OFFSET
1,2
COMMENTS
The numbers 2^2^n-1 for n=0,1,...,5 are in the sequence because 2^2^n-1=(2^2^0+1)*(2^2^1+1)*(2^2^2+1)*...*(2^2^(n-1)+1); 2^2^k+1 for k=0,1,2,3 & 4 are primes (Fermat primes); sigma(2^k)=2^(k+1)-1 and phi is a multiplicative function. Hence if p is a known Fermat prime (p=2^2^n+1 for n=0,1,2,3 & 4) then p-2 is in the sequence, note that this is not true for unknown Fermat primes if they exist. - Farideh Firoozbakht, Aug 27 2004
LINKS
Graeme L. Cohen, On a conjecture of Makowski and Schinzel, Colloquium Mathematicae, Vol. 74, No. 1 (1997), pp. 1-8. See Notes p. 7.
FORMULA
sigma(A001229), sorted.
MATHEMATICA
Select[Range[10^6], DivisorSigma[1, EulerPhi[#]] == # &] (* Amiram Eldar, Dec 10 2020 *)
PROG
(PARI) is(n)=sigma(eulerphi(n))==n \\ Charles R Greathouse IV, Nov 27 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Wilson's search was complete only through a(19) = 50319360. Jud McCranie reports Jun 15 1998 that the terms through a(24) are certain.
a(26)-a(28) added. Verified sequence is complete through a(28) by Donovan Johnson, Jun 30 2012
STATUS
approved