Is it possible that the number of $\mathcal{T}$-algebras is an arithmetic progression?

Question

For a single-sorted algebraic theory $\mathcal{T}$ denote by $t_n$ the number of $\mathcal{T}$-algebras with $n$ elements (up to isomorphism). Is there an example for $\mathcal{T}$ such that eventually $t_n$ is a non-constant arithmetic progression? By this I mean that there are $N,a,b \in \mathbb{N}$ with $a > 0$ such that $t_n = a \cdot n + b$ for all $n \geq N$. In particular, $t_n$ must be eventually strictly increasing. Is it maybe even possible to have $a=1$?

Some context. The sequence of the numbers of groups of order $n$ (up to isomorphism) is a quite complicated sequence, it is not increasing in particular. The sequence of the numbers of posets of order $n$ (up to isomorphism) is strictly increasing for $n \geq 1$, as can be checked easily by adjoining a smallest element to a given poset. There is also an algebraic theory with this property, namely $\mathbb{Z}$-sets, or equivalently, sets equipped with a bijection. Here $t_n$ is the number of conjugacy classes in $S_n$, which is equal to the number of integer partitions of $n$. This sequence is eventually strictly increasing, but of course not an arithmetic progression. I have excluded constant sequences ($a=0$) since otherwise sets would provide a boring answer.

I believe that such an algebraic theory must be very very nice, hence my curiosity if it exists. Maybe some mathematical logic will be helpful to answer this question.

Answer 1

Is it possible that the number of $\mathcal{T}$-algebras is an arithmetic progression?

Here is a near miss: Let $\mathcal{V}$ be the variety of $\mathbb Z_2$-sets. These may be thought of as algebras $\langle X; \alpha(x)\rangle$ with a single unary operation $\alpha(x)$ such that $\alpha^2(x)=x$. The number of isomorphism types of algebras of size $n$ in this variety is $t_n=\big\lfloor\frac{1}{2} n\big\rfloor+1$.

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Edit. I will leave the original answer above, but edit it to make it more complete.

This will be an example of a variety of algebras in which there are exactly $n$ isomorphism types of algebras of size $n$ for each $n$. The language is the language of one unary operation $\alpha(x)$ and one constant $c$. We axiomatize the variety with the single axiom $\alpha^2(x)=x$. Each algebra in the variety is a disjoint union of $1$-element orbits under $\alpha$ and $2$-element orbits under $\alpha$, and one of the orbits is distinguished by the fact that one of its elements is named $c$.

In the original answer, the constant was missing. To count the number of isomorphism types of structures of this type which have size $n$ it suffices to specify the number of $2$-element orbits in a model. This can be any number $k$ chosen from $k=0,1,\ldots,\lfloor \frac{1}{2}n\rfloor$, so there are $\lfloor \frac{1}{2}n\rfloor+1$ possible isomorphism types of constant-free structures of size $n$.

James Hanson suggested adding the constant $c$. This distinguishes one of the orbits, say $O_c$. If $O_c$ is a $1$-element orbit, then the isomorphism type of the model is determined by the isomorphism type of the remaining constant-free structure of size $n-1$, so there will be $\lfloor \frac{1}{2}(n-1)\rfloor+1$ of these algebras. If $O_c$ is a $2$-element orbit, then the isomorphism type of the model is determined by the isomorphism type of the remaining constant-free structure of size $n-2$, so there will be $\lfloor \frac{1}{2}(n-2)\rfloor+1$ of these algebras. Altogether, this yields $\lfloor \frac{1}{2}(n-1)\rfloor+1+\lfloor \frac{1}{2}(n-2)\rfloor+1 = n$ isomorphism types.