TITLE: Is $L(\mathbb{Z}*\mathbb{Z}_{2})$ a free group factor?
QUESTION [8 upvotes]: This is a reference request for something that is likely to be well-known to operator algebraists. I will not, therefore, include the technical definition of free product of finite von Neumann algebras, but instead refer the reader to Ching - Free products of von Neumann algebras for the definition.
Theorem 3.5 of Dykema - Interpolated free group factors (letting $A=L(\mathbb{Z})$ and $B=\mathbb{C}$) gives that $M_{2}(L(\mathbb{\mathbb{Z}}))*L(\mathbb{Z}_{2})\cong M_{2}(L(\mathbb{F}_{3}))$. Is it known whether or not $L(\mathbb{Z}*\mathbb{Z}_{2})\cong L(\mathbb{Z})*L(\mathbb{Z}_{2})$ is a free group factor, or interpolated free group factor?
I am, of course, interested in related results like the one quoted above, as well, if the original question is still unknown. Please feel free to provide references as answers.

REPLY [9 votes]: This should just be a comment- but for some reason I couldn't add a comment.
It seems to me that using Corollary 5.3 of this paper by Dykema, we indeed get a positive answer to your question.
Corollary 5.3 states that $L(G \ast H) \cong L(F(2-|G|^{-1}-|H|^{-1}))$, if $G$ and $H$ are nontrivial amenable groups, with $|G|+|H| \geq 5$.( $\infty ^{-1}=0$).
So $L(\mathbb Z \ast \mathbb Z_2)= L(F(1.5))$ according to the above formula (provided I subtracted correctly).
EDIT: I also found Theorem 1.1 in this paper to be very interesting. It relates to reduced $C^{\ast}$-algebras of free product groups.