OFFSET
1,2
COMMENTS
Subsequence of A236693. Proof: if n is in this sequence, then 2^phi(n) - 1 is divisible by n and 2^sigma(n) - 1 is divisible by 2^phi(n) - 1. Therefore, 2^sigma(n) == 1 (mod n) and n is in A236693. - Jinyuan Wang, Mar 13 2020
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (calculated using data from Jud McCranie; terms 1..1000 from Donovan Johnson)
MATHEMATICA
Select[Range[1, 151001, 2], Divisible[DivisorSigma[1, #], EulerPhi[#]]&] (* Harvey P. Dale, Sep 16 2016 *)
PROG
(PARI) isok(m) = (m % 2) && !(sigma(m) % eulerphi(m)); \\ Michel Marcus, Mar 14 2020
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
EXTENSIONS
Offset corrected by Donovan Johnson, Jan 18 2012
STATUS
approved