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For $d=1$, let $M_1 = 1$. For $d>1$, define $M_d$ recursively by $$M_d = d(d!) - \sum_{i=1}^{d-1} (d-i)! M_i.$$

Is $M_d$ bounded by a polynomial (of some high degree) in $d$?

Note that $M_d$ is the number of subgroups of index $d$ in the free group of rank $2$.

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up vote 2 down vote accepted

$M_d$ forms the sequence A003319.

Notice that the sequence $n_d = \frac{M_d}{d!} - d$ satisfies: $$ n_d = - \sum_{k=1}^{d-1} \frac{n_k}{\binom{d}{k}} - \sum_{k=1}^{d-1} \frac{k}{\binom{d}{k}} $$ The sum $\sum_{k=1}^{d-1} \frac{k}{\binom{d}{k}}$ approaches $1$ for large $d$.

Numerical experiments suggest that $n_d \to -1$ as $d \to + \infty$, therefore indicating that $M_d$ grows as $M_d \sim (d-1) (d!)$, that is faster than a polynomial.

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