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  • Let $a(n)$ be A005225 i.e. number of permutations of length $n$ with equal cycles. Here $$ a(n)=n!\sum\limits_{d|n}\frac{1}{d!(\frac{n}{d})^d} $$
  • Let $$ R(n,q,z)=(q+1)R(n-1,q+1,z)+\sum\limits_{j=0}^{q}z^{q-j}[z^{q-j}]R(n-1,j,z), \\ R(0,q,z)=z^q $$

I conjecture that $$R(n,0,1)=a(n+1).$$

Here is the PARI/GP prog to check it numerically:

a(n)=n!*sumdiv(n, d, 1/(d!*(n/d)^d))
R_upto(n)=my(v1, v2, v3); v1=vector(n+1, i, z^(i-1)); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, (n-i), v2[q+1]=(q+1)*v1[q+2]+sum(j=0, q, z^(q-j)*polcoeff(v1[j+1], q-j))); v1=v2; v3[i+1]=v1[1]); v3=vector(n+1, i, subst(v3[i], z, 1))
test(n)=vector(n+1, i, a(i))==R_upto(n)

Is there a way to prove it?

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    $\begingroup$ So, for $k | n$, we have that $[z^{k-1}] R(n-1, 0, z)$ is the number of permutations of cycle type $(n/k, \dots, n/k)$. $\endgroup$
    – Martin Rubey
    Commented Aug 4, 2023 at 7:43

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