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?
Best Answer
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.
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*}