Let $n\geq 100$ be some positive integer in base $10$. Take its consecutive digits in groups of $k$ digits each. I have the following
Conjecture: If the digits of some positive integer $n\geq 100$ in base 10, taken consecutively in groups of $k$ digits each, form an increasing or decreasing arithmetic progression of three or more elements, then $n$ is neither a prime number nor a perfect power.
For instance, $125$ is a perfect power, but $1,2,5$ do not form an arithmetic progression. $1,2,3,4$ form an increasing arithmetic progression, but $1234$ is neither a prime nor a perfect power.
I have verified the conjecture for all three-digits numbers (which only admit groups of $1$ digit each) and all four-digits numbers (which only admit groups of $1$ digit each, as groups of $2$ digits can not form an arithmetic progression of three or more elements).
I must admit that I am a bit lost about (i) how to check efficiently for numbers with a bigger number of digits, and (ii) why the conjecture could be true, apart from the scarcity argument of the sets that need to intersect to have a counterexample.
Any help on checking and / or proving / disproving the conjecture, or understanding better how to approach the problem, would be welcomed.
Thanks!