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 injective. 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 $m=1$.

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 $m=1$.

The generating function is $$ \sum_{n\geq 0} sq(n) z^n = \prod_{k\geq 1} (1+z^{k^2}). $$ Using complex integration you can use this to get an asymptotic formula for $sq(n)$. 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 $sq(n)\sim e^{n^{1/3}}$ times some minor terms, hence $sq$ is ultimately increasing at a pretty fast rate. In particular, $sq(n)$ is not injective. For $sq^{-1}$ you can either derive an asymptotic series or compute the proportion of all partitions not containing the summand $1^2$ to find that $sq$ is strictly increasing from some point onwards. Hence you will most likely obtain that $sq^{-1}(\{m\})$ is infinite if and only if $m=1$.

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 injective. 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 $m=1$.