Suppose $F_n$ is a free group of rank $n$. It is a rather well known fact, that $b_4(n) = [F_n : V_{\{x^4\}}(F_n)]$ is finite for all $n \in \mathbb{N}$. Is there a some sort of formula for $b_4(n)$? Here $V_Q$ is the verbal subgroup for the collection of group words $Q$.
I know the solutions of the similar problems for $b_2(n) = [F_n : V_{\{x^2\}}(F_n)]$ and $b_3(n) = [F_n : V_{\{x^3\}}(F_n)]$. They are:
$$b_2(n) = 2^n$$
$$b_3(n) = 3^{n + C_n^2 + C_n^3}$$
The first equality is the consequence of all groups of exponent $2$ being elementary abelian (which is itself a consequence of the First Pruefer Theorem). The second equality was proved by Bartel van der Waerden and Friedrich Wilhelm Levi in "Über eine besonderen Klasse von Gruppen" in 1933.
I also know, that $$b_4(n) = [F_n : V_{\{x^4\}}(F_n)] \geq [F_n : V_{\{x^4, [x,y]\}}(F_n)] = 4^n$$
Here my current knowledge of the subject ends.