$R$-recursion for the A307389

Question

• Let $a(n)$ be A307389 i.e. an integer sequence such that its exponential generating function $A(x)$ satisfies $$ A(x)=\exp\left(\frac{\exp(2x)-2\exp(x)+2x+1}{2}\right) $$ The sequence begins with $$ 1, 1, 2, 7, 29, 136, 737, 4537, 30914, 229831, 1850717, 16036912, 148573889, 1463520241 $$
• Let $$ R(n,q) = (q+1)R(n-1,q) - R(n-1,q+1) + R(n-1,q+2), \\ R(0,q) = 1 $$

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

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

a_upto(n) = my(t='t+O('t^(n+1))); Vec(serlaplace(exp((exp(2*t)-2*exp(t)+2*t+1)/2)))
R_upto(n) = my(v1, v2, v3); v1 = vector(2*n + 1, i, 1); v2 = v1; v3 = vector(n + 1, i, 0); v3[1] = 1; for(i = 1, n, for(q = 0, 2*(n - i), v2[q + 1] = (q + 1) * v1[q + 1] - v1[q + 2] + v1[q + 3]); v1 = v2; v3[i + 1] = v1[1]); v3
test(n) = R_upto(n) == a_upto(n)

Is there a way to prove it?

Answer 1

Let \begin{equation*} A(x,q) = e^{qx}A(x)=e^{(q+1)x+(e^x-1)^2/2} \end{equation*} and define $a(n,q)$ by \begin{equation*} A(x,q) = \sum_{n=0}^\infty a(n,q) \frac{x^n}{n!}. \end{equation*} Then $a(n,0) = a(n)$. We find that \begin{align*} \frac{d\ }{dx}A(x,q)&=\bigl((q+1) - e^x +e^{2x}) A(x,q)\\ &=(q+1)A(x,q) -A(x,q+1) +A(x,q+2). \end{align*} Equating coefficients of $x^{n-1}/(n-1)!$ for $n\ge 1$ gives \begin{equation*} a(n,q) = (q+1)a(n-1,q) -a(n-1,q+1) + a(n-1, q+2). \end{equation*} Since $a(0,q)=1$, it follows that $R(n,q) = a(n,q)$, and in particular, $R(n,0)=a(n,0)=a(n)$.