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Let $f(n)=1+x^n+x^{2n}+...+x^{n^2}.$

Let $p(x)$ be $1+x+x^2+x^5+x^7+...$ where the exponents are the pentagonal numbers.

Let $a(n)$ be the sequence of integers such that the coefficients of the series $f(a(1)) f(a(2)) f(a(3))...$ are congruent mod $2$ to the coefficients of $p(x)$

The first few values of $a(n)$ are: $1,2,3,5,6,7,9,11,12,13,15,17,19,21,23,25,27$.

Describe the sequence a(1),a(2),a(3),...

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I asked George Andrews about this problem and this was part of his reply:

"Define $g_1(n)=2n-1$ and $g_m(n)=g_{m-1}(n)(g_{m-1}(n)+1)$

Thus

$g_1(n):1,3,5,7,9,11,13,15,...$

$g_2(n):2,12,30,56,90,132,182,...$

$g_3(n):6,156,930,3192,8190,17556,...$

$g_4(n):42,24492,865830,...$

I claim that the sequence for $a$ consists of all the values of $g_m(n)$ for $m\geq 1, n\geq 1$."

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    $\begingroup$ Claim = conjecture? Is "m-1" all in a subscript? The general consensus on MO is that answers (unlike comments) should not be telegraphic. $\endgroup$
    – darij grinberg
    Commented Dec 6, 2018 at 3:41
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    $\begingroup$ @darijgrinberg this is less or more straightforward: by Pentagonal theorem mod 2 we have $p(x)=\prod (1+x^n)=\prod 1/(1-x^{2n-1})$, and each multiple is represented as a telescopic product of ratios $(1-x^{m(m+1)})/(1-x^m)$ starting with $m=2n-1$ through the orbit of the map $x\to x(x+1)$ $\endgroup$
    – Fedor Petrov
    Commented Jan 5, 2019 at 7:08
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    $\begingroup$ @FedorPetrov: Nice argument (and not one I'd call straightforward)! $\endgroup$
    – darij grinberg
    Commented Jan 5, 2019 at 13:38

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