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How can we calculate the number of subgroups of odd index in the free product of a group of order $3$ with itself?

Is there any algorithm to find this out in the general case? I know that it is given by $s_n$, where $a^2_0s_n+a^2_{1}s_{n-1}+\cdots+(n-1)!a^2_{n-1}s_1=n(n!)a^2_n$ with $a_n$ defined recursively by $na_n=a_{n-1}+a_{n-3}, n\ge3$ and $a_0=1,a_1=1$ and $a_2=\frac1{2}$.

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    Do you know if there are only finitely many? (This is not obvious, at least not to me! I know that there are only finitely many of index $3^n$ though...) Do you know about the Kurosh subgroup theorem? – 2017-02-21

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