OFFSET
1,1
COMMENTS
If n = P*q, where P is a multiple perfect number and q is prime so that gcd(P,q) = 1, then sigma(n) = kP(q+1) = kn + kP. Consequently sigma(n) == kP (mod n). Such values of n are regular solutions to this congruence. Here P = 6, and we disallow regular solutions and some additional solutions divisible by 6. Cf. A076496 and A234238.
a(6) > 10^11. - Donovan Johnson, Sep 20 2012
If p = 2^k - 13 > 3 is a prime number, then 2^(k-1)*p is a term. This happens for k = 5, 9, 13, 17, 57, 105, 137, 3217, ... (A096818). - Giovanni Resta, Apr 01 2014
a(6) > 10^24. - Max Alekseyev, Mar 16 2026
EXAMPLE
n = 33501184 = 4096*8179; sigma(n) = 2n + 12 = 67002380.
MATHEMATICA
Do[s=Mod[DivisorSigma[1, n], n]; If[IntegerQ[n/100000], Print[{n}]]; If[Equal[s, 12]&&!Equal[Mod[n, 6], 0], Print[n]], {n, 1, 100000000}]
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Labos Elemer, Jun 11 2003
EXTENSIONS
a(5) from Donovan Johnson, Sep 20 2012
Edited by Max Alekseyev, Jan 16 2026
STATUS
approved
