The standard model of PA has order type $\omega$. By compactness PA has a model of order type $\omega+n$ for any $n$, since every finite subset of the following set of statements is provable:
"$\exists$ $x_1, ...,x_n$ $>1$ s.t. $x_1>...>x_n$"
"$\exists$ $x_1, ...,x_n$ $>2$ s.t. $x_1>...>x_n$"
$\vdots$
"$\exists$ $x_1, ...,x_n$ $>m$ s.t. $x_1>...>x_n$"
$\vdots$
You can get $\omega * 2$ by taking, for all $m \in \mathbb{N} $:
- "$\exists$ $x_1, ..., x_m$ $>m$ s.t. $x_1>...>x_m$.
To get $\omega * n$, take for all $m \in \mathbb{N} $:
- "$\exists$ $x^1_1,...,x^1_m, ...,x^{n-1}_1, ..., x^{n-1}_m$ $>m$ s.t. $x^1_1>...>x^1_m>...>x^{n-1}_1>...>x^{n-1}_m$"
Etc. This leads me to the question:
True or false: PA has a model of order type $\alpha$ for every ordinal $\alpha$.
You can prove that every element except $0$ has an immediate predecessor, so you cannot get any of the order types $\omega+n$ for $n>0$ or $\omega\cdot 2$. Every model has an order type of the form $\omega+(\omega^*+\omega)\cdot\eta$ for some dense linear order type $\eta$ without endpoints;. In particular, if the model is countable, $\eta$ is the order type of the rationals.
Added: Each of the $\omega^*+\omega$ blocks is a $\Bbb Z$-chain. To see that $\eta$ must be densely ordered, suppose that $x$ and $y$ are from different $\Bbb Z$-chains. In $\mathsf{PA}$ you can define $\left\lfloor\frac{x+y}2\right\rfloor$, and it cannot be in the same $\Bbb Z$-chain as either $x$ or $y$. Similarly, if $\left\lfloor\frac{x}2\right\rfloor$ and $2x$ much be in different $\Bbb Z$-chains from $x$, so there is neither a first nor a last $\Bbb Z$-chain.