A136541 - OEIS

%I #8 Mar 13 2015 22:47:13

%S 33,2889,235953,19129689

%N Numbers n such that sum of the proper divisors of n is equal to (3/4)*phi(n).

%C If m>0 and p=4*3^m-1 is prime(m is in the sequence A005540) then n=3^m*p is in the sequence. Because sigma(n)-n=(1/2)*(3^(m+1)-1) *4*3^m-3^m*(4*3^m-1)=3^m*(2*3^m-1)=(3/4)*(2*3^(m-1))*((4*3^m-1)-1) =(3/4)*phi(3^m)*phi(p)=(3/4)*phi(3^m*p)=(3/4)*phi(n). The first four terms of the sequence are of such form if the 5th term is also of such form then it is equal to 823564514029689. Next term is greater than 2*10^9. Is it true that all terms are of the mentioned form?

%C a(5) > 10^12. - _Giovanni Resta_, Nov 03 2012

%F For n=1,2,3 & 4 a(n)=3^(2n-1)*(4*3^(2n-1)-1).

%e sigma(33)-33=48-33=15=(3/4)*20=(3/4)*phi(33).

%t Do[If[DivisorSigma[1,n]-n==3/4*EulerPhi@n,Print[n]],{n,2000000000}]

%Y Cf. A005540, A076373.

%K more,nonn

%O 1,1

%A _Farideh Firoozbakht_, Jan 08 2008