Positivity of a certain sum of Stirling numbers 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.
3) With $k=1$, we get simply the Stirling numbers of the first kind.
4) With $m=1$ the sum is always positive. 
Probably someone with more experience working on this kind of sums can give me a hand. I bet that there may be even a way to understand this sum combinatorially.
 A: For the moment at least, I can just individuate the first step of an approach which
might be possibly interesting.   
The sum can be rewritten as
$$
\eqalign{
  & S(q,n,m) = \sum\limits_{\left( {0\, \le } \right)\,\,i\,\,\left( { \le \,n - m - 1} \right)}
 {\;\sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1} {\left( { - 1} \right)^{\,i + j} \left( \matrix{  n \cr   j \cr}  \right)\left( {q - j} \right)^{\,m} 
 \left[ \matrix{  j \cr   j - i \cr}  \right]\left[ \matrix{  n - j \cr   m + 1 + i - j \cr}  \right]} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,m + 1} \right)}
  {\;\sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1} {\left( { - 1} \right)^{\,k} \left( \matrix{  n \cr   j \cr}  \right)\left( {q - j} \right)^{\,m}
  \left[ \matrix{  j \cr   k \cr}  \right]\left[ \matrix{  n - j \cr   m + 1 - k \cr}  \right]} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1}
  {\left( \matrix{  n \cr   j \cr}  \right)\left( {q - j} \right)^{\,m} \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,m + 1} \right)} {\left( { - 1} \right)^{\,k}
 \left[ \matrix{  j \cr   k \cr}  \right]\left[ \matrix{  n - j \cr   m + 1 - k \cr}  \right]} }  \cr} 
$$
where putting the bounds in parentheses is meant to underline that they are implicit in the binomial / Stirling n. ,
which is a useful indication for dealing with convolutions.   
Since
$$
x^{\,\overline {\,n\,} } x^{\,\overline {\,m\,} }  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n + m} \right)} {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)}
  {\left[ \matrix{ n \cr 
  j \cr}  \right]\left[ \matrix{
  m \cr 
  k - j \cr}  \right]x^{\,k} } } 
$$
where $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $ represent respectively the 
Falling and Rising Factorial
then the inner sum above can be written as
$$
\eqalign{
  & \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,m + 1} \right)}
 {\left( { - 1} \right)^{\,k} \left[ \matrix{  j \cr   k \cr}  \right]\left[ \matrix{  n - j \cr   m + 1 - k \cr}  \right]}
  = \left[ {x^{\,m + 1} } \right]\left( {\left( { - x} \right)^{\,\overline {\,j\,} } x^{\,\overline {\,n - j\,} } } \right) =   \cr 
  &  = \left[ {x^{\,m + 1} } \right]\left( {\left( { - 1} \right)^j x^{\,\underline {\,j\,} } x^{\,\overline {\,n - j\,} } } \right)
 = \left[ {x^{\,m + 1} } \right]\left( {\left( { - 1} \right)^j x^{\,\underline {\,j\,} } \left( {x + n - 1 - j} \right)^{\,\underline {\,n - j\,} } } \right)
 \quad \left| \matrix{  \;1 \le n \hfill \cr   \;j \le n \hfill \cr}  \right. \cr} 
$$
thus giving
$$ \bbox[lightyellow] {  
S(q,n,m) = \left[ {x^{\,m + 1} } \right]\sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1} {\left( { - 1} \right)^j
  \left( \matrix{ n \cr   j \cr}  \right)
 \left( {q - j} \right)^{\,m} x^{\,\underline {\,j\,} } x^{\,\overline {\,n - j\,} } } \quad \left| {\;1 \le n} \right.
 }$$
The function on RHS can be further rewritten as
$$
\eqalign{
  & F(q,n,m,x) = \sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1}
 {\left( { - 1} \right)^j \left( \matrix{  n \cr   j \cr}  \right)\left( {q - j} \right)^{\,m} x^{\,\underline {\,j\,} } x^{\,\overline {\,n - j\,} } }  =   \cr 
  &  = n!\sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1}
 {\left( { - 1} \right)^j \left( {q - j} \right)^{\,m} \left( \matrix{  x \cr   j \cr}  \right)\left( \matrix{  x + n - 1 - j \cr   n - j \cr}  \right)}  =   \cr 
  &  = n!\sum\limits_{\left( {0\, \le } \right)\,j\, \le \,q - 1}
 {\left( {q - j} \right)^{\,m} \left( \matrix{  j - x - 1 \cr   j \cr}  \right)\left( \matrix{  x + n - 1 - j \cr   n - j \cr}  \right)}  \cr} 
$$
