Let $ped(n)$ denote the number of partition of a positive integer $n$ wherein the even parts are distinct and the odd parts are unrestricted. I follow this paper:

Andrews, G. E., Hirschhorn, M. D., & Sellers, J. A. (2010). Arithmetic properties of partitions with even parts distinct. The Ramanujan Journal, 23(1-3), 169-181.

and found some confusions there. They are:

  1. $ped(n)$ is divisible by $6$ at least $\frac16$ of the time. What is the meaning of "at least..."? Is it such a probability or what?
  2. The proof by induction of third congruence in the Theorem 3.4. : $ped(3^{2\alpha}n+\frac{3^{2\alpha}-1}{8})q^n \equiv (-1)^{\alpha-1}\psi(-q)\varphi(-q^3) \pmod3$.

Anyone can help me to understanding these, especially about number 2? Many thanks in advanced.


1 Answer 1

  1. Their result about $ped(9k+7)$ and Corollary 3.6 mean they can guarantee $1/6$ of $ped(n)$ values are divisible by 6. There are certainly other values of $n$ for which $ped(n)$ is a multiple of 6, thus the "at least" statement. Compare Ramanujan's result that $p(5k+4) \equiv 0 \bmod 5$ for every $k$. That doesn't give every value for which $p(n) \equiv 0 \bmod 5$, e.g., $p(7) = 15$. So Ramanujan's result can be stated as "at least 1/5 of $p(n)$ values are multiples of 5."

  2. The $ped(9n+1)$ result is the $\alpha=1$ base case. A related paper from the same journal in the same year goes through a very similar induction argument in detail: Look at Hirschhorn and Sellers, Arithmetic Properties of Partitions with Odd Parts Distinct, Ramanujan Journal 22(3) (2010) 273-284.

Edit: More detail for #1. Knowing $ped(9k+7) \equiv 0 \bmod 6$ for every nonnegative $k$ establishes that $ped(n)$ is a multiple of 6 for at least 1/9 of all values of $n$. The $\alpha = 1$ cases of Corollary 3.6 say $ped(27k + 19)$ and $ped(81k + 64)$ are multiples of 6, giving another 1/27 and 1/81 of all values of $n$ (one should confirm that none of these sequences overlap). The $\alpha = 2$ cases of Corollary 3.6 say $ped(243k + 172)$ and $ped(729k + 577)$ are multiples of 6, giving another 1/243 and 1/729 of all values of $n$, etc. Thus the partial geometric series 1/9 + 1/27 + 1/81 + 1/243 + ...

  • $\begingroup$ Very nice explanation! But, I still confuse about the 1. What about the sentence "These sequences account for..."? $\endgroup$
    – math404
    Apr 27 at 5:44
  • $\begingroup$ I mean, where did the numbers $1/9+1/27+...=1/6$ came from? $\endgroup$
    – math404
    Apr 27 at 6:01

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