OFFSET
1,1
COMMENTS
Is the number of solutions finite? Do solutions to n+k*phi(n)=sigma(n) exist for all values of k? For k=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 the number of solutions I know below 1000000 is {1, 7, 2, 2, 1, 5, 3, 3, 0, 1, 1}). Not more for larger k.
If 3*2^n-1 is prime for n>0, then 2^n(3*2^n-1) belongs to the sequence; therefore this sequence is infinite if the sequence of primes of the form 3*2^n-1 (A007505) is infinite. - Matthew Vandermast, Jul 31 2004
3796=4.13.73 and 60610624=64.199.4759 do not belong to the class of numbers mentioned above by Vandermast.
a(20) > 10^12. - Donovan Johnson, Feb 29 2012
a(20) > 10^13. - Giovanni Resta, Apr 24 2016
EXAMPLE
n=44, phi(n)=20, sigma(44)=1+2+4+11+22+44=84=44+2*20
MATHEMATICA
ta={{0}}; k=2; Do[g=n; If[Equal[n+k*EulerPhi[n], DivisorSigma[1, n]], ta=Append[ta, n]; Print[n]], {n, 1, 182000000}]; {ta, g}
PROG
(PARI) is(n)=2*eulerphi(n)==sigma(n)-n \\ Charles R Greathouse IV, Feb 19 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Oct 15 2002; more terms Aug 04 2004.
EXTENSIONS
a(10)-a(19) from Donovan Johnson, Feb 29 2012
STATUS
approved