7
$\begingroup$

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.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ 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? $\endgroup$ Commented Dec 19, 2010 at 15:42
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Dec 19, 2010 at 17:17
  • $\begingroup$ 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 . $\endgroup$ Commented Dec 19, 2010 at 19:55
  • $\begingroup$ @ 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. $\endgroup$ Commented Dec 20, 2010 at 12:46
  • $\begingroup$ @Johann: it's what they call "forward substitution" here: books.google.com/books?id=mlOa7wPX6OYC&pg=PA88 ; one thing that may ease things is that most of the $b_i$ (to use the notation of that book) are zero. $\endgroup$ Commented Dec 20, 2010 at 14:45

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.