4
$\begingroup$

Past days I've been trying to prove that certain polynomials have positive coefficients. After a lot of thinking, I came up with a formula for each coefficient individually, and they are not that ugly. I think maybe someone can give me a hand on this.

Synthetically, what I want to prove is that the following sum is positive:

$$S(k,n,m)=\sum_{i=0}^{n-m-1} \sum_{j=0}^{k-1} (-1)^{i+j} \binom{n}{j}(k-j)^m {j \brack {j-i}} {{n-j}\brack {m+1+i-j}}$$

Where the symbol ${x \brack y}$ stands for the Stirling numbers of the first kind (without sign).

I'm interested in the case $1\leq m,k\leq n-1$.

I have already proven the following:

1) If in the sum we set $m=n-1$, we get just the well known recurrence for Eulerian numbers, so it is positive. For $m=n-2$, the result is a sum of two Eulerian numbers.

2) If we replace $k$ by $n-k$, the sum remains the same (the proof of this fact is somewhat abstract in the sense one has to understand the context on which this sum arises).

3) With $k=1$, we get simply the Stirling numbers of the first kind.

4) With $m=1$ the sum is always positive.

I don't mind if the proof is strictly combinatoric or involves inequalities of Stirling numbers or even uses the exponential generating function of some of the numbers inside. However, I strongly suspect that this alternating sum can be rewritten as a sum of products of stirling, eulerian, or binomial numbers (I couldn't manage to guess such a formula in any case not listed before).

I posed this question in MSE, but I think it fits a lot better here

Improve this question
$\endgroup$
5
  • $\begingroup$ Would you please write the polynomials also? $\endgroup$
    – Fedor Petrov
    Commented Oct 30, 2019 at 15:07
  • 1
    $\begingroup$ Let $P_{k,n}(t)$ the coefficient of $x^{kt}$ in $(1+x+\ldots+x^t)^n$. You can easily prove this is indeed a polynomial. The number $S(k,n,m)$ would be the coefficient of $t^m$ in $P_{k,n}(t)$. $\endgroup$
    – Luis Ferroni
    Commented Oct 30, 2019 at 15:25
  • 1
    $\begingroup$ I forgot to add, to avoid carrying denominators, $S(k,n,m)$ is multiplied by $(n-1)!$. $\endgroup$
    – Luis Ferroni
    Commented Oct 30, 2019 at 15:35
  • $\begingroup$ This is repost from math.stackexchange $\endgroup$
    – René Gy
    Commented Oct 31, 2019 at 15:46
  • $\begingroup$ Hi Rene, indeed in the original post it already says it is a repost. It happens that the difficulty of the question I think fits better here rather than MSE. $\endgroup$
    – Luis Ferroni
    Commented Oct 31, 2019 at 16:33

0

Reset to default

You must log in to answer this question.