File Edit Options Buffers Tools Help
File Edit Options Buffers Tools Help

Here is an idea for a proof, but I don't know if it can actually be carried out. Suppose that $\Sigma p^{a_i}x^i$ is rational. Then for $n$ sufficiently large, we have a linear recurrence $$ c_0 p^{a_n} + c_1 p^{a_{n-1}} +\cdots+c_k p^{a_{n-k}}=0, $$ where each $c_i$ is an integer. Write each $c_i$ in base $p$ and distribute, so now the terms have the form $d_{ij} p^{a_{n-i}+j}$, where $0\leq d_{ij}\leq p-1$. Move all the terms with minus signs to the other side. Compare the base $p$ expansions of both sides. This will involve some care since there could be "carrying," i.e., after collecting the powers of $p$ some coefficients could exceed $p-1$. But it seems as if the possibilities are limited and that a proof may be possible.

File Edit Options Buffers Tools Help
File Edit Options Buffers Tools Help
Here is an idea for a proof, but I don't know if it can actually be carried out. Suppose that $\Sigma p^{a_i}x^i$ is rational. Then for $n$ sufficiently large, we have a linear recurrence $$ c_0 p^{a_n} + c_1 p^{a_{n-1}} +\cdots+c_k p^{a_{n-k}}=0, $$ where each $c_i$ is an integer. Write each $c_i$ in base $p$ and distribute, so now the terms have the form $d_{ij} p^{a_{n-i}+j}$, where $0\leq d_{ij}\leq p-1$. Move all the terms with minus signs to the other side. Compare the base $p$ expansions of both sides. This will involve some care since there could be "carrying," i.e., after collecting the powers of $p$ some coefficients could exceed $p-1$. But it seems as if the possibilities are limited and that a proof may be possible.