16
$\begingroup$

This is again a question that I asked at Stack Exchange, but got no answer so far, so I am trying here.
Let:

$$ a_n=\sum_{k\ge0}(k+1) {n+2\brack k+2}(n+2)^kB_k$$ $B_k$ is the Bernoulli number. ${n\brack {k}}\;$ is the unsigned Stirling number of first kind, $\left( {0\brack {0}}\;=1 \text{ and }{{n}\brack {k}}\;=(n-1){{n-1}\brack {k}}\;+{{n-1}\brack {k-1}}\;\right)$.

From $n=0$, the first terms are: $ \ \ 1\ ,\ 0\ ,\ -5\ ,\ 0\ ,\ 238\ ,\ 0\ ,\ -51508\ ,\ 0\ ,\ 35028576\ , ..$

The $a_n$ are all integers, and the odd-indexed $a_{2n+1}$ vanish.

A generating function should be even, but I could not find it.

Also, any possible combinatorial interpretation (when removing the sign)?

I would welcome any help or indication on this. Thank you in advance.

EDIT(01/09/18): Actually, this is a particular case of: $$ a_{n,h}=\sum_{k\ge0}{k+h-1\choose k} {n\brack h+k}n^k B_k$$ Here is a table for $a_{n,h}$, for $1\le n,h \le 9$.

\begin{matrix} n&|&a_{n,1}&a_{n,2}&a_{n,3}&a_{n,4}&a_{n,5}&a_{n,6}&a_{n,7}&a_{n,8}&a_{n,9}\\ -&&---&---&---&---&---&---&---&---&---\\ 1&|&1&0&0&0&0&0&0&0&0\\ 2&|&0&1&0&0&0&0&0&0&0\\ 3&|&-1&0&1&0&0&0&0&0&0\\ 4&|&0&-5&0&1&0&0&0&0&0\\ 5&|&24&0&-15&0&1&0&0&0&0\\ 6&|&0&238&0&-35&0&1&0&0&0\\ 7&|&-3396&0&1281&0&-70&0&1&0&0\\ 8&|&0&-51508&0&4977&0&-126&0&1&0\\ 9&|&1706112&0&-408700&0&15645&0&-210&0&1\\ \end{matrix}

I know now how to show that $a_{n,h}$ is integer, and that it is zero when $n-h$ is odd. But the proofs that I have found are quite lengthy, technical, and not really enlightening about the mathematical signification of these numbers.

Any idea?

Improve this question
$\endgroup$
10
  • 3
    $\begingroup$ I note that OP has put this sequence up on the Online Encyclopedia of Integer Sequences, oeis.org/A286483 $\endgroup$
    – Gerry Myerson
    Commented Jul 30, 2017 at 0:56
  • 2
    $\begingroup$ @Gerry, I did add it on the OEIS, because I was told it was not there, and I was hoping to advertise about it. $\endgroup$
    – René Gy
    Commented Jul 30, 2017 at 12:53
  • 2
    $\begingroup$ Might have been a good idea to mention that, so as to prevent people from wasting their time looking it up. $\endgroup$
    – Gerry Myerson
    Commented Jul 30, 2017 at 13:07
  • $\begingroup$ They seem to have alternating signs. $\endgroup$
    – F. C.
    Commented Jan 12, 2018 at 20:08
  • 2
    $\begingroup$ The diagonals $a_{n+k,n}$ for fixed even $k$ seem to be polynomials in $n$ of the form $p_k(n)(n+k+1)\cdots (n+1)n$ where $p_k(n)$ is a polynomial of degree $k-2$. $\endgroup$
    – Martin Rubey
    Commented Sep 1, 2018 at 20:08

1 Answer 1

Reset to default
5
$\begingroup$

I've got that $a_{n,h}$ is the coefficient of $x^{n-1}$ in \begin{split} &(-1)^{n+h+1} \frac{n!}{2\cdot (h-1)!}\frac{\log(1+x)^{h}\left(\coth(-\frac{n}2\log(1+x))-1\right)}{1+x} \\ =\ &(-1)^{n+h} \frac{n!}{(h-1)!}\frac{\log(1+x)^{h}}{(1-(1+x)^{-n})(1+x)}. \end{split} It's yet unclear if the dependency on $n$ can be eliminated to turn this expression into a generating function.

Improve this answer
$\endgroup$
1
  • $\begingroup$ That is some progress. Can this expression be used to show that $a_{n,h}$ is zero when $n-h$ is odd? $\endgroup$
    – René Gy
    Commented Nov 21, 2019 at 18:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .