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.