I am interested in the sequence
$$a(n)=\sum_{k=0}^n {p(n-k)-1 \choose k}$$$$a(n)=\sum_{k=0}^n {p(n-k) \choose k}$$
where $p(n)=(r-1)n^2+(2r-1)n+r$ for some $r \in \mathbb{N}$ or more generally any$p(n)$ is a polynomial equation.
When $r=1$$p(n)=n$ this reduces to the Fibonacci sequence which has exponential growth., but what about when $p(n)$ is quadratic?
ItFor example when $p(n)=n^2$, it can be seen that $a(n)$ has superexponential growth when $r \ge 2$ by considering only one of the central termterms of the binomial sum $$a(n) \ge {p(n-(n/2))-1 \choose n/2}={p(n/2)-1 \choose n/2}\ge\left(\frac{p(n/2)-1}{n/2}\right)^{n/2}=\left(\sqrt{(r-1)\frac{n}{2}+(2r-1)+\frac{2n-2}{n}}\right)^n$$$$a(n) \ge {p(n-(n/2)) \choose n/2}={p(n/2) \choose n/2}\ge\left(\frac{p(n/2)}{n/2}\right)^{n/2}=\left(\sqrt{\frac{n}{2}}\right)^n$$ But I would like to know more information than just this lower bound - an asymptotic formula would be great. Any ideas?
I found a related sequence here (which is equivalent to the case when $p(n)=n^2$) along with its generating function but I am not familiar enough with generating functions to extractif that is any asymptotic information from ithelp to anyone.