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.



Evaluation of Hankel determinants for the reverse Bessel polynomials
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](https://arxiv.org/abs/1708.03227). 


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](https://oeis.org/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](https://golem.ph.utexas.edu/category/2017/08/schrder_paths_and_reverse_bessel_polynomials.html) 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](https://en.wikipedia.org/wiki/Lindstr%C3%B6m%E2%80%93Gessel%E2%80%93Viennot_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.


