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Let $\mathbb{N}$ denote the positive integers and let $S = \{n^2: n\in \mathbb{N}\}$. For any positive integer $k$ we define $$\text{sq}(k) = |\{F\subseteq S: F\neq \emptyset, F\text{ is finite and } k = \sum_{n\in F} n\}|.$$

Questions:

  1. Is the function $\text{sq}:\mathbb{N}\to \mathbb{N}\cup\{0\}$ surjective?
  2. Is there $m_0>1$ such that $\text{sq}^{-1}(\{m\})$ is infinite for all $m\geq m_0$?
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  • $\begingroup$ This is A033461 in the OEIS. $\endgroup$ – Charles Sep 3 '15 at 17:56
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The generating function is $$ \sum_{n\geq 0} \text{sq}(n) z^n = \prod_{k\geq 1} (1+z^{k^2}). $$ Using complex integration you can use this to get an asymptotic formula for $\text{sq}(n)$. This involves quite some work, but the path is well described in Andrews, The theory of partitions, chapter 6. You will arrive at something like $\text{sq}(n)\sim e^{n^{1/3}}$ times some minor terms, hence $\text{sq}$ is ultimately increasing at a pretty fast rate. In particular, $\text{sq}(n)$ is not surjective. For $\text{sq}^{-1}$ you can either derive an asymptotic series or compute the proportion of all partitions not containing the summand $1^2$ to find that $\text{sq}$ is strictly increasing from some point onwards. Hence you will most likely obtain that $\text{sq}^{-1}(\{m\})$ is infinite if and only if $m=1$.

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