OFFSET
1,1
COMMENTS
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.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..500
EXAMPLE
MATHEMATICA
Select[Range[10^5], (e = Max @@ Last /@ FactorInteger[#]) > 1 && Divisible[# -e, CarmichaelLambda[#]] &] (* Amiram Eldar, Dec 05 2019 *)
PROG
(PARI) isok(n) = ! issquarefree(n) && (Mod(n, lcm(znstar(n)[2])) == vecmax(factor(n)[, 2])); \\ Michel Marcus, Dec 05 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Thomas Ordowski, Dec 05 2019
EXTENSIONS
More terms from Amiram Eldar, Dec 05 2019
STATUS
approved