Property of a recursive integer sequence

Solution 1:

Here are some partial results. I reduce your conjecture to identities involving only polynomials, which are hopefully easier to show.

Observing the rightmost part of the triangle, it seems that $b(n-1,n)=(-1)^n (n!)$, and indeed this fact is easily checked by induction on $n$. Going leftwards by just one step, it also seems that $b(n-2,n)=(-1)^{n-1}n!\frac{(n-1)(n-8)}{12}$, and indeed this fact is also easily checked by induction on $n$. More generally, let us define for integers $0\leq d \lt n$

$$ t(d,n)=\frac{b(n-1-d,n)}{(-1)^{n+d}n!\prod_{k=1}^d (n-k)} \tag{3} $$

So, the two remarks above tell us that $t(0,n)=1$ and $t(1,n)=\frac{n-8}{12}$. In fact, I have checked for $n\leq 10$ the following :

Conjecture 2. When nonzero (i.e. for $n \gt d$), $t(d,n)$ is a polynomial of degree $d$ in $n$.

When translated in terms of $t$, the recursive relation governing $b$ becomes

$$ n(n+1)t(d+1,n+1)-(n-d-1)(n+2d+3)t(d+1,n)=(n-2d-2)t(d,n) \tag{4} $$

Let $h(j,n)= 32j^3-32(2n-1)j^2 +2(22n^2-30n+13)j-(n-1)(2n-1)(5n-6)$. Your sum $S_n=\sum_{j=0}^{n-1} b(j,n)h(j,n)$ can be rewritten as $S_n=\sum_{d=0}^{n-1} h(n-1-d,n)b(n-1-d,n)=(-1)^n n!U_{n}(n)$ where $U_{n}(m)=\sum_{d=0}^{n-1} (-1)^d h(m-1-d,m)t(d,m)\prod_{k=1}^d (m-k)$. Now, I have checked for $n\leq 10$ the following

Conjecture 3. $U_{n}$ is a polynomial (in $m$) divisible by $\prod_{k=1}^{n} (m-k)$.

In particular, we have $U_{n}(n)=0$, which implies your conjecture.