$R$-recursion for unsigned Genocchi numbers (of first kind) of even index

Question

• Let $G_n$ be A036968 (i.e., Genocchi numbers). Here $$ \frac{2t}{1+e^t}=\sum\limits_{n=0}^{\infty}G_n\frac{t^n}{n!}. $$ Also $$ t\tan\left(\frac{t}{2}\right)=\sum\limits_{n=1}^{\infty}(-1)^n G_{2n}\frac{t^{2n}}{(2n)!}. $$
• Let $$ R(n, q) = \sum\limits_{j=0}^{q+1}\binom{q+2}{j}R(n-1, j), \\ R(0, q) = 1. $$

I conjecture that $$ R(n, 0) = (-1)^n G_{2(n+2)}. $$

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

upto1(n) = my(v1, v2, v3, v4); v1 = vector(n + 1, i, 1); v2 = v1; v3 = vector(n + 1, i, 0); v3[1] = 1; v4 = vector(n+2, i, vector(i, j, j==1 || i==j)); for(i=2, n+2, for(j=2, i-1, v4[i][j] = v4[i-1][j] + v4[i-1][j-1])); for(i=1, n, for(q=0, n-i, v2[q + 1] = sum(j=0, q+1, v4[q+3][j+1]*v1[j + 1])); v1 = v2; v3[i + 1] = v1[1]); v3
upto2(n) = my(v1); v1 = vector(n+1, i, i++; 2*(-1)^i*(1 - 4^i)*bernfrac(2*i))
test(n) = upto1(n) == upto2(n)

Is there a way to prove it?

Answer 1

From the recurrence it follows that the generating function $F(x,y) := \sum_{n,k\geq 0} R(n,k) \frac{x^{2n}}{(2n)!} \frac{y^k}{k!}$ satisfies a neat PDE: $$\frac{\partial^2}{\partial^2 x} F(x,y) = \frac{\partial^2}{\partial^2 y}(e^y-1)F(x,y)$$ with the boundary conditions $F(0,y)=e^y$ and $F_x(0,y)=0$. The question is then equivalent to showing that $$F(x,0) = \frac{\partial^4}{\partial^4 x} x\tan(\tfrac{x}2).$$

Unfortunately, the "standard" route of solving this PDE and plugging in $y=0$ in the solution seems to be infeasible here. At least both Mathematica and Maple fail to solve this PDE. So, we need to take another route.

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Let's focus on generating functions for each row: $$G_n(y) := \sum_{k\geq 0} R(n,k) \frac{y^k}{k!}.$$ We have $G_0(y) = e^y$ and for $n\geq1$, $$G_{n+1}(y) = \frac{\partial^2}{\partial^2 y}(e^y-1)G_n(y).$$ Clearly, $G_n(y)$ represents a polynomial of degree $n+1$ in $e^y$, and with a bit of effort we can derive the following formula for the coefficient of $e^{ky}$ in $G_n(y)$: $$(-1)^{n+1-k} [z^{n+1}]\ \prod_{j=1}^k \frac{j^2z}{1-j^2z}.$$ It further implies that \begin{split} R(n,0) = G_n(0) &= (-1)^{n+1} [z^{n+1}] \sum_{k\geq 0} k!^2 (-z)^k \prod_{j=1}^k \frac{1}{1-j^2z} \\ &=(-1)^n [z^{n+2}] \sum_{k\geq 0} k!^2 (-z)^{k+1} \prod_{j=1}^k \frac{1}{1-j^2z}. \end{split} The series in the right-hand side can be recognized as the generating function for Genocchi numbers listed in OEIS A001469:

O.g.f.: Sum_{n>=0} n!^2*(-x)^(n+1) / Product_{k=1..n} (1-k^2*x). - Paul D. Hanna, Jul 21 2011

I'm not sure what is the source of Hanna's formula, but the above analysis essentially shows that the question is equivalent to it.