Let $S_k$ be a recursive sequence defined by
$S_k = (x+1)S_{k-1}+S_{k-2}$ with $k>1$
with initial conditions:
$S_0 = 2, S_1=x+1$.
Then the first few terms are
$2, x + 1, x^2 + 2x + 3, x^3 + 3x^2 + 6x + 4, x^4 + 4x^3 + 10x^2 + 12x + 7,\\ x^5 + 5x^4 + 15x^3 + 25x^2 + 25x + 11, x^6 + 6x^5 + 21x^4 + 44x^3 + 60x^2 + 48x + 18$
For simplicity, the first term $S_0$ will be set to $1$ instead of $2$. We can arrange the coefficients of all the terms of $S_k$ in a triangle like format (Similar to Pascals Triangle).
$[1]$
$[1, 1]$
$[1, 2, 3]$
$[1, 3, 6, 4]$
$[1, 4, 10, 12, 7]$
$[1, 5, 15, 25, 25, 11]$
$[1, 6, 21, 44, 60, 48, 18]$
$...$
Let $P_{t,n}$ denote the $t$-th coefficient of $S_k$ (starts at $t=0$). For example, $P_{3,3} = 4$.
If $t$ is fixed, then $P_{t,n}$ is a polynomial in $n$.
$P_{0,n} = 1$, $P_{1,n} = n$, $P_{2,n} = \frac{n^2 + n}{2}, P_{3,n} = \frac{n^3 + 3n^2 - 10n}{6}$
and so on.
If $t > 1, n > 2$, then $P_{t,n} = P_{t,n-1}+P_{t-1,n-1}+P_{t-2,n-2}$. This is similar to Pascal's Identity for binomial coefficients.
What is the generating function for the power series
$$G(x,n) = \displaystyle\sum_{t=0}^{\infty}P_{t,n}x^t = 1 + nx + \frac{n^2 + n}{2}x^2 + \frac{n^3 + 3n^2 - 10n}{6}x^3 \\+ \frac{n^4 + 6n^3 - 37n^2 + 30n}{24}x^4 + \frac{n^5 + 10n^4 - 85n^3 + 50n^2 + 264n}{120}x^5 + O(x^6)$$
Is there a general formula for generating functions involving other polynomial recursive functions?
Edit:
$$\displaystyle\sum_{t=0}^{\infty}P_{t,n}x^t = (\sum_{j=0}^{\lceil{t/2}\rceil} \frac{n}{n-j}\binom{n-j}{j}\binom{n-2j}{t-2j})x^t $$
The solutions to $X^2 -(x+1)X-1=0$ are $X_1= \frac{x+1+\sqrt{(x+1)^2+4}}{2}$ and $X_2= \frac{x+1-\sqrt{(x+1)^2+4}}{2}$. Then, with the given initial values, we have
$$S_k(x)= X_1^k+X_2^k.$$ Since by definition $$S_n(x):=\sum_{t\ge 0}x^{n-t}P_{t,n},$$ we have $$S_n(x^{-1})=\sum_{t\ge 0}x^{t-n}P_{t,n},$$ that is $$\sum_{t\ge 0}P_{t,n}x^{t}=x^nS_n(x^{-1}),$$ that is $$\sum_{t\ge 0}P_{t,n}x^{t}=\big(\frac{1+x+\sqrt{5x^2+2x+1}}{2}\big)^{n}+\big(\frac{1+x-\sqrt{5x^2+2x+1}}{2}\big)^{n},$$ if I did not make a mistake.