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OEIS is the online encyclopedia of integer sequences, Here is the link to the sequence $A002326$: https://oeis.org/A002326

For $n\geq 0$, the $n$th term in the sequence is defined as: $a(n)$ equals the smallest natural number $m$ such that $(2n+1)\mid (2^m-1)$.

There is a conjecture that was posed by Thomas Ordowski at the page of the mentioned sequence, and I posed a generalization of his conjecture. The non-generalized conjecture states that if $p$ is an odd prime; then $a((p^3-1)/2)=p\cdot a((p^2-1)/2)$.

The generalized conjecture states that if $p$ is an odd prime and $k$ is a natural number larger than $1$; then $a((p^{k+1}-1)/2)=p\cdot a((p^k-1)/2)$.

Computer testing of the generalized conjecture shows that there is no counterexample for $k$ and $p$ both up to $1000$. My question is to find a proof or a disproof to the generalized conjecture, and if that was not possible then at least finding a proof or a disproof to the non-generalized conjecture. Thank you.

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    $\begingroup$ Your conjecture is equivalent to non-existence of order-$(k+1)$ Wieferich primes (satisfying $p^{k+1} \mid 2^{p-1}-1$). As we do not even know if order-2 Wieferich primes exist besides the known two, there is little hope to prove your or Ordowski's conjectures. On the other hand, if $p$ is not a Wieferich prime, then both statements are more or less trivial. $\endgroup$
    – Max Alekseyev
    Commented Oct 24, 2020 at 14:12
  • $\begingroup$ @Max Alekseyev I must say that Michel Marcus found some counterexamples if we assume p is not prime. you can see the history. $\endgroup$
    – Ahmad Jamil Ahmad Masad
    Commented Oct 24, 2020 at 14:16
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    $\begingroup$ If $p$ is not prime, then anything can happen :) $\endgroup$
    – Max Alekseyev
    Commented Oct 24, 2020 at 14:21

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