cp's OEIS Frontend

The condition e(k) > 1 excludes primes and Carmichael numbers.

Numbers n such that e(k) > 1 and b^k == b^e(k) (mod k) for all b.

These are numbers k such that A276976(k) = e(k) > 1.

Are there infinitely many such numbers? Are all such numbers even?

A number k is a term if and only if k is e(k)-Knödel number with e(k) > 1. So they may have the name nonsquarefree e(k)-Knodel numbers k.

It seems that if k is in this sequence, then e(k) = A007814(k) and k/2^e(k) is squarefree.

Conjecture: there are no composite numbers m > 4 such that m == e(m) (mod phi(m)). By Lehmer's totient conjecture, there are no such squarefree numbers.

Problem: are there odd numbers n such that e(n) > 1 and n == e(n) (mod ord_{n}(2)), where ord_{n}(2) = A002326((n-1)/2)? These are odd numbers n such that 2^n == 2^e(n) (mod n) with e(n) > 1.

Numbers k for which A051903(k) > 1 and A219175(k) = A329885(k). - Antti Karttunen, Dec 11 2019