<s>[INST] Closed form for series of multiple sums I have a series that goes like that:
\begin{align}
(n=1)\qquad &\sum_{i=0}^x\bigg(\sum_{j=1}^i 1\bigg)\\
(n=2)\qquad  &\sum_{i=0}^x\bigg(\sum_{j=0}^i \bigg(1+\sum_{k=1}^j 2\bigg)\bigg)\\
(n=3)\qquad  &\sum_{i=0}^x\bigg(\sum_{j=0}^i \bigg(1+\sum_{k=1}^j \bigg(2+\sum_{l=2}^k 3\bigg)\bigg)\bigg)\\
(n=4)\qquad  &\sum_{i=0}^x\bigg(\sum_{j=0}^i \bigg(1+\sum_{k=1}^j \bigg(2+\sum_{l=2}^k \bigg(3+\sum_{m=3}^l 4\bigg)\bigg)\bigg)\bigg)
\end{align}
and so on.
For each $n$, it is easy to write the sums in closed expressions:
\begin{align}
(n=1)\qquad &\frac{1}{2} (x+1)\,x \\
(n=2)\qquad &\frac{1}{6} (x+1)\,(x+2)\,(2x+3)\\
(n=3)\qquad &\frac{1}{24} (x+1)\,(x+2)\,(3x^2+5x+12)\\
(n=4)\qquad &\frac{1}{120} (x+1)\,(x+2)\,(4x^3+3x^2+33x+60)
\end{align}
What I would like to have is a closed or more compact expression that can give me the polynomials in $x$ for arbitrary $n$. I tried OEIS for the coefficients with little success, and also couldn't find if the polynomials (or parts thereof) are known.
Has someone seen such a construction for a sequence and has an idea how I could go about finding a closed or more compact expression?
 [/INST] Here we give a somewhat more compact expression. We look at the  case $n=3$ again and derive from it a representation for the general case.

We have
  \begin{align*}
\color{blue}{\sum_{j_0=0}^x}&\color{blue}{\left(\sum_{j_1=0}^{j_0}\left(1+\sum_{j_2=1}^{j_1}\left(2+\sum_{j_3=2}^{j_2}3\right)\right)\right)}\\
&=\sum_{j_0=0}^x\sum_{j_1=0}^{j_0}1+2\sum_{j_0=0}^x\sum_{j_1=0}^{j_0}\sum_{j_2=1}^{j_1}1+3\sum_{j_0=0}^x\sum_{j_1=0}^{j_0}\sum_{j_2=1}^{j_1}\sum_{j_3=2}^{j_2}1\tag{1}\\
&=\sum_{0\leq  j_1\leq   j_0\leq  x}1+2\sum_{1\leq j_2\leq j_1\leq   j_0\leq  x}1+3\sum_{2\leq j_3\leq     j_2\leq j_1\leq   j_0\leq  x}1\tag{2}\\
&\,\,\color{blue}{=\binom{x+2}{2}+2\binom{x+2}{3}+3\binom{x+2}{4}}\tag{3}\\
&=\frac{1}{2}(x+2)(x+1)+\frac{1}{3}(x+2)(x+1)x+\frac{1}{8}(x+2)(x+1)x(x-1)\\
&=\frac{1}{24}(x+2)(x+1)(3x^2+5x-3)
\end{align*}

Comment:


*

*In (1) we multiply out and factor out the constants.

*In (2) we use  convenient representation to better see the index range.

*In (3) we note the number of summands given by the index range $2\leq j_3\leq j_2\leq  j_1\leq j_0\leq x$  is the number of ordered $4$-tupel $(j_0,j_1,j_2,j_3)$ between $2$ and $x$ with repetition. This number is given by the binomial coefficient
\begin{align*}
\binom{4+(x-1)-1}{4}=\binom{x+2}{4}
\end{align*}

In  general we have in the case $n=N$:
  \begin{align*}
\color{blue}{\sum_{j_0=0}^x}&\color{blue}{\left(\sum_{j_1=0}^{j_0}\left(1+\sum_{j_2=1}^{j_1}\left(2+\cdots+\sum_{j_N=N-1}^{j_{N-1}}N\right)\cdots\right)\right)}\\
&=\sum_{0\leq  j_1\leq   j_0\leq  x}1+2\sum_{1\leq j_2\leq j_1\leq   j_0\leq  x}1+\cdots+N\sum_{N-1\leq j_N\leq\cdots\leq     j_2\leq j_1\leq   j_0\leq  x}1\\
&=\binom{x+2}{2}+2\binom{x+2}{3}+\cdots+N\binom{x+2}{N+1}\\
&\,\,\color{blue}{=\sum_{k=1}^N k\,\binom{x+2}{k+1}}\\
\end{align*}

</s>