How can I show that $$\prod_{k \ge 1}(1+z^{2k}) = \prod_{k \ge 1}(1+z^k+z^{2k}+z^{3k}) \quad ?$$

I have worked on this for a while and I am even doubting that maybe both are not equal

  • $\begingroup$ Compare the coefficients of $z$, or for that matter any $z^{2n+1}$, on both sides... $\endgroup$ – Sam Streeter Oct 24 '18 at 20:10
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    $\begingroup$ If we divide both sides by $\prod_{k\geq 1}(1+z^{2k})$ we get $$ 1 = \prod_{k\geq 1}(1+z^k)$$ which is certainly not true. $\endgroup$ – Jack D'Aurizio Oct 24 '18 at 20:14
  • $\begingroup$ @JackD'Aurizio . I got The LHS as the generating function for the set of partions of $n$ which every even part appear at most once and the RHS as the generating function for the set of partitions of $n$ in which every part appears at most three times. Did I interpret this wrong? My goal is to show that both are equal. $\endgroup$ – Jaynot Oct 24 '18 at 20:18
  • $\begingroup$ @Jaynot: I just added an answer addressing your original problem. $\endgroup$ – Jack D'Aurizio Oct 24 '18 at 20:37
  • $\begingroup$ @JackD'Aurizio Thank you $\endgroup$ – Jaynot Oct 24 '18 at 20:44

The generating function for the partitions in which each part appears at most three times is $\prod_{k\geq 1}(1+x^k+x^{2k}+x^{3k})$, correct, but the generating function for the partitions in which every even part appears at most once is $$ \prod_{k\geq 1}(1+x^{2k})\prod_{k\geq 1}\frac{1}{1-x^{2k-1}} $$ since the odd parts may appear as many times as they want. Then we have to check that

$$\prod_{k\geq 1}\frac{1+x^{2k}}{1-x^{2k-1}}=\prod_{k\geq 1}\frac{1-x^{4k}}{1-x^k} $$ which is equivalent to $$ \prod_{k\geq 1}\frac{1-x^k}{1-x^{2k-1}}=\prod_{k\geq 1}\frac{1-x^{4k}}{1+x^{2k}} $$ or to $$ \prod_{k\geq 1}\frac{1-x^k}{1-x^{2k-1}}=\prod_{k\geq 1}(1-x^{2k}) $$ which is trivial since $$ \prod_{k\geq 1}(1-x^k) = \prod_{k\geq 1}(1-x^{2k})(1-x^{2k-1})$$ (the $k$ on the left is either odd or even).


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