Let $F\_n$ be a free profinite group of finite rank $n$ and let $V\_k$ denote the intersection of all open subgroups of $F\_n$ of rank at most $k$ ($k \in \mathbb{N}$).


My questions are:



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> Can I explicitly compute the index of $V\_k$ in $F\_n$ for all $k \in \mathbb{N}$?
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> Or if not, maybe for some special $k$, e.g. $k = p$, $p$ prime?
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I came to that question when I looked up the proof that the automorphism group of a finitely generated free profinite group is profinite (see e.g. Proposition 4.4.3 in Profinite Groups by Ribes-Zalesskii).



Intersection growth of free profinite groups
Ok, I think I understand the question now. To clarify, the free profinite group of rank $n$ may be regarded the profinite completion $\hat{K}\_n$ of $K\_n=\mathbb{Z}^{\ast n}$ the free group of rank $n$. An open subgroup $H< F\_n$ must be finite index since $F\_n$ is compact, and the cosets of $H$ cover $F\_n$. Thus, $H$ induces a finite-index subgroup $J< K\_n$. Then $J$ is a free group of rank $$rank(J)=1+[K\_n:J](n-1)=1+[F\_n:H](n-1).$$ Since $J$ is finite-index in $K\_n$, $H=\hat{J}$, the profinite completion of $J$, and therefore $H$ is a free group of rank $rank(H)=k=1+[F\_n:H](n-1)$. Thus, your question is equivalent to asking for the index of the intersection of all subgroups of $K\_n$ of index $\leq \frac{k-1}{n-1}$. I'm not sure this has been worked out, except possibly for $k=2n-1$ corresponding to index $2$ subgroups, where of course the answer is index $2^n$. One may obtain a crude upper bound by counting homomorphisms of $K\_n$ to $S\_j$ (where $j=\frac{k-1}{n-1}$), which is at most $j!^n$. So one has a bound on the index of the intersection of stabilizers of all of these homomorphisms of $j^{j!^n}$ :). 


