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Let $a(n,k)=(-1)^k {{2n-k}\choose k}$ for $0 \le k \le n$ and $a(n,k)=0$ else. Then it is known (cf. OEIS A005439 and A098435) that the first column of the inverse matrix of $(a(i,j))_{i,j\ge0}$ is the sequence $1,1,2,8,56,608,9440,…$ of median Genocchi numbers.

Now let $b(n,k)=(-1)^k {{2n+1-k}\choose k}$ for $0 \le k \le n$ and $b(n,k)=0$ else.

My question is: Is a formula known for the entries of the first column of the inverse matrix of $(b(i,j))_{i,j\ge0}?$ The first terms of this sequence are $1, 1/2, 1/3, 1/3, 8/15, 4/3, 512/105, 368/15…$. Multiplying the $n-$th term with $n!$ gives the sequence $1,1,2,8,64,960,24576,989184,…$.

Edit. Motivated by the comments below and analogous results by D. Dumont and J. Zeng about Genocchi numbers I found the following connections with Bernoulli numbers $B_{2n}.$

1) Let $\left( {c(n)} \right)_{n \ge 0}=\left({1, 1/2, 1/3, 1/3, 8/15, 4/3, 512/105, 368/15…}\right).$ Then there is an expansion into a formal power series $\sum\limits_{n \ge 0} {( - 1)^n c(n)\frac{{z^{2n} }}{{(1 - z)^{n + 1} }}} = z + \sum\limits_{n \ge 0} {(2n + 1)B_{2n} z^{2n} } .$

2) ` $c(2n + 1) = \sum\limits_{j = 0}^n {{2n+1}\choose{2j+1} } (2n + 2j + 3)B_{2n + 2j + 2} $ and $c(2n ) = -\sum\limits_{j = 0}^{n-1} {{2n}\choose{2j+1} } (2n + 2j + 3)B_{2n + 2j + 2}. $

There remains the question whether the sequence $(c(n))$ also occurs in a natural way in other contexts.

share|cite|improve this question
1. The 1,1,2,8,64,960... sequence isn't in the OEIS. 2. Have you tried using the formulae for forward elimination to attempt to derive a series expression for those inverse entries? – J. M. Dec 19 '10 at 15:42
This is a shot in the dark, but try looking at the sequences 1, 1/3, 8/15, 512/105 and 1/2, 1/3, 4/3, 368/15 (the odd- and even-numbered terms of your original sequence). These sequences "look easier" to me in some sense, although I still can't identify them. – Michael Lugo Dec 19 '10 at 17:17
Denote the coefficients of the first columns of $B^-1$ by $c_k$ and the quotient of consecutive coefficients $q_k=c_k/c_{k-1}$ where $q_0=1 $ . Denote the forward differences of 2'nd order as $d_{2,k}$. Then the $d_k$ seem to converge to something like 0.202642045678 . – Gottfried Helms Dec 19 '10 at 19:55
@ J.M. I did not quite understand what you mean with forward elimination. But your advice to derive a series expression was helpful. @ Michael Lugo. You are right. This becomes easier. See edit of my question. – Johann Cigler Dec 20 '10 at 12:46
@Johann: it's what they call "forward substitution" here: ; one thing that may ease things is that most of the $b_i$ (to use the notation of that book) are zero. – J. M. Dec 20 '10 at 14:45

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