OFFSET
1,2
COMMENTS
Converse does not necessarily hold: phi(k) may have prime factors not dividing k.
Numbers k for which phi(k)*lambda(k) == 0 (mod k), where lambda(k) = A002322(k) is the Carmichael function. - Michel Lagneau, Nov 18 2012
Pollack and Pomerance call these numbers "phi-abundant numbers". Numbers k such that rad(k) | phi(k), where rad(k) is the squarefree kernel of k (A007947). - Amiram Eldar, Jun 02 2020
If p is the largest prime divisor of a term k, then p^2 divides k. - Max Alekseyev, Aug 27 2024
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
Paul Pollack and Carl Pomerance, Prime-Perfect Numbers, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 12a, Paper A14, 2012.
EXAMPLE
98 = 2*7^2 and phi(98)=2*3*7 so if p divides 98 then p divides phi(98), hence 98 is in the sequence.
MATHEMATICA
Select[Range[700], And@@Divisible[EulerPhi[#], Transpose[FactorInteger[#]] [[1]]]&] (* Harvey P. Dale, Nov 02 2011 *)
PROG
(Magma) [n: n in [1..620] | IsZero(EulerPhi(n)^NumberOfDivisors(n) mod n)]; // Bruno Berselli, Jul 27 2012
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, Aug 27 2002
STATUS
approved