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Let $\pi$ be a permutation in $S_n$ with cycle type $\lambda$.

How many factorizations into two factors $\pi=\sigma_1\sigma_2$ are there, such that the subgroup $\langle \sigma_1,\sigma_2\rangle$ generated by $\sigma_1$ and $\sigma_2$ has $k$ orbits?

(I impose no condition at all on the factors.)

For $k=1$, I am asking for the total number of transitive factorizations of $\pi$ into two factors. At least this case should be known, I hope.

I present the numerical results, as example, for $n=5$:

$$ \left[ \begin {array}{ccccc} 24&50&35&10&1\\ 48&52& 18&2&0\\ 80&36&4&0&0\\ 72&42&6&0&0 \\ 108&12&0&0&0\\ 96&24&0&0&0 \\ 120&0&0&0&0\end {array} \right] $$

Columns are labelled by number of orbits, $k$, increasing; rows are labelled by cycle type $\lambda$, starting from $1^n$. So a permutation with cycletype $(2,2,1)$ has 80 transitive factorizations and 36 factorizations with 2 orbits.

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  • $\begingroup$ This is a kind of Hurwitz number. The most typical Hurwitz numbers count the number of factorizations of $\lambda$ as a product of a specified number of transpositions (the parity of the number is determined by the sign of $\lambda$). Those numbers are already very complicated, cf. the ELSV formula. I doubt the numbers you mention are any simpler (perhaps you came at this from the problem of enumerating Belyi curves). $\endgroup$ – Jason Starr Mar 15 '18 at 13:21
  • $\begingroup$ @JasonStarr Papers I have seen on related problems usually deal with refined counting where the cycle types of the factors are specified. I was hoping that my question would be simpler. $\endgroup$ – Marcel Mar 15 '18 at 13:36
  • $\begingroup$ Well, maybe my instinct is wrong. $\endgroup$ – Jason Starr Mar 15 '18 at 13:52

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