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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 $t_n=\big\lfloor\frac{1}{2} n\big\rfloor+1$.

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 algebras of 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.

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. 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. Altogether, this yields $\lfloor \frac{1}{2}(n-1)\rfloor+1+\lfloor \frac{1}{2}(n-2)\rfloor+1 = n$ isomorphism types.

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 $t_n=\big\lfloor\frac{1}{2} n\big\rfloor+1$.

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.

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 $t_n=\big\lfloor\frac{1}{2} n\big\rfloor+1$.

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 $t_n=\big\lfloor\frac{1}{2} n\big\rfloor+1$.

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 algebras of 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.

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. 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. Altogether, this yields $\lfloor \frac{1}{2}(n-1)\rfloor+1+\lfloor \frac{1}{2}(n-2)\rfloor+1 = n$ isomorphism types.

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 are the 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 $t_n=\big\lfloor\frac{1}{2} n\big\rfloor+1$.

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 $t_n=\big\lfloor\frac{1}{2} n\big\rfloor+1$.