Finding generating function for the recurrence $a_0 = 1$, $a_n = {n \choose 2} + 3a_{n - 1}$ I am trying to find generating function for the recurrence:


*

*$a_0 = 1$,

*$a_n = {n \choose 2} + 3a_{n - 1}$ for every $n \ge 1$.


It looks like this:


*

*$a_0 = 1$

*$a_1 = {1 \choose 2} + 3$

*$a_2 = {2 \choose 2} + 3{1 \choose 2} + 9$

*$a_3 = {3 \choose 2} + 3{2 \choose 2} + 9{1 \choose 2} + 27$

*$a_4 = {4 \choose 2} + 3{3 \choose 2} + 9{2 \choose 2} + 27 {1 \choose 2} + 81$


I know what the generating function of the sequence $3 ^n = (1, 3, 9, 27, 81, \dots)$ is, as well as what the generating functions for some sequences of combinatorial numbers are, but how do I split the sequence up into these pieces I know?
(The problem is those combinatorial numbers "move right" every time. If they were growing left-to-right along with their coefficients, it would be much easier. And there is no constant difference between $a_i$ and $a_{i + 1}$.)
 A: Let $A(x)=\sum_{n=0}^\infty a_nx^n$. Then
\begin{eqnarray}
A(x)&=&1+\sum_{n=1}^\infty a_{n}x^n\\
&=&1+\sum_{n=1}^\infty (3a_{n-1}+\frac{1}{2}n(n-1))x^n\\
&=&1+3xA(x)+\frac{1}{2}\sum_{n=1}^\infty n(n-1)x^n.
\end{eqnarray}
Note $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ for $|x|<1$. Differentiating this twice, you can give
$$ \sum_{n=2}^\infty n(n-1)x^{n-2}=\frac{2}{(1-x)^3}. $$
Thus
$$ A(x)=1+3xA(x)+\frac{x^2}{(1-x)^3} $$
from which you can get $A(x)$.
A: A related problem. Assume $ F(x) = \sum_{n=0}^{\infty}a_n x^n $, then
$$ a_n = {n \choose 2} + 3a_{n - 1} \implies a_{n+1} = {n+1 \choose 2} + 3a_{n}$$
$$ \sum_{n=0}^{\infty} a_{n+1} x^n = \frac{1}{2}\sum_{n=0}^{\infty}n(n+1)x^n + 3\sum_{n=0}^{\infty}a_{n}x^n  $$
$$ \implies \sum_{n=1}^{\infty} a_{n} x^{n-1} = \frac{1}{2}\sum_{n=1}^{\infty}nx^{n}+\frac{1}{2}\sum_{n=1}^{\infty}n^2x^{n} +3F(x) $$
$$ \implies \frac{1}{x}F(x)-\frac{a_0}{x}-3F(x) = \frac{1}{2}\sum_{n=1}^{\infty}nx^{n}+\frac{1}{2}\sum_{n=1}^{\infty}n^2x^{n}  $$
$$\implies \left(\frac{1}{x}-3 \right)F(x)=\frac{1}{x}+\frac{1}{2}\frac{x}{(x-1)^2}-\frac{1}{2}\frac{x(x+1)}{(x-1)^3} $$
$$ \implies \left(\frac{1}{x}-3 \right)F(x)=\frac{1}{x}-\frac{x}{(x-1)^3} $$
$$ \implies F(x)=\frac{x}{1-3x}\left(  \frac{1}{x}-\frac{x}{(x-1)^3} \right). $$
A: Hint Let $F:=\sum_0^\infty a_n x^n$. Consider $a_n x^n= 3 a_{n-1}x^{n}+C_n^2 x^n$.
A: As I said in my comment, you have:
$$\displaystyle a_n=3^n+\sum_{k=0}^{n-1}3^k{n-k \choose 2}$$
You can rewrite this as:
$$\displaystyle a_n=3^n+\sum_{k=1}^{n}3^{n-k}{k \choose 2}=3^n+3^n\sum_{k=1}^{n}3^{-k}{k \choose 2}=3^n\left(1+\sum_{k=1}^{n}\left(\frac{1}{3}\right)^{k}{k \choose 2}\right)$$
Also you have:
$$\displaystyle\sum_{k=1}^\infty \left(\frac{1}{3}\right)^{k}{k \choose 2}=\sum_{k=0}^\infty \left(\frac{1}{3}\right)^{k}{k \choose 2}=\dfrac{\left(\frac{1}{3}\right)^{2}}{\left(1-\frac{1}{3}\right)^{3}}=\frac{3}{8}$$
A: Sneaky. Write your recurrence without subtractions in indices, i.e.:
$$
a_{n + 1} = 3 a_n + \binom{n + 1}{2}
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply by $z^n$, sum over $n \ge 0$ and recognize the resulting sums, particularly:
\begin{align}
\sum_{n \ge 0} \binom{n + 1}{2} z^n
  &= z \sum_{n \ge 0} \binom{n + 1}{2} z^{n - 1} \\
  &= z \sum_{n \ge 0} \binom{n + 2}{2} z^n \\
  &= \frac{z}{(1 - z)^3}
\end{align}
so that:
$$
\frac{A(z) - a_0}{z} = 3 A(z) + \frac{z}{(1 - z)^3}
$$
Using $a_0 = 1$ and solving as partial fractions:
$$
A(z) = \frac{11}{8 (1 - 3 z)}
         - \frac{1}{2 (1 - z)^3}
         + \frac{1}{4 (1 - z)^2}
         - \frac{1}{8 (1 - z)}
$$
We can read off the coefficients here:
\begin{align}
a_n &= \frac{11}{8} \cdot 3^n
         - \frac{1}{2} \binom{-3}{n} (-1)^n
         + \frac{1}{4} \binom{-2}{n} (-1)^n
         - \frac{1}{8} \\
    &= \frac{11 \cdot 3^n - 1}{8}
         - \frac{1}{2} \binom{n + 3 - 1}{3 - 1}
         + \frac{1}{4} \binom{n + 2 - 1}{2 - 1} \\
    &= \frac{11 \cdot 3^n - 1}{8}
         - \frac{1}{2} \cdot \frac{(n + 2) (n + 1)}{2!}
         + \frac{1}{4} \cdot \frac{n + 1}{1!} \\
    &= \frac{11 \cdot 3^n - 2 n^2 - 4 n - 3}{8}
\end{align}
