Coefficients of $(2+x+x^2)^n$ from trinomial coefficients I would like to be able to express the coefficients of $(2+x+x^2)^n$ in terms of the trinomial coefficients studied by Euler, ${n \choose \ell}_2 = [x^\ell](1+x+x^2)^n$ where $[x^\ell]$ denotes the coefficient of $x^\ell$.  The triangle of these numbers is given in OEIS A027907 and begins
\begin{matrix}
1 \\
1 & 1 & 1 \\
1 & 2 & 3 & 2 & 1 \\
1 & 3 & 6 & 7 & 6 & 3 & 1\\
1 & 4 & 10 & 16 & 19 & 16 & 10 & 4 & 1
\end{matrix}
The triangle $t(n,\ell) = [x^\ell](2+x+x^2)^n$ I want to relate to the ${n \choose \ell}_2$ begins
\begin{matrix}
1 \\
2 & 1 & 1 \\
4 & 4 & 5 & 2 & 1 \\
8 & 12 & 18 & 13 & 9 & 3 & 1\\
16 & 32 & 56 & 56 & 49 & 28 & 14 & 4 & 1
\end{matrix}
I'm hoping for a general result of the form $t(n,\ell) = \left(\text{function of ${m \choose k}_2$}\right)$ with $m \le n$ and $k \le \ell$.  I see patterns for certain columns and diagonals, and recurrence relations within the triangle, but not yet a general expression in terms of trinomial coefficients.
One note: The trinomial coefficients can be worked out in terms of binomial coefficients, but I'd like an expression in ${n \choose \ell}_2$ instead, as this is the first step in a larger program: Eventually I want to relate the coefficients of $(2+x+\cdots+x^k)^n$ to ${n \choose \ell}_k = [x^\ell](1+x+\cdots+x^k)^n$.
 A: Using Abdelmalek's tip in the comments, here's a solution to a more general version of the "larger program" mentioned at the end.  For an arbitrary constant $c$,
\begin{align}
[x^\ell](c+x+\cdots+x^k)^n & = [x^\ell] \left((c-1) + (1+x+\cdots+x^k)\right)^n \\
& = \sum_{m=0}^n {n \choose m}(c-1)^{n-m}[x^\ell](1+x+\cdots+x^k)^m \\
& = \sum_{m=0}^n {n \choose m}(c-1)^{n-m}{m \choose \ell}_k
\end{align}
where we use the binomial theorem in the second line.
In the case of the original question, $c=2$ means the $(c-1)^{n-m}$ factor is always 1.  You can think of the row $t(4,\ell)$ coming from dot products of $(1,4,6,4,1)$ with each column in the first five rows of the ${n \choose k}_2$ triangle:
\begin{gather}
(1,4,6,4,1)\cdot(1,1,1,1,1) = 16,\\
(1,4,6,4,1)\cdot(0,1,2,3,4) = 32,\\
(1,4,6,4,1)\cdot(0,1,3,6,10) = 56,\\
(1,4,6,4,1)\cdot(0,0,2,7,16) = 56,\\
(1,4,6,4,1)\cdot(0,0,1,6,19) = 49,\\
(1,4,6,4,1)\cdot(0,0,0,3,16) = 28,\\
(1,4,6,4,1)\cdot(0,0,0,1,10) = 14,\\
(1,4,6,4,1)\cdot(0,0,0,0,4) = 4,\\
(1,4,6,4,1)\cdot(0,0,0,0,1) = 1.
\end{gather}
Thanks for putting up with what ended up being an elementary question.
