* Let $a(n)$ be [A001515][1], i.e., the Bessel polynomial $y_n(x)$ evaluated at $x=1$. Here
$$
a(n) = (2n-1)a(n-1) + a(n-2), \\
a(0) = 1, a(1) = 2
$$
The closed form is
$$
a(n)=\sum\limits_{k=0}^{n}\binom{n+k}{2k}\frac{(2k)!}{k!2^k}
$$
Also exponential generating function is
$$
\frac{\exp(1-\sqrt{1-2x})}{\sqrt{1-2x}}
$$
and generating function using continued fraction is $\frac{1}{G(0)}$ where
$$
G(j)=1-x-\frac{(j+1)x}{G(j+1)}
$$
* Let
$$
R(n,q)=(q+1)R(n-1,q+1)+\sum\limits_{j=0}^{q}(j+1)R(n-1,j), \\
R(0,q)=1
$$

I conjecture that
$$R(n,0)=a(n).$$

Here is the PARI/GP prog to check it numerically:

    a_upto(n)=my(v1); v1=vector(n+1, i, 0); v1[1]=1; v1[2]=2; for(i=2, n, v1[i+1]=(2*i-1)*v1[i] + v1[i-1]); v1
    R_upto(n)=my(v1, v2, v3); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, n-i, v2[q+1]=(q+1)*v1[q+2] + sum(j=0, q, (j+1)*v1[j+1])); v1=v2; v3[i+1]=v1[1]); v3
    test(n)=a_upto(n)==R_upto(n)


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**UPD:**

* Let $T(n,k)$ be [A001498][2], i.e., triangle of coefficients of Bessel polynomials $y_n(x)$. Here
$$
T(n,k)=\frac{(n+k)!}{2^k(n-k)!k!}
$$
* Let
$$
R_1(n,q,z)=(q+1)zR_1(n-1,q+1,z)+\sum\limits_{j=0}^{q}(jz+1)R_1(n-1,j,z), \\
R_1(0,q,z)=1
$$

I conjecture that
$$T(n,k)=[z^k]R_1(n,0,z)$$

Here is the PARI/GP prog to check it numerically:

    T(n, k)=if(k<0||k>n, 0, binomial(n, k)*(n+k)!/2^k/n!)
    y(n)=sum(j=0, n, x^j*T(n,j))
    R1_upto(n)=my(v1, v2, v3); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, n-i, v2[q+1]=(q+1)*x*v1[q+2] + sum(j=0, q, (j*x+1)*v1[j+1])); v1=v2; v3[i+1]=v1[1]); v3
    test1(n)=vector(n+1, i, y(i-1))==R1_upto(n)


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Is there a way to prove it?


  [1]: https://oeis.org/A001515
  [2]: https://oeis.org/A001498