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find this value $$B_{n\times n}=\begin{vmatrix} a_{1}&a_{0}&\cdots&\cdots&0\\ a_{2}&\cdots&&\cdots&\cdots\\ \cdots&\cdots&&\cdots&\cdots\\ a_{n-1}&&\cdots&a_{1}&a_{0}\\ a_{n}&a_{n-1}&\cdots&a_{2}&a_{1} \end{vmatrix}$$ where $$a_{0}=1,a_{n}=\dfrac{(2n-1)!!}{(2n)!!}(n\ge 1)$$ Thank you everyone, Later I post my idea.

math110
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1 Answers1

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It's possible to derive a recurrence. What you have here is a Toeplitz-Hessenberg matrix. Moreover, $a_0 = 1$, so it has the special form

$$ \left[\begin{matrix} a_1&1&0&\cdots&0\\ a_2&\ddots&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&0\\ a_{n-1}&\ddots&\ddots&a_1&1\\ a_{n}&a_{n-1}&\cdots&a_2&a_1 \end{matrix}\right] $$

for which Inselberg [1] gives the elegant recurrence:

$$\sum_{i=0}^n (-1)^{n-1} a_i B_{(n-i)\times(n-i)} = 0$$

(For a less elegant form, see the revision history for my derivation...)

Although he doesn't state it, I believe this only holds for $n > 0$, as otherwise we could trivially prove that if all of the $a_i$ are non-zero then all of the $B_{i\times i}$ are zero.

He also provides a link to generating functions: if the $a_i$ have generating function $f(z) = \sum_{n=0}^\infty a_n z^n$ then $\frac 1{f(z)} = \sum_{n=0}^\infty (-1)^n B_{n\times n}z^n$. I don't know why he doesn't fix the signs by stating instead that $g(z) = \frac 1{f(-z)} = \sum_{n=0}^\infty B_{n\times n}z^n$

Unfortunately I don't have Maple to try using convert/ratpoly to guess a generating function for your specific $a_i$. However, I can compute some values:

$B_{2\times 2} = -\frac{1}{8};\; B_{3\times 3} = \frac{1}{16}; \; B_{4\times 4} = -\frac{5}{128}; \; B_{5\times 5} = \frac{7}{256}; \; B_{6\times 6} = -\frac{21}{1024}; \; B_{7\times 7} = \frac{33}{2048}; \; B_{8\times 8} = -\frac{429}{32768}; \; B_{9\times 9} = \frac{715}{65536}; \; B_{10\times 10} = -\frac{2431}{262144} $

OEIS turns up two matches for the unsigned numerators, with links to double-factorials and Catalan numbers. The comments there lead me to speculate that a generating function for (signed) $B_{n\times n}$ is $g(z) = \frac 1{1+\sqrt{1+z}}$ and that $$ B_{n \times n} = \frac{(-1)^{n-1} (2n-3)!!}{n! \; 2^n} = \frac{(-1)^{n-1} (2n-3)!!}{(2n)!!} = \frac{(-1)^{n-1} a_n}{2n-1} $$

There may be a combinatorial proof, but I'm really a compsci and hence I automatically look for inductive proofs.

We can consider $B_{0\times 0}$ to be $1$ (the determinant of an empty matrix is a zero-element product), and the recurrence gives us $B_{1\times 1} = a_1$, so we're ok for the base case.

Rearranging the recurrence and using $a_0 = 1$ we get

$$B_{n\times n} = \sum_{i=1}^n (-1)^{i-1} a_i B_{(n-i)\times(n-i)}$$

Assuming our inductive hypothesis and substituting the definition of $a_i$, $$B_{n\times n} = (-1)^n \sum_{i=1}^n \frac{(2i-1)!!(2n-2i-3)!!}{(2i)!!(2n-2i)!!}$$

This is Gosper-summable, with indefinite sum $$\sum_i \frac{(2i-1)!!(2n-2i-3)!!}{(2i)!!(2n-2i)!!} = \frac{i(2i-1)!!(2n-2i-1)!!}{n(2i)!!(2n-2i)!!}$$ so we get

$$\begin{eqnarray} B_{n\times n} & = & (-1)^n \left[\frac{i(2i-1)!!(2n-2i-1)!!}{n(2i)!!(2n-2i)!!}\right]_1^{n+1} \\ & = & (-1)^n \left[\frac{(n+1)(2n+1)!!(-3)!!}{n(2n+2)!!(-2)!!} - \frac{(1)!!(2n-3)!!}{n(2)!!(2n-2)!!}\right] \\ & = & (-1)^{n-1} \frac{(2n-3)!!}{(2n)!!} \end{eqnarray} $$

as desired (where we use the fact that there's a pole at $(-2)!!$).

To close the loop on that generating function digression: $$f(z) = (1-z)^{-1/2} = 1 + \frac{\frac{1}{2} z}{1!} + \frac{\frac{1}{2} \frac{3}{2} z^2}{2!} + \dots = \sum_{i=0}^\infty \frac{(2i-1)!! z^i}{2^i i!} = \sum_{i=0}^\infty a_i z^i$$

So the generating function for $B_{i\times i}$ is $\frac{1}{f(-z)}$ which agrees with my speculation, although it is more elegantly written as $g(z) = \frac{1}{f(-z)} = \sqrt{1+z}$.

Reference

[1] On determinants of Toeplitz-Hessenberg matrices arising in power series, Alfred Inselberg, Journal of Mathematical Analysis and Applications, Volume 63, Issue 2, April 1978, pp 347–353.

Peter Taylor
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