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  • Let $B_n$ be the Bernoulli number.
  • Let $a(n)$ be A110501 (i.e., unsigned Genocchi numbers (of first kind) of even index). Here $$ a(n) = 2(4^n-1)B_{2n}(-1)^{n-1}. $$
  • Let $s(n,k)$ be the (signed) Stirling number of the first kind.
  • Let $n \brace k$ be the Stirling number of the second kind.
  • Let $T(n,m)$ be an integer coefficients such that $$ T(n,m) = \sum\limits_{i=1}^{m+1}(-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace j}. $$
  • Let $b(n)$ be an integer sequence such that $$ b(n) = \sum\limits_{k=0}^{2n} T(2n-k, k)(-1)^{n-k}. $$

I conjecture that $$ b(n) = a(n+1). $$

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

a(n) = 2*(4^n-1)*bernfrac(2*n)*(-1)^(n-1)
T(n,m) = sum(i=1, m+1, (-1)^(m-i+1)*sum(j=i, m+1, j^n*stirling(j,i,1)*stirling(m+1,j,2)))
b(n) = sum(k=0, 2*n, T(2*n-k,k)*(-1)^(n-k))
test(n) = b(n) == a(n+1)

Is there a way to prove it?

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1 Answer 1

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Changing the order order of summation, we have \begin{split} T(n,m) &= \sum\limits_{i=1}^{m+1}(-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace j} \\ &= (-1)^{m+1}\sum_{j=1}^{m+1} j^n {m+1 \brace j} \sum_{i=1}^i (-1)^i s(j,i) \\ &= (-1)^{m+1}\sum_{j=1}^{m+1} j^n {m+1 \brace j} (-1)^j j! \\ &= (-1)^{m+1} n! [z^n] \sum_{j=1}^{m+1} e^{jz} {m+1 \brace j} (-1)^j j! \\ &= (-1)^{m+1} (m+1)! n! [y^{m+1}z^n] \sum_{j=1}^{m+1} (-e^z(e^y - 1))^j \\ &= (m+1)! n! [y^{m+1}z^n] (1-e^z + e^{z-y})^{-1}. \end{split}

Correspondingly, \begin{split} b(n) &= \sum\limits_{k=0}^{2n} T(2n-k, k)(-1)^{n-k} \\ &= \sum_{k=0}^{2n} (k+1)! (2n-k)! [y^{k+1}z^{2n-k}] (1-e^z + e^{z-y})^{-1} (-1)^{n-k} \\ &= (2n+2)! \sum_{k=0}^{2n} \int_{0}^1 {\rm d}t (1-t)^{k+1} t^{2n-k} [y^{k+1}z^{2n-k}] (1-e^z + e^{z-y})^{-1} (-1)^{n-k} \\ &= (-1)^{n+1} (2n+2)! \int_{0}^1 {\rm d}t \sum_{k=0}^{2n} [y^{k+1}z^{2n-k}] (1-e^{zt} + e^{zt-y(t-1)})^{-1} \\ &=(-1)^{n+1} (2n+2)! \int_{0}^1 {\rm d}t [z^{2n+1}] (1-e^{zt} + e^z)^{-1}\\ &=(-1)^{n+1} (2n+2)! [z^{2n+2}] \frac{2z}{1+e^z} \\ &=a(n+1). \end{split}

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