What is the asymptotic growth of the sequence
$$a\_n:=\sum\_{k\geq 0} 3^k c\_{n,k},$$
as $n\rightarrow\infty$, where $c\_{n,k}$ denotes the number of integer compositions of $n$ with exactly $k$ many 2s?


A composition of $n$ is a sum $n=c\_1+c\_2+\cdots+c\_p$, with all the $c\_i$ positive.
The first values of the sequence $a\_n$ are $1,1,4,8,22,52,135,\ldots$ (not in the OEIS). [Edit: As pointed out by Somos below, the value 135 is wrong, and must be corrected to 132, and then the sequence is in the OEIS.]


So far, I was only able to prove the following bounds: As $\sum\_{k\geq 0}c\_{n,k}=2^{n-1}$, it follows that
$$2^{n-1}\leq a\_n \leq (2\sqrt{3})^n=(3.464...)^n.$$



Count weighted integer compositions
When corrected, $a\_n$ is the [OEIS sequence A052528](https://oeis.org/A052528) whose first nine values are
 $\,1,1,4,8,22,52,132,324,808,\dots\,$ and it has a linear recurrence.


The combinatorial recurrence from its combination definition leads immediately to
$$ a\_n = 2a\_{n-2} + \sum\_{k=0}^{n-1} a\_k $$
where the $\, 2a\_{n-2}\,$ comes from the combination part $2.$ 
The ordinary generating function is
$$ (1 - x)/(1 - 2 x - 2 x^2 + 2 x^3). $$
The growth rate of $\,a\_n\,$ depends on the reciprocal of the smallest root
 $\,\alpha\,$ of $\, 2x^3 -2x^2 -2x +1.$ Thus
$\,1/\alpha \approx 2.4811943\,$ so that $\,a\_n \propto 1/\alpha^n.$


**EDIT**: The sequence $\,c\_{n,k}\,$ is the triangular [OEIS sequence A105114](https://oeis.org/A105114)



> 
> Triangle read by rows: T(n,k) is the number of compositions of n having exactly k parts equal to 2.
> 
> 
> 


Again, its combination definition leads immediately to $$ c\_{n,k} = \sum\_{j=1}^n c\_{n-j,k-[j=2]} $$
where $\,[j=2] := 1\,$ if $j=2$, else $0$.


