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Let $k\geq 1$ be an integer and let $P(n)$ be the polynomial $\binom{n+k}{k}$. Consider the series

$$ L_k(s) = \sum_{n \geq 0} \frac{P'(n)}{P(n)^s}. $$

It is known (by previous work of myself and collaborators) that this series shares some properties of the Riemann's $\zeta$ function (which is the case $k=1$). Namely

  • analytic continuation to $\mathbb{C}$ with a single simple pole at $1$,
  • rational values at negative integers with an explicit description.

Nothing is known about

  • functional equation,
  • Riemann's hypothesis.

But one can not seriously expect them, for lack of arithmetic context.

Still, I would like to be able to compute with good precision the values in the "critical band" and on the real negative axis. One expects an oscillatory behaviour on the real negative axis, with frequency proportional to $k$. Nothing is known about zeros outside of the real line.

How to compute the analytic continuation of these functions ?

Our proof of analytic continuation gives us some formulas, but they do not seem to be really useful for concrete computations.

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The binomial series gives a meromorphic continuation of any series of the form $\sum_{n=N}^\infty g(n) f(n)^{-s}$ for some polynomials $g(x) = \sum_{j=0}^t c_j x^j$ and $f(x) = x^r \prod_{k=1}^d (x-a_k)$

$$f(n)^{-s} = n^{-(d+r)s} \prod_{k=1}^d (1-a_kn^{-1})^{-s} = n^{-(d+r)s} \prod_{k=1}^d \left(\sum_{m=0}^\infty {-s \choose m} (-a_k)^m n^{-m}\right) \\= \sum_{l=0}^\infty h_l(s) n^{-s(d+r)-l}$$ $$\sum_{n=N}^\infty g(n) f(n)^{-s} =\sum_{j=0}^t c_j \sum_{l=0}^\infty h_l(s) (\zeta(s(d+r)+l-j)-\sum_{n=1}^{N-1} n^{-s(d+r)-l+j})$$ Where for $N > \max_k \frac{1}{|a_k|}$ the last series converges locally uniformly (away from the poles) for every $s$

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  • $\begingroup$ This is the kind of series that I used in my previous tentatives, but I did not try to find good bounds for the summation order. I was hoping for some alternative way to compute the values. Also, what do you mean by "away from the poles". There is only one pole at 1. $\endgroup$ – F. C. Sep 3 '17 at 6:17
  • $\begingroup$ @F.C. Do you see how changing $N$ changes the rate of convergence ? Also, how do you see the number of poles ? $\endgroup$ – reuns Sep 3 '17 at 6:23
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Maple (2018 edition) can compute this sum numerically quite fast. I don't know the method it uses.

> dP := unapply(diff(binomial(n+k,k),n),n,k);
> Lsum := (k,s) -> evalf(Sum(dP(n,k)/binomial(n+k,k)^s,n=0..infinity));

Now, values like Lsum(3,-0.43+1.7*I) take about 0.3s for 20 digits and 3.3s for 100 digits.

Note that 'Sum' is the inert form of 'sum'; it stops Maple from spending time trying to find an analytic formula for the sum before jumping into the numerics.

It starts to have trouble when $k$ gets larger than 5, though.

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  • $\begingroup$ I am surprised. Do you mean that somehow an analytic continuation is being made automagically ? This sum is divergent when the real part of s is smaller than 1. For k=2, the values at negative integers are known to be (0: -1/3), (-1: -1/60), (-2: 1/315), (-3: -1/840), (-4: 1/1155). Do you get those ? $\endgroup$ – F. C. Apr 5 '19 at 18:09
  • $\begingroup$ Actually, no. It only seems to be able to evaluate the series when it converges. Sorry about the false alarm. If there is some form for the analytic continuation, such as an integral, it can probably compute that. $\endgroup$ – Brendan McKay Apr 6 '19 at 0:47

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