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I am reading Richard Stanley's "Topics in Algebraic Combinatorics" and just before the notes for Chapter 8, he was discussing generating functions for plane partitions and solid partitions. It is claimed there that:

It is easy to see that for any integer sequence $a_0 = 1, a_1, a_2, \dots$, there are unique integers $b_1, b_2, \dots$ for which $$\sum\limits_{n\ge 0} a_n x^n = \prod\limits_{i\ge 1}(1-x^i)^{-b_i}$$

Not sure if I am missing something obvious, but this is certainly not "easy to see" for me. Any help will be appreciated.

suncup224
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    This is best understood by trying a few examples. Try it with the first terms of $$1 + 4x -12x^2 + 6x^3 + \dots.$$ (Not worrying about $\dots$) See if you can inductively find the $b_i$'s. What's $b_1$? Given $b_1$, what's $b_2$? etc. There's always an inductive process. And btw the $b_i$'s might have to be negative! – Dzoooks Jul 01 '18 at 04:02
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    Ok cool, I see it now. – suncup224 Jul 01 '18 at 04:34

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Expaning the factors on the right-hand side as power series shows that they affect the coefficients $a_n$ only for $n\ge i$. Thus the $b_i$ can be iteratively determined for $i=1,2,3,\ldots$ , with each $b_i$ chosen such $a_i$ comes out right, without messing up $a_n$ for $n\lt i$.

joriki
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