OFFSET
1,1
COMMENTS
This sequence was inspired by A269454 submitted by Marina Ibrishimova.
It seems that if for an odd prime p > 3 the order(2, p*3) < phi(p*3)/2 = p-1 then p is in this sequence.
Note that 2^(phi(p*q)/2) == 1 (mod p*q) for distinct odd primes p and q, due to Nagell's corollary on Theorem 64, p. 106. The products of distinct primes considered in the present sequence have order of 2 modulo p*q smaller than phi(p*q)/2.
Up to and including prime(100) = 541 the only odd primes p such that for all odd primes q smaller than p the order of 2 modulo p*q equals phi(p*q)/2 are 5, 7, and 11.
Complement of A216371 = A001122 U A105874 in the set of odd primes. Composed of the primes modulo which neither 2 nor -2 is a primitive root. Also, prime(n) is a term iff A376010(n) > 2. - Max Alekseyev, Sep 05 2024
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..1000
Wolfdieter Lang, On the Equivalence of Three Complete Cyclic Systems of Integers, arXiv:2008.04300 [math.NT], 2020.
EXAMPLE
n=1: Order(2, 17*3) = 8, and 8 is a proper divisor of phi(17*3)/2 = 16;
order(2, 17*5) = 8, and 8 is a proper divisor of phi(17*5)/2 = 32;
order(2, 17*7) = 24, and 24 is a proper divisor of phi(17*7)/2 = 48;
order(2, 17*11) = 40, and 40 is a proper divisor of phi(17*11)/2 = 80;
order(2, 17*13) = 24, and 24 is a proper divisor of phi(17*13)/2 = 96.
MATHEMATICA
Select[Prime@ Range[3, 157], Function[p, AllTrue[Prime@ Range[2, PrimePi@ p - 1], Function[q, With[{e = EulerPhi[p q]/2}, And[Divisible[e, #], # != e]] &@ MultiplicativeOrder[2, p q]]]]] (* Michael De Vlieger, Apr 01 2016, Version 10 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Wolfdieter Lang, Apr 01 2016
EXTENSIONS
More terms from Michael De Vlieger, Apr 01 2016
STATUS
approved