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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

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

Now, I have two questions about this function:

  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!

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  • $\begingroup$ To save future readers some effort, I note that at present this sequence is not in OEIS. $\endgroup$ – Peter Taylor Dec 20 '18 at 13:02
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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 $$ \prod_{k=1}^n \left(kn - \frac{(k-1)(k+2)}{2}\right). $$

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

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

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