Consider a power series $p = \sum_{i=0}^\infty a_i z^i \in \mathbb{Q}[[z]]$, which corresponds to a rational function. (That means, that there exist $q,r \in \mathbb{Q}[z]$ such that $p$ is the unique solution to the equation $q \cdot x = r$ in $\mathbb{Q}[[z]]$.)
Now let's take a natural number $k \geq 2$ and define $\bar p := \sum_{i=0}^\infty a_{k \cdot i} z^i$.
I wonder if $\bar p$ is a rational function, too.
I am grateful for every support. I already tried to apply some rationality-criterions I know, but have not been successful so far. Perhaps anyone knows a criterion which suits to this situation, or a counterexample.
The answer is yes. $\bar{p}$ is rational: Let $\gamma=\exp(\frac{2\pi i}{k})$ and set $w_j=z \gamma^j$, $1\leq j\leq k$. Then $$ \bar{p}(z^k)=\frac{1}{k} \sum_{j=1}^{k} p(w_j) = \frac{1}{k} \frac{\sum_j r(w_j) \prod_{n\neq j} q(w_n)}{\prod_m q(w_m)}=\frac{R(w_1,\ldots,w_k)}{Q(w_1,\ldots,w_k)}$$ where $R$ and $Q$ are symmetric polynomials in the $k$ variables. Since the $w_j$'s are the solutions to $w^k-1=0$ (so only the $0$'th and the $k$'th coefficients are non-zero) the only non-zero symmetric polynomials are those with powers that are multiples of $k$ and the coefficients will be rational as well. Therefore, (setting $Z=z^k$): $$ \bar{p}(Z) = \frac{\tilde{R}(Z)}{\tilde{Q}(Z)}$$ for some $\tilde{Q},\tilde{R}\in {\Bbb Q}[Z]$