# The fraction $\frac{g_{\mu}}{f_{\lambda}}$ is an integer

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-$$1$$.

• Did you check this for small values of $n$? Commented Apr 27, 2022 at 16:14
• Yes, I did check for a few values. Commented Apr 27, 2022 at 16:30
• I am not sure whether I understand the question. Are you asking for a bijective map $\phi$ such that $\frac{g_{\phi(\lambda)}}{f_\lambda}$ is integral? Commented Apr 27, 2022 at 16:34
• Essentially, yes. I just wish to avoid and answer like: there is one $\mu$ that works for each $\lambda$. Something trivial like that. Of course, the bijection is not unique. Commented Apr 27, 2022 at 16:36
• Going through all bijections $\phi$ satisfying your constraint up to size 4 (there are 80 such maps, if I made no mistake), none of them is a composition of at most 9 maps known to findstat. This makes a constructive answer to your question quite interesting to me! Commented Apr 27, 2022 at 17:22

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.