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 $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ Commented Jul 30, 2017 at 13:07
  • $\begingroup$ They seem to have alternating signs. $\endgroup$ 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 $k-2$. $\endgroup$ 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? $\endgroup$ Commented Nov 21, 2019 at 18:37

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.