I am reading this Kolberg's article, where he proofs that the partition function takes both even an odd values infinitely often.


Although I'm sure it's simple (the proof is very short), I can't understand how he gets the contradiction, assuming the partition funtion takes odd (or even, in the other case) values for all $n\geq a$

Can anyone help me with this?

Thanks a lot.

  • 1
    $\begingroup$ What part of the contradiction is giving you problems? $\endgroup$ – Tobias Kildetoft Oct 15 '13 at 9:22

By assumption, say, we have $p(n)\equiv 0 \mod 2$ for all $n\ge a$, but $p(0)=1$ is not congruent $0$ modulo $2$. This is a contradiction to Euler's identity $$ p(a(3a-1)/2)\pm \cdots \pm p(2a-1) \pm p(0)=0, $$ because all terms except the last one are congruent $0$ modulo $2$. (For $p(0)=1$ see the first line of Kolberg's paper.)

  • $\begingroup$ Thanks a lot. Now it's clear. I had problems because for some reason i was assuming $p(0)=0$ $\endgroup$ – Mark_Hoffman Oct 15 '13 at 10:46

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