# Product of Stirling Numbers of the first kind

I have been messing around with coefficients of various polynomials and was wondering if there was a way to reduce the following stuff.

Let polynomial, $b(x,t_i)=\sum\limits_{k=0}^{t_i}s(t_i,k)x^{t_i-k}$

Which was generated from the manipulating the falling factorial $(x+t_i-1)(x+t_i-2)...(x)$

Now let the polynomial $\prod\limits_{i=1}^{q}b(x,t_i)=\sum\limits_{k=0}^{m-q}D_k x^{m-k}$ And Where $m=\sum\limits_{i=1}^{q}t_i$

From what I have figured out each $D_k$ resembles a product of stirling numbers of the first kind.

$D_k=\sum\limits_{e_1+e_2...+e_q=k}^{}s(t_1, e_1)s(t_2, e_2)...s(t_q, e_q)$

This is kind of an unwieldy form so I was wondering if there was some way to simplify and write this in better terms. Since $\sum t_i=m$ and $\sum e_i=k$ is it possible to write this in terms of $m,k$?

Thanks guys, I know this problem is a long one and was just wondering if there was some identity to help simplify this mess.

• Just to be sure about notations: usually Stirling numbers of the first kind are defined such that $$\sum_{k=0}^ns(n,k)x^k=x(x-1)\dots(x-n+1).$$ Using this identity I find that $$b(x,n)=\sum_{k=0}^ns(n,k)x^{n-k}=x^n(x^{-1})(x^{-1}-1)\dots(x^{-1}-n+1)\\=(1-x)(1-2x)\dots(1-(n-1)x).$$ Is that correct ? – Tom-Tom Apr 23 '15 at 8:53

If the values of $t_i$ are not specified, there cannot be any closed formula for $D_k$ depending only on $m$ and $k$, because the product of $b(x,n)$ rewrites as $$(1-x)^{p_1}(1-2x)^{p_2}\dots(1-rx)^{p_r}$$ where $p_k=\text{#}\{t_i\leq k\,|i=1,\dots,q\}$ and $r$ is the largest value among the $t_i$'s.

• I really like the form you expressed it in. Would it be possible for you to provide a link to where you got the equations that led to your final form? – user122523 Apr 23 '15 at 16:44
• There is no reference, this is a simple counting from the expression of $b(x,n)$ I wrote in commenting your question. – Tom-Tom Apr 23 '15 at 20:31