# Closed form for the A110501 (unsigned Genocchi numbers (of first kind) of even index)

Let $$a(n)$$ be A110501 (i.e., unsigned Genocchi numbers (of first kind) of even index). Here

$$a(n) = \sum\limits_{i=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n}{2i}a(n-i)(-1)^{i-1}, \\ a(1) = 1$$

Let $$T(n, k)$$ be A036969. Here

$$T(n, k) = k^2T(n-1, k) + T(n-1, k-1), \\ T(n, 1) = 1$$

Let $$b(n)$$ be an integer sequence such that

$$b(n) = \sum\limits_{i=1}^{n-1} (i!)^2T(n-1, i)(-1)^{n-i-1}, \\ b(1) = 1$$

I conjecture that $$b(n) = a(n).$$

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

a(n) = if(n<1, 0, 2 * (-1)^n * (1 - 4^n) * bernfrac(2*n))
b_upto(n) = {my(x='x, v1, v2, A, B);
v1 = vector(n, i, 0); v1[1] = 1;
for(i=2, n+1,
if(i<(n+1), v2 = v1[i-1];
v1[i] = x*(v2 + v2' + x*v2''));
if(i>2, v2 = Vecrev(v1[i-1]/x);
A = 0; B = 1;
forstep(j=i-2, 1, -1,
A = A*(j+1)^2 + B*v2[j];
B *= -1);
v1[i-1] = A));
v1}
test(n) = b_upto(n) == vector(n, i, a(i))


Is there a way to prove it?

• Thank you for answer! What do you think about $$a(n) = \sum\limits_{i=1}^{n}i((i-1)!)^2T(n, i)(-1)^{n-i}.$$? Commented May 24 at 18:03