Partial sums of binomial coefficients and related family of polynomials

Question

• Let $a(n)$ be A302117. Here $$ a(n) = 4(n-1)a(n-1) - \frac{1}{3}\prod\limits_{k=0}^{n-1}(2k-3), \\ a(0) = 0. $$
• Let $$ T(n,k) = \sum\limits_{i=0}^{k} \binom{n}{i}. $$
• Let $P_n(z)$ be the family of polynomials of degree $n$ such that $P_k(n)=T(n+k-1, k)$.
• Let $$ b(n) = n!\sum\limits_{k=0}^{n}2^{n-k}[z^k]P_n(z). $$

I conjecture that $$ b(n)=a(n+1). $$

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

upto1(n) = my(A = -3, v1); v1 = vector(n+1, i, 0); for(i=1, n, v1[i+1] = 4*(i-1)*v1[i] - (1/3)*A; A *= (2*i - 3)); v1
b(n) = my(v1, M1, M2); M1 = matrix(n+1, n+1, i, j, i^(j-1)); M2 = matrix(n+1, 1, i, j, sum(s=0, n, binomial(i+n-1, s))); M2 = matsolve(M1, M2); v1 = vector(n+1, i, M2[i, 1]); n!*sum(i=0, n, 2^(n-i)*v1[i+1])
test(n) = concat(0, vector(n, i, b(i-1))) == upto1(n)

Is there a way to prove it?

Answer 1

We have $$b(n) = n!2^nP_n(\frac12) = n!2^n\sum_{i=0}^n\binom{n-\frac12}i.$$

Now, for $i>0$ by Pascal's rule $\binom{n-\frac12}i = \binom{n-1-\frac12}i + \binom{n-1-\frac12}{i-1}$, and thus $$b(n) = 4n b(n-1) + n!2^n\binom{n-1-\frac12}n,$$ which directly translates into the recurrence formula for $a(n)$.