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.

EDIT: I have not been able to prove this statement yet, but I tried to prove an even stronger statement, which in the context of the polynomials I briefly mentioned, it would amount to prove that $P(t-1)$ has positive coefficients. So, summarizing, if one proves the following assertion, then the original one follows directly.

$$F(k,n,m) = \left(\sum_{i=0}^{k-1} (-1)^i {k \brack{k-i}} {{n-k}\brack{m+1-k+i}}\right)\cdot\left(\sum_{j=0}^{k-1} (-1)^j \binom{n}{j} (k-j)^m\right) \geq 0$$

In particular, one must prove that each parentheses have the same sign as $k,n,m$ vary. This sounds easier to prove than the original statement, although it is even much stronger. By now I haven't worked so much with it, but understanding how the sign of each of these expressions depend on $k,n,m$ doesn't seem easy.

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

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.

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.

EDIT: I have not been able to prove this statement yet, but I tried to prove an even stronger statement, which in the context of the polynomials I briefly mentioned, it would amount to prove that $P(t-1)$ has positive coefficients. So, summarizing, if one proves the following assertion, then the original one follows directly.

$$F(k,n,m) = \left(\sum_{i=0}^{k-1} (-1)^i {k \brack{k-i}} {{n-k}\brack{m+1-k+i}}\right)\cdot\left(\sum_{j=0}^{k-1} (-1)^j \binom{n}{j} (k-j)^m\right) \geq 0$$

In particular, one must prove that each parentheses have the same sign as $k,n,m$ vary. This sounds easier to prove than the original statement, although it is even much stronger. By now I haven't worked so much with it, but understanding how the sign of each of these expressions depend on $k,n,m$ doesn't seem easy.