Let $\lambda=(\lambda\_1\geq\lambda\_2\geq\cdots\geq\lambda\_{\ell(\lambda)}>0)$ be an integer partition of $n\in\mathbb{N}$; i.e., $\lambda\_1+\cdots+\lambda\_{\ell(\lambda)}=n$.


One may now associate $f\_{\lambda}=\dim(\lambda)=\#SYT(\lambda)$ which is computed by $f\_{\lambda}=\frac{n!}{H\_{\lambda}}$ where $H\_{\lambda}=\prod\_{u\in\lambda}h\_u$ is the product of the hook-lengths $h\_u$ of cells $u$ in the Young diagram of $\lambda$. On the other hand, for the symmetric group $\frak{S}\_n$ of permutations on $n$ letters $\{1,2,\dots,n\}$, there is the cycle index formula $g\_{\lambda}=\frac{n!}{z\_{\lambda}}$ counting the numbers of permutations indexed by cycle-type $\lambda$. If $\lambda$ is expressed in frequency notation as $\lambda=1^{a\_1}2^{a\_2}\cdots n^{a\_n}$ then $z\_{\lambda}=1^{a\_1}2^{a\_2}\cdots n^{a\_n}a\_1!a\_2!\cdots a\_n!$ as a product.


Now, consider the two data of multisets (items may be repeated)
$$\mathcal{F}\_n=\{f\_{\lambda}: \lambda\vdash n\} \qquad \text{and} \qquad
\mathcal{G}\_n=\{g\_{\lambda}: \lambda\vdash n\}.$$


Observe $\#\mathcal{F}\_n=\#\mathcal{G}\_n=p(n)$, the number of partitions of $n$.


I would like to ask whether the following is true or not:



> 
> **QUESTION.** For any $f\_{\lambda}\in \mathcal{F}\_n$, there exists $g\_{\mu}\in \mathcal{G}\_n$ such that the fraction $\frac{g\_{\mu}}{f\_{\lambda}}=\frac{H\_{\lambda}}{z\_{\mu}}$ is actually an integer. We insist the map $\lambda\rightarrow\mu$ to be $1$-to-$1$.
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The fraction $\frac{g_{\mu}}{f_{\lambda}}$ is an integer
With computer search one finds that the premise is first violated at $n = 19$. The obstruction is as follows: consider $\mu\_1 = 1^{19}$, $\mu\_2 = 1^{17}2$, $\mu\_3 = 1^{16} 3$. We have $g\_{\mu\_1} = 1$, $g\_{\mu\_2} = {19 \choose 2} = 9 \cdot 19$, $g\_{\mu\_3} = 2{19 \choose 3} = 2 \cdot 3 \cdot 17 \cdot 19$. There are only two partitions $\lambda$ such that $f\_{\lambda}$ divides *any* of $g\_{\mu\_i}$, namely $f\_{1^{19}} = f\_{19} = 1$. I'm not currently able to present a short proof of the latter fact.


To provide some insight, here are all partitions of $19$ with $f\_{\lambda} \leq \max g\_{\mu\_i}$:


* $f\_{19} = f\_{1^{19}} = 1$,
* $f\_{1^{17}2} = f\_{1, 18} = 18 = 2 \cdot 3^2$,
* $f\_{1^{16}3} = f\_{1^2 17} = 153 = 3^2 \cdot 17$,
* $f\_{1^{15}2^2} = f\_{2, 17} = 152 = 2^3 \cdot 19$,
* $f\_{1^{15}4} = f\_{1^3 16} = 816 = 2^4 \cdot 3 \cdot 17$,
* $f\_{1^{14}, 2, 3} = f\_{1, 2, 16} = 1615 = 5 \cdot 17 \cdot 19$,
* $f\_{1^{13} 2^3} = f\_{3, 16} = 798 = 2 \cdot 3 \cdot 7 \cdot 19$.


Next bad value is $n = 25$, with the same $1^n$, $1^{n - 2} 2$, $1^{n - 3}3$ vs $1^n, n$ obstruction. $n = 31$ is the same.


