Free spectrum of a variety

Remarks

The values are all finite exactly when $\mathcal{V}$ is a locally finite variety. When $\mathcal{V}$ is generated by a finite algebra $A$, the free algebra $F_{\mathcal{V}}(n)$ embeds in $A^{A^n}$, so $f_{\mathcal{V}}(n) \le \abs{A}^{\abs{A}^n}$. Example: for Boolean algebras, $f(n) = 2^{2^n}$, since $F(n)$ is the algebra of all Boolean functions $\set{0, 1}^n \to \set{0, 1}$. Example: for semilattices, $f(n) = 2^n - 1$, since $F(n)$ is the set of nonempty subsets of the generators under union. Non-example: the variety of all groups has no finite spectrum value for $n \ge 1$, since the free group on one generator is $\Z$. Some authors start the spectrum at $n = 1$, or use the related function $\log_2 f_{\mathcal{V}}(n)$.