OFFSET
1,1
COMMENTS
This sequence is infinite because for all natural numbers n, 2^n is in the sequence. 28, 3100 & 118458 are in the sequence but they aren't of the form 2^n. Are there numbers n such that sum of the proper divisors of n is equal to phi(sigma(n))?
FORMULA
a(5)=28; a(13)=3100; a(19)=118458 & If n<5 then a(n)=2^n, if 5<n<13 then a(n)=2^(n-1), if 13<n<19 then a(n)=2^(n-2) and if 19<n<30 then a(n)=2^(n-3).
EXAMPLE
m=2^n(n>0), phi(m)=2^(n-1) so sigma(phi(m))=2^n-1=(2^(n+1)-1)-2^n=
sigma(m)-m hence m is in the sequence.
MATHEMATICA
Do[If[m + DivisorSigma[1, EulerPhi[m]] == DivisorSigma[1, m], Print [m]], {m, 70000000}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht, Mar 31 2005
EXTENSIONS
a(30)-a(33) from Donovan Johnson, Jul 29 2009
a(34)-a(35) from Donovan Johnson, Jan 25 2013
STATUS
approved