OFFSET
1,2
COMMENTS
Comments from Farideh Firoozbakht, Dec 01 2005: (Start)
I. All numbers of the form 2^(4m-1)*5^n where m & n are natural numbers are in the sequence. Because if s=2^(4m-1)*5^n then phi(s)=2^(4m-2)*4*5^(n-1); sigma(s)=(2^(4m)-1)*(5^(n+1)-1)/4 so phi(s)*sigma(s)=6*((16^m-1)/15)*((5^(n+1)-1)/4)*(2^(4m-1)*5^n)= 6*((16^m-1)/15)*((5^(n+1)-1)/4)*s, note that (16^m-1)/15 and (5^(n+1)-1)/4 are integers, hence s divides phi(s)*sigma(s).
II. All numbers of the form 2^(2m-1)*3^n where m & n are natural numbers (A228104) are in the sequence. Because if s=2^(2m-1)*3^n then phi(s)=2^(2m-2)*2*3^(n-1); sigma(s)=(2^(2m)-1)*(3^(n+1)-1)/2 so phi(s)*sigma(s)=((3^(n+1)-1)/2)*((4^m-1)/3)*(2^(2m-1)*3^n) =((3^(n+1)-1)/2)*((4^m-1)/3)*s, note that ((3^(n+1)-1)/2 and (4^m-1)/3 are integers, hence s divides phi(s)*sigma(s).
LINKS
Donovan Johnson, Table of n, a(n) for n = 1..10000
Richard K. Guy, Divisors and desires, Amer. Math. Monthly, 104 (1997), 359-360.
MATHEMATICA
Select[Range[1770], IntegerQ[DivisorSigma[1, # ]*EulerPhi[ # ]/# ] &] (* Farideh Firoozbakht, Dec 01 2005 *)
PROG
(PARI) is(n)=sigma(n)*eulerphi(n)%n==0 \\ Charles R Greathouse IV, Nov 27 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Corrected and extended by David W. Wilson
STATUS
approved