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Let

  • $B_n$ be the $n$-th Bernoulli number with $B_1 = -\frac{1}{2}$.
  • $T_m(n,k)$ be the family of integer coefficients such that $$T_m(n,k) = \sum\limits_{j=k}^{n} {n \brace j} {j \brack k} j^m. $$
  • $S_1(n,m)$ be an integer coefficients such that $$ S_1(n,m) = \sum\limits_{k=1}^{n} T_m(n,k). $$
  • $S_2(n,m)$ be an integer coefficients such that $$ S_2(n,m) = \sum\limits_{k=1}^{n} T_m(n,k) B_{k-1}. $$

I conjecture that $S_1(n,m) = S_2(n,m+2)$ for all $n \in \mathbb{N}, m \in \mathbb{Z}$.

Here is the PARI/GP program to check it numerically:

T(n,k,m) = sum(j=k, n, stirling(n, j, 2)*abs(stirling(j, k, 1))*j^m)
S1(n,m) = sum(k=1, n, T(n,k,m))
S2(n,m) = sum(k=1, n, T(n,k,m)*bernfrac(k-1))
test(n,m) = S1(n,m) == S2(n,m+2)

Is there a way to prove it?

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We have: \begin{split} S_2(n,m) &= \sum_{k=1}^n B_{k-1} \sum_{j=k}^n {n \brace j} {j \brack k} j^m = \sum_{j=1}^n {n \brace j} j^m \sum_{k=1}^j B_{k-1} {j \brack k} \\ &= \sum_{j=1}^n {n \brace j} j^{m-2} j! = \sum_{j=1}^n {n \brace j} j^{m-2} \sum_{k=1}^j {j \brack k}\\ &= \sum_{k=1}^n T_{m-2}(n,k) = S_1(n,m-2), \end{split} where we used the identity DLMF (24.15.8): $$\sum_{k=1}^j B_{k-1} {j \brack k} = \frac{j!}{j^2}.$$

PS. In other words, the relation between $S_1$ and $S_2$ essentially follows from the identity: $$\sum_{k=1}^j B_{k-1} {j \brack k} = \frac1{j^2}\sum_{k=1}^k {j \brack k}.$$

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