Consider sums of the form


$S(n)=\sum^{n}\_{m=0}\sum^{m}\_{k=1}2^{2k+m+1}{n-m+k+1 \choose 2k+2}{m \choose k}$


I am interested in the asymptotics of $S(n)$ as $n\to \infty$.


More precisely I would like to know whether it grows purely exponential in $n$ i.e.
whether there are numbers $c>0$ and $R>0$ such that


$R^{-1}\leq \lim \inf \frac{S(n)}{c^{n}}\leq \lim \sup \frac{S(n)}{c^{n}}\leq R$ 



Asymptotic of a certain double sum involving binomial coefficients
The inner summation from $k=1$ looks a bit suspicious, since starting with $k=0$ appears to be more natural. So I assume that the inner summation starts with $k=0$ (the case of $k=1$ easily follows) and to avoid ambiguity I denote this sum by $S'(n)$. 


The generating function for $S'(n)$ is
$$\sum\_{n=0}^{\infty} S'(n)\cdot x^n = \frac{2x}{(1-x)(1-4x-3x^2-2x^3)}.$$


It follows that $S'(n)$ grows proportionally to $\alpha^{-n}$, where $\alpha$ is the smallest (by absolute value) zero of the polynomial $1-4x-3x^2-2x^3$, that is
$$S'(n) \sim 4.7246^n.$$
More precisely, $S'(n) = 0.4580\cdot 4.7246^n +$ smaller terms.


UPDATE. For the sake of completeness, here is the generating function for $S(n)$:
$$\sum\_{n=0}^{\infty} S(n)\cdot x^n = \frac{2x}{(1-x)(1-4x-3x^2-2x^3)} - \frac{2x}{(1-x)^3 (1-2x)}.$$
As Brendan McKay pointed out, the asymptotics for $S(n)$ is the same as for $S'(n)$.


