= Solution
Use an <involution with exactly one fixed point>:
$$
\boxed{\phi=\forall x\,f(f(x))=x\ \land\ \exists z\bigl(f(z)=z\land\forall x\,(f(x)=x\to x=z)\bigr).}
$$
The first conjunct makes every orbit of the <function> $f$ have size one or two. The second says precisely one orbit has size one. A finite model therefore has $1+2m$ elements, an odd number.
Conversely, on any set of size $2m+1$, fix one element and partition the remaining $2m$ elements into pairs; let $f$ swap the two members of each pair. This defines a model of $\phi$. The case $m=0$ gives a one-element model. Thus the finite model sizes are \b[exactly the positive odd integers].
Back to article page
