cp's OEIS Frontend

Conjecture: a(n) is the smallest integer k > 1 such that 2^(k-1) == 1 (mod a(0)*...*a(n-1)*k), with a(0) = 1. - Thomas Ordowski, Mar 13 2019

Either a(n) > a(n-1), or a(n) = a(n-1) is a Wieferich prime (A001220). - Max Alekseyev, Sep 29 2024

For n > 0, a(n) is prime or pseudoprime (a Fermat pseudoprime to base 3).

Conjecture: a(n) is prime for every n > 0, namely a(n) is the smallest prime p > a(n-1) different from 3 such that 3^(p-1) == 1 (mod a(n-1)), with a(0) = 1.

Generally: for a fixed integer base b > 1, a(n) is the smallest k > a(n-1) such that b^(k-1) == 1 (mod a(n-1)*k), with a(0) = 1. For n > 0, a(n) is prime or pseudoprime (a Fermat pseudoprime to base b). If for a base b, a(n) is a prime for every n > 0, then a(n) is the smallest prime p > a(n-1) that does not divide b such that b^(p-1) == 1 (mod a(n-1)), with a(0) = 1. For any integer base b > 1, a(n) is prime for almost all n. Seems that at most finitely many terms are composite.

a(n) = smallest k > 2^n such that k == 1 (mod n) and 2^(k-1) == 1 (mod k), so a(n) is an odd prime or a pseudoprime (Fermat pseudoprime to base 2).

Conjecture: a(n) is composite if and only if n = 2^j and 2^(2^j) + 1 is composite (presumably for all j > 4).

Note that a(2^j) = 2^(2^j) + 1 = A000215(j), the Fermat numbers.

For n <> 2^j, a(n) is the smallest k = 2^n - (2^n mod n) + m*n + 1 for m > 0 such that 2^(k-1) == 1 (mod k).

The last definition, also without the condition n <> 2^j, probably gives only primes.