 6).  Based on odd p, it is easy to assume that the next has the form,

\begin{aligned} & a_0\sum_{n=1}^\infty \frac{1}{n^9\,\binom {2n}n} = a_1\,\zeta(9) +a_2\,\zeta(2)\zeta(7)+a_3\, \zeta(3)\zeta(6)+a_4\, \zeta(4)\zeta(5)+a_5\,\pi\sqrt{3}\,H_8\\ \end{aligned}

where,

$H_8 = \left(\zeta(8,\tfrac{1}{3})-\zeta(8,\tfrac{2}{3})\right)$

and the six $a_i$ are integers.  One can use Mathematica’s LatticeReduce function (which employs an integer relations algorithm) to find them, if any exists. Unfortunately, it didn’t find any exact relation, nor for analogous forms for prime p = 11 or 13.  Either my old version of Mathematica is just not powerful enough, or odd p > 7 do not have analogous forms to the ones above.

Can you find the next in the family?