I'm working with a sequence of summations, as follows:
$$ S_1(n) = \sum_{i=1}^n i $$
$$ S_2(n) = \sum_{j=1}^{n-1} \left( j \sum_{i=j+1}^n i \right) $$
$$ S_3(n) = \sum_{k=1}^{n-2} \left( k \sum_{j=k+1}^{n-1} \left( j \sum_{i=j+1}^n i \right) \right) $$
and the pattern continues. Obviously, as the index of the sequence increases, the more summations we have in the expression. But is there a way to express this compactly, as in with only one summation for each index, or even a closed formula? For example, suppose we want to know the expression for $S_{100}$. By the examples I've given, that would include 100 $\sum$ symbols. Is there a shorter way to write this? I'd appreciate any help.
Update (5/1/2022):
I've been working on this ever since I originally posted this question. So far, I've gotten to:
$$ S_p \left( n \right) = \frac{f \left( n, p \right)}{g \left( n, p \right)} \cdot \frac{\Gamma \left( n + 1 \right)}{\Gamma \left( n - p + 1 \right)} $$
where $f \left( n, p \right)$ is some polynomial of $n$ of highest degree $p$ and $g \left( n, p \right)$ is an integer given by the following function:
$$ g \left( n, p \right) = \prod_i^n i^{\sum_{j=0}^{\infty} \left \lfloor \frac{n-1}{i^j \left( i - 1 \right)} \right \rfloor} $$
where $i$ is prime. I'm still trying to decipher the sequence of polynomials given by $f \left( n, p \right)$.
Update (5/13/2022):
I'm still working on this problem. I do realize that since both $f$ and $g$ are dependent on $n$ and $p$, one could combine them into one function, but I prefer working with integers rather than rationals. Plus, we already have a function for $g$ (although, not a pleasant one). And the coefficients of $f$ are integers. In fact, I did figure out the first ten polynomials of $f$ in terms of $n$. There's not much of a pattern, except if we separate the sequence into subsequences where $p$ is odd or even, then the coefficient of the leading term never decreases. Also, the constant term for when $p$ is odd (except when $p = 1$), tends to always be zero. When $p = 1$, the constant term is $1$.
Update (6/3/2022):
As I do more research on this problem, I came across Faulhaber's formula for the sum of the same powers of consecutive integers from $1$ to $n$. This was helpful in some ways, but not in other ways. It allowed me to convert a sum of integers to the power $p$ into a polynomial in terms of $n$, but the issue here is that it still uses the $\sum$ symbol and includes Bernoulli numbers, which have no closed formula (to my knowledge). Hence, even if I used Faulhaber's formula, the formula is still not closed in terms of $n$ and $p$. On the plus side, I have noticed a pattern on how to find a formula for power $p$, but it relies on knowing the closed formula for power $p - 1$ in terms of $n$ and is very convoluted, even using Faulhaber's formula. I'm hoping I can construct something using linear algebra maybe, like with the closed formula for the Fibonacci sequence. In fact, while doing so, I derived a closed formula for what I like to call the "product of consecutive Gaussian sums", which is the following:
$$ \prod_{j=0}^{p-1} \sum_{i=1}^{n-j} i = \frac{\left( p! \right)^2}{2^p} \binom{n}{p} \binom{n+1}{p} $$
I'm not sure if this has been figured out before, but I thought it was an interesting consequence that came from this research. Again, any input would greatly be appreciated.
Basically, you're facing Stirling numbers of the first kind: $$S_p(n)=\sum_{1\leqslant n_1<\ldots<n_p\leqslant n}n_1\cdots n_p={n+1\brack n+1-p};$$ the connection is seen as $\sum_{p=0}^n S_p(n)x^{n-p}=\prod_{k=1}^n(x+k)$ if we put $S_0(n)=1$.