Finding the Explicit Formula of a Recursion Using Generating Functions I have to fid the explicit formula of the following recursion:
$$a_0 = 1, a_n = \sum_{i=0}^{n-1} (i+1)a_i.$$
The summation reminds me a lot of derivatives and I feel like I am going to have to take a derivative somewhere while solving this problem, but I can't exactly figure out how to solve it.
Any help would be greatly appreciated!
 A: For all $n \in \mathbb{N}$
$$
a_{n+1}=\sum_{i=0}^{n}\left(i+1\right)a_i=\underbrace{\sum_{i=0}^{n-1}\left(i+1\right)a_i}_{=a_n}+\left(n+1\right)a_n=\left(n+2\right)a_n
$$
Can you take it from there ?
A: Let $A(z)=\sum_{n \ge 0} a_n z^n$ be the ordinary generating function.  The recurrence implies that
\begin{align}
A(z) 
&= 1+\sum_{n \ge 1} \sum_{i=0}^{n-1} (i+1)a_i z^n \\
&= 1+\sum_{i\ge 0} (i+1)a_i \sum_{n \ge i+1} z^n \\
&= 1+\sum_{i\ge 0} (i+1)a_i \frac{z^{i+1}}{1-z} \\
&= 1+\frac{z}{1-z}\sum_{i\ge 0} (i+1)a_i z^i \\
&= 1+\frac{z}{1-z}\cdot\frac{d}{dz} (z A(z)) \\
&= 1+\frac{z}{1-z}(z A'(z) + A(z))
\end{align}
Hence you need to solve the differential equation
$$A(z) = \frac{1-z + z^2 A'(z)}{1-2z}$$
with initial conditions $A(0)=a_0=1$ and $A'(0)=a_1=1$.

If you instead let $B(z)=\sum_{n \ge 0} a_n \frac{z^n}{n!}$ be the exponential generating function, the recurrence implies that
\begin{align}
B(z) 
&= 1+\sum_{n \ge 1} \sum_{i=0}^{n-1} (i+1)a_i \frac{z^n}{n!} \\
&= 1+\sum_{i\ge 0} (i+1)a_i \sum_{n \ge i+1} \frac{z^n}{n!} \\
&= 1+\sum_{i\ge 0} (i+1)a_i \sum_{n \ge 0} \frac{z^{n+i+1}}{(n+i+1)!}
\end{align}
Not sure where to go from there, but I wanted to point out the difference from what you tried in the comments.

The correct formula, obtained by using the approach of @Atmos but starting from $n=2$, turns out to be $a_n=\begin{cases}1 &\text{if $n=0$}\\(n+1)!/2&\text{otherwise}\end{cases}$
Working backwards from this formula means that $$B(z)=\frac{1}{2}\left(1+\frac{1}{(1-z)^2}\right)$$
More info here: https://oeis.org/A001710
