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In my recent work I have been led to consider the following type of permutation factorizations.

Let $\pi_1$ and $\pi_2$ be two fixed permutations, with disjoint support, i.e. $\pi_1$ acts on $\{1,...,n\}$ and $\pi_2$ acts on $\{n+1,...,n+m\}$.

Find all pairs $(\tau_1,\tau_2)$, such that $\pi_1\pi_2=\tau_1\tau_2$, with the condition that every cycle of the factors $\tau_1,\tau_2$ has exactly one element in $\{1,...,n\}$.

Notice that the last condition implies that $\ell(\tau_1)=\ell(\tau_2)=n$.

For example, $ (123)(45678)=(1 5)(2 4)(3 8 7 6)\cdot (1 4)(2 6 8)(3 5 7)$ is a case with $n=3$, $m=5$.

I just want to know if anyone around here has seen a similar problem before. If this is really a new problem, does anyone have a suggestion for naming these things?

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  • $\begingroup$ Not being a combinatorialist, consider the following as a crazy idea with a high probability of failure: wlog n less than m, in which case the first tau "pushes up" the integers 1 through n into different cycles, and the second " pushes them down" again, meaning each tau has at least n cycles. This should make it easy to count. For some reason the names zigzag permutation and juggling jump to mind. If Richard Stanley doesn't have a suggestion for you, you might try looking up zigzag and juggling in the combinatorics literature. Gerhard "Could Be Wrong, Of Course" Paseman, 2016.01.18. $\endgroup$ Jan 18, 2016 at 19:59
  • $\begingroup$ Actually each tau always has exactly $n$ cycles. I don't really think that makes it easy to count, though. $\endgroup$
    – Marcel
    Jan 18, 2016 at 22:17

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