# Closed formula of $(1+x_1)(1+x_1+x_2)^2\dots(1+x_1+\dots+x_n)^n$

I am working on a larger problem which I have managed to reduce to a pretty neat generating function. The answer will be given by the coefficient of $$x_1x_2\dots x_n$$ in

$$F(x_1,\dots,x_n)=(1+x_1)(1+x_1+x_2)^2\dots (1+x_1+x_2+\dots+x_n)^n$$

1. I have been trying desperately to find a closed formula for this function, but without success. Am I missing something very obvious or is this as good as it gets?
2. Is there a nice expression for the coefficient (call it $$a_n$$) of $$x_1x_2,\dots,x_n$$? Here I might have made some progress. By writing each $$(1+x_1+\dots+x_i)^i$$ as a multinomial I have been able to come up with this expression (hopefully correct...): \begin{align} \sum_{k_1+\dots+k_n=n}\left[\binom{1}{k_1}\binom{2}{k_2}\dots \binom{n}{k_n}(2-k_1)(3-(k_1+k_2))\dots(n-(k_1+\dots+k_{n-1})) \right] \end{align} but I don't seem to be able to reduce it to something nicer.

Any hints would be very much appreciated!

• To save future readers some effort, I note that at present this sequence is not in OEIS. – Peter Taylor Dec 20 '18 at 13:02

To obtain the monomial $$x_1x_2\cdots x_n$$ from the product $$p = (1+x_1)(1+x_1+x_2)^2\cdots (1+x_1+\cdots+x_n)^n$$ you have to pick the $$x_n$$ from one of the $$(1+x_1+\cdots+x_n)$$-factors and there are $$n$$ factors to choose from. Now you are left with the coefficient of $$x_1x_2\cdots x_{n-1}$$ in $$(1+x_1)(1+x_1+x_2)^2\cdots (1+x_1+\cdots+x_n)^{n-1}$$ which is the same as the coefficient in $$(1+x_1)(1+x_1+x_2)^2\cdots (1+x_1+\cdots+x_{n-2})^{n-2}(1+x_1+\cdots+x_{n-1})^{2n-2}.$$ Now you have to pick the $$x_{n-1}$$ from one of the $$(1+x_1+\cdots+x_{n-1})$$-factors and there are $$2n-2$$ to pick from.

Continuing this line of thought you get that the coefficient of $$x_1x_2\cdots x_n$$ in $$p$$ is $$n(2n-2)(3n-5)(4n-9)...(nn-*).$$ Note that by construction the constants $$0,2,5,9,\dots$$ in this product are the sums $$0, 2, 2+3, 2+3+4, \dots$$ which are given by $$\sum_{i=2}^k i = \frac{k(k+1)}{2} - 1 = \frac{(k-1)(k+2)}{2}.$$ Hence, the desired coefficient is $$\prod_{k=1}^n \left(kn - \frac{(k-1)(k+2)}{2}\right).$$

• Aah, of course. I can't believe I couldn't figure this out myself. Thank you very much! – KurtKnödel Dec 20 '18 at 12:56

Write out all $$n(n+1)/2$$ factors in a row.
The $$x_n$$ comes from one of the final $$n$$ factors, the $$x_{n-1}$$ from one of the final $$n+(n-1)$$ factors but not the one used for $$x_n$$, and so on.
The product is
$$n(2n-2)(3n-5)(4n-9)(5n-14)...$$ But I can't help you with that.

• Thank you, it all makes sense now:) – KurtKnödel Dec 20 '18 at 12:56