Consider the sequence $(\varphi_i)$ of reverse Bessel polynomials which begins as follows. \begin{align*} \varphi_0&=1\\ \varphi_1&=x\\ \varphi_2&=x^2 + x\\ \varphi_3&=x^3 + 3x^2 + 3x\\ \varphi_4&=x^4 + 6x^3 + 15x^2 + 15x \end{align*}

In general we have $$\varphi_0=1;\qquad\varphi_i = \sum_{k=1}^{i} \frac{(2i-k-1)!\,x^{k} }{(i-k)!\,(k-1)!\,2^{i-k}} \quad \text{for}\ i>0. $$ This is a sequence of binomial type (I don't know if that's relevant) with exponential generating function: $$ \sum_{i=0}^{\infty} \varphi_i(x)\frac{z^i}{i!} = \exp(x(1-\sqrt{1-2z})). $$ I want to know if there's going to be an explicit formula for the Hankel determinants: $$ H_n= \det\left([\varphi_{i+j}]_{i,j=0}^{n}\right). $$ Here are the first few. \begin{align*} H_0 & = 1 \\ H_1 & = x \\ H_2 & = 2 \cdot x^{2} \cdot(x + 3) \\ H_3 & = 12 \cdot x^{3} \cdot(x^{3} + 12 x^{2} + 48 x + 60) \\ H_4 & = 288 \cdot x^{4} \cdot (x^{6} + 30 x^{5} + 375 x^{4} + 2475 x^{3} + 9000 x^{2} + 16920 x + 12600) \end{align*} There's clearly a factor of $\left(\prod_{k=1}^{n}k!\right) \cdot x^n$ in $H_n$.

I've had a look in the Hankel determinant literature but it seems extensive and much of it seem to be relevant to sequences of integers rather than polynomials. So I'd be interested in any pointers.

I'm also interested in calculating the Hankel determinants of the sequence offset by two: $$ H^{[2]}_n= \det\left([\varphi_{i+j+2}]_{i,j=0}^{n}\right). $$ \begin{align*} H^{[2]}_0 & = x \cdot (x + 1) \\ H^{[2]}_1 & = x^{2} \cdot (x^{3} + 6 x^{2} + 12 x + 6) \\ H^{[2]}_2 & = 2 \cdot x^{3} \cdot (x^{6} + 18 x^{5} + 135 x^{4} + 525 x^{3} + 1080 x^{2} + 1080 x + 360) \\ H^{[2]}_3 & = 12 \cdot x^{4} \cdot (x^{10} + 40 x^{9} + 720 x^{8} + 7620 x^{7} + 52080 x^{6} + 238140 x^{5} + 730800 x^{4} + 1467900 x^{3} + 1814400 x^{2} + 1209600 x + 302400) \end{align*}

EDIT: I hadn't realised earlier that my polynomials are the reverse Bessel polynomials. It looks like knowing the Hankel determinants of the usual Bessel polynomials could help.

  • $\begingroup$ Have you read Krattenthaler's Advanced Determinant Calculus? Theorem 11 should work for polynomials instead of numbers, too. $\endgroup$ – Nicolas Malebranche Jan 15 '16 at 12:17
  • $\begingroup$ @NicolasMalebranche I have now: it's a good read! I had looked at it before, but the reason I was put off using Theorem 11 is that each b_i is a rational function that seems to be far more complicated than the polynomial H_i, and I don't see an obvious way that would help me. $\endgroup$ – Simon Willerton Jan 20 '16 at 9:29
  • $\begingroup$ The last coefficient of $\varphi_n$ is $(2n-1)!! =: f_n$, and $\sum_n f_nt^n = \frac{1}{1- \frac{t}{1-\frac{2t}{1-\ldots}} }$ has a very easy continued fraction expansion. This might be a good starting point to find the general one. $\endgroup$ – Nicolas Malebranche Jan 20 '16 at 10:29

There is a combinatorial formula for the Hankel determinant $H_n= \det\left([\varphi_{i+j}]_{i,j=0}^{n}\right)$ in terms of weighted sums of disjoint collections of Schröder paths (and also for the offset version). This is given in Theorem 26 of my paper The magnitude of odd balls via Hankel determinants of reverse Bessel polynomials.

The key observation was pointed out to me be Alan Sokal, this is that there is the following Thron-type continued fraction expansion for the generating function of the reverse Bessel polynomials.

\begin{equation*} \sum_{i=0}^\infty t^i \phi_i(x) = \frac{1}{1- \frac{x t}{1- \frac{t}{1-xt - \frac{2t}{1-xt- \frac{3t}{1-xt- \frac{4t}{1-\dots}}}}}} \end{equation*}

(A similar result was noted by Paul Barry in the formula section of OEIS:A001497.)

One can then use Flajolet's fundamental lemma (relating generalized continued fractions to lattice path enumeration) to give a combinatorial interpretation of the reverse Bessel polynomials as certain weighted counts of Schröder paths.

[I now know that you can prove the combinatorial interpretation of the reverse Bessel polynomials without knowing about the continued fraction interpretation: see the post on Schröder Paths and Reverse Bessel Polynomials at the $n$-Category Café.]

Using this Schröder path enumeration interpretation of the reverse Bessel polynomials, you can apply the Karlin-McGregor-Lindström-Gessel-Viennot Lemma relating determinants to counting disjoint paths to obtain a nice formula for the Hankel determinant required. In fact, you can obtain a combinatorial (path enumeration) formula for each of the coefficients in the determinant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.