$101$ is the only prime in the sequence $1,101,10101,\ldots$ as shown in this Putnam question.

I also know from studying the Collatz conjecture that $101_2$ is also the only prime in the same sequence considered as a sequence of base $2$ numbers.

In what bases is this true, and is there a general proof for some class of bases?

  • 3
    $\begingroup$ There is currently no known general solution to your question. This is since it's solution requires knowledge of primes of the form $101_n=n^2+1$. If the bases can be written $n=2^{2^{k-1}}$, then $2^{2^k}+1$ must be prime. This is called a Fermat prime, and it is currently unknown for what of values of $k$ this number becomes prime, and also if there are finitely/infinitely many of them. See en.m.wikipedia.org/wiki/Fermat_number for more. $\endgroup$ – Alex S Apr 18 at 2:38

The proof in the Putnam question goes through regardless of the base. Replace $10$ by $b$, $9$ by $b-1$ and $11$ by $b+1$. It is true in all bases $b$ where $b^2+1$ is prime.


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