Is there a simple way to expand the following Product Sum?

Question

$$ F(n) = \prod_{i = 1}^{n} \sum_{j = 1}^{i} v_j $$

$$ v_k \in \mathbb R $$

I would like to convert this from a Product-Sum to just a Sum of the following form:

$$ F(n) = \sum_{i=1}^{C_n} g(i) $$

Unfortunately, the $v$'s do not have any additional structure to them and are empirically discovered values. They are discrete velocities in a system I am simulating and the above Product-Sum has emerged in a calculation.

I was hoping there was some sort of multinomial-coefficient-like function that would allow me to simplify the Product-Sum in any way, or eat least find the coefficients of the fully expanded polynomial.

$$ F(1) = v_1\\ F(2) = v_1^2 + v_1v_2\\ F(3) = v_1^3 + 2v_1^2v_2 + v_1^2v_3 + v_1v_2^2 + v_1v_2v_3 $$

It gets rather unwieldy quite quickly. I honestly have trouble even figuring out where to start.

Answer 1

The closed form is given by $$F_n = \sum_{\sum p_i = n } a(p_1, p_2, \ldots , p_n) v_1^{p_1} v_2 ^{p_2} \ldots v_n^{p_n}$$ $$a(p_1, p_2, \ldots , p_n) = \prod_{i=1}^n { {n-i+1 - \sum_{j=i+1}^{n} p_i} \choose { p_i }}. $$

As to why that works, we look at how many ways we can pick the $v_n$, then the $v_{n-1}$, then the $v_{n-2}$, so on and so forth.

It's easier to explain with reference to an explicit calculation:
Suppose we wanted to find the coefficient $a(2, 0, 2, 1, 0)$ of $a^2 c^2d$ in $F(5) = (a)(a+b)(a+b+c)(a+b+c+d)(a+b+c+d+e)$.

There is 1 way of choosing 0 $e$'s.: There are 1 terms with $e$ to choose from -> ${ 1 \choose 0 }$ ways.
After that, there are 2 ways to choose 1 $d$'s: There are 2 terms with $d$ to choose from, and exactly 0 of them have been used up by the $e$ -> ${ 2 - 0 \choose 1 }$ ways.
After that, there are 1 ways to choose 2 $c$'s: There are 3 terms with $c$ to choose from, and exactly 0+1 of them have been used by the $e, d$ -> ${ 3 - 0-1 \choose 2}$ ways.
After that, there are 1 ways to choose 0 $b$'s: There are 4 terms with $b$ to choose from, and exactly 0+1+2 of them have been used up by the $e, d, c$ -> ${ 4 - 0-1-2 \choose 0}$ ways.
After that, there are 1 ways to choose 2 $a$'s: There are 5 terms with $a$ to choose from, and exactly 0+1+2+0 of them have been used up by the $e, d, c, b$ -> ${ 5 - 0-1-2-0 \choose 2}$ ways.

Note:

• We're going with the convention that
  • ${ 0 \choose n } = 1$ for non-negative integers $n$ and
  • $ { k \choose n } = 0 $ for negative integers $k$ and non-negative integers $n$.

Answer 2

Hint: The sequence of coefficients in \begin{align*} &v_1=\color{blue}{1}\,v_1\\ &v_1(v_1+v_2)=\color{blue}{1}\,v_1^2+\color{blue}{1}\,v_1v_2\\ &v_1(v_1+v_2)(v_1+v_2+v_3) =\color{blue}{1}\,v_1^3+\color{blue}{2}v_1^2v_2+\color{blue}{1}\,v_1^2v_3+\color{blue}{1}\,v_1v_2^2 +\color{blue}{1}\,v_1v_2v_3\\ &v_1(v_1+v_2)(v_1+v_2+v_3)(v_1+v_2+v_3v_4)\\ &\qquad=\color{blue}{1}\,v_1^4+\color{blue}{3}v_1^3v_2+\color{blue}{2}v_1^3v_3+\color{blue}{1}\,v_1^3v_4+\color{blue}{3}v_1^2v_2^2+\color{blue}{4}v_1^2v_2v_3+\color{blue}{2}v_1^2v_2v_4+\color{blue}{1}\,v_1^2v_3^2\\ &\qquad\quad+\color{blue}{1}\,v_1^2v_3v_4+\color{blue}{1}\,v_1v_2^3+\color{blue}{2}v_1v_2^2v_3+\color{blue}{1}\,v_1v_2v_3^2+\color{blue}{1}\,v_1v_2^2v_4+\color{blue}{1}\,v_1v_2v_3v_4\\ &\cdots \end{align*} is stored as A347917 in OEIS. It starts with \begin{align*} &\color{blue}{1,}\\ &\color{blue}{1, 1,}\\ &\color{blue}{1, 2, 1, 1, 1,}\\ &\color{blue}{1, 3, 2, 1, 3, 4, 2, 1, 1,1,2,1,1,1}\\ &\color{blue}{\ldots} \end{align*}

A cited paper On the coefficients of the distinct monomials in the expansion of $x_1(x_1+x_2)\cdots(x_1+x_2+\cdots +x_n)$ by S. Fried states the formula of coefficients for the terms $v_1^{a_1}v_2^{a_2}\cdots v_n^{a_n}$ as \begin{align*} \prod_{k=1}^{n-1}\binom{n-k+1-\sum_{i=k+1}^n a_i}{a_{k}} \end{align*} in accordance with @CalvinLin's result.