In the following summation, I need the current sum to be a part of the computation in the actual sigma, represented as,
$$a_n = \sum_{i=1}^n {[a_{n-1}(i+1) + i]}$$
for example, $$a_1 = \sum_{i=1}^1 {[a_{0}(i+1) + i]}$$
and, $$a_2 = \sum_{i=1}^2 {[a_{1}(i+1) + i]}$$ $$...$$ $$...$$
I need the answer to be in terms of $a_0$. Any ideas?
there are a few simplifications we can make: $$S=\sum_{i=1}^n\left[a_{n-1}(i+1)+i\right]=\sum_{i=1}^na_{n-1}(i+1)+\sum_{i=1}^ni$$ $$=a_{n-1}\sum_{i=1}^n(i+1)+\sum_{i=1}^ni$$ $$a_n=\frac{n(n+3)}{2}a_{n-1}+\frac{n(n+1)}{2}$$
EDIT 1:
If we looks at the formulas for each of the following terms we could try and estimate how it follows $a_0$:
This chart shows the iteration its is in column 1 (i.e. $a_1$,$a_2$ etc.), the coefficient of $a_0$ in column two and the added second part in the third column. For example, $a_8=1047816000a_0+917080508$. This appears to follow a near-exponential pattern so if many more terms were calculated a fairly accurate formula could be made, although I have one other idea.
EDIT 2:
If you could find a way of generating a general formula for a $a_{n+k}$ and prove this is true, then you could work back and find a formula for $a_n$. We can start by saying: $$a_{n+1}=\frac{(n+1)(n+4)}{2}a_n+\frac{(n+1)(n+2)}{2}$$ $$a_{n+1}=\frac{n(n+1)(n+3)(n+4)}{2^2}a_{n-1}+\frac{n(n+1)^2(n+4)}{2^2}+\frac{(n+1)(n+2)}{2}$$
A similar example of a case is the Fibonacci sequence, which can be simplified as follows: $$F_{n+1}=F_n+F_{n-1}$$ $$F_n=\frac{\varphi^n-(-\varphi)^n}{\sqrt{5}}$$
