Is there a closed form expression for the "generalized" addition of the first $n$ numbers? Firstly, I will explain what I am trying to do intuitively. We take the sum of the first $n$ positive integers. Let's say this sum is equal to $q$. Then you add that sum to the sum of the first $q$ positive integers. Let's say this new sum is equal to $m$. Then you add that to the sum of the first $m$ positive integers. The number of times this process is iterated is specified by some $k$, which, along with $n$, is the independent variable for this function. 
Formally, suppose we define a function $\sigma: \mathbb{N}\times\mathbb{N} \rightarrow \mathbb{N}$ recursively as follows. 
$$\sigma(0,n) = n$$
and if $k>0$, $$\sigma(k,n) = \sum_{j = 0}^{k-1}\sum_{i = 1}^{\sigma(j,n)} i$$  
For example, $$\sigma(1,n) = \sum_{j = 0}^{0}\sum_{i = 1}^{\sigma(j,n)} i = \sum_{i = 1}^{\sigma(0,n)} i = \frac{n(n+1)}{2}$$
$$\sigma(2,n) = \sum_{j = 0}^{1}\sum_{i = 1}^{\sigma(j,n)} i = \sum_{i = 1}^{\sigma(0,n)} i + \sum_{i = 1}^{\sigma(1,n)} i = \sum_{i = 1}^{n} i + \sum_{i = 1}^{\frac{n(n+1)}{2}} i$$
Is there a "closed form" expression, or simply any general formula, for $\sigma(k,n)$?
 A: The recurrence can be written as:
$$
\begin{aligned}
\sigma(k + 1, n) &= \sigma(k, n) + \frac{\sigma(k, n)^2 + \sigma(k, n)}{2}\\
&= \frac{\sigma(k, n)^2 + 3\sigma(k, n)}{2}\\
\end{aligned}
$$
Since $n$ doesn't matter for this analysis, I'll rewrite with $\sigma(n, k) = \sigma_k$ for conciseness.
$$
\sigma_{k+1} = \frac{\sigma_k^2 + 3\sigma_k}{2}
$$
This is a quadratic recurrence relation. I'll put it in standard form $S_{k+1} = aS_k^2 + bS_k + c$.
$$
\sigma_{k+1} = \frac{1}{2}\sigma_k^2 + \frac{3}{2}\sigma_k
$$
Now we're just trying to find a closed form for a quadratic recurrence relation. Following Will Jagy's answer here, we first try to create another sequence $T_k$ with a recurrence of the form $T_{k+1} = T_k^2 + c$. It turns out this sequence is just:
$$
T_k = \frac{1}{2}\sigma_k + \frac{3}{4}
$$
with the recurrence
$$
T_{k+1} = T_k^2 + \frac{3}{16}
$$
(Check me!). Will Jagy then says that there are only two cases with a closed form: when $c = 0$ and when $c = -2$. Neither case holds, so there isn't a closed form.
I would love to know more about this problem in generality. Why are those two cases the only ones that can be solved?
A: Let $s_n$ denote $\sigma(n, i)$.
Using Eli Rose's (now deleted) comment as guidance:
$$s_{n+1} = \sum_{i=1}^{s_n}i + s_n + s_{n-1} + \cdots + s_1$$
$$s_{n} = \sum_{i=1}^{s_{n-1}}i + s_{n-1} + \cdots + s_1$$
Subtracting them, we see
$$s_{n+1}-s_n = \sum_{i=1}^{s_n}i + s_n - \sum_{i=1}^{s_{n-1}}i = \frac{s_n^2+s_n}{2} + s_n - \frac{s_{n-1}^2+s_{n-1}}{2}$$
So we get a recursive formula for
$$2s_{n+1} = s_n^2 + 3s_n - s_{n-1}^2 - s_{n-1}$$
Since we know $s_1=i$ and $s_2=\frac{i(i+1)}{2}$ as given in the problem, we can use this formula to calculate $s_n$ for all $n$.  (I apologize, I reversed the $n$ and the $i$ in my calculations, as compared to the original problem.)
Unfortunately, a closed form did not give very nice coefficients using several approaches.
