A136541 Numbers n such that sum of the proper divisors of n is equal to (3/4)*phi(n).

33, 2889, 235953, 19129689

Offset: 1

Author

Farideh Firoozbakht, Jan 08 2008

Keywords

• more
• nonn

Comments

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?

a(5) > 10^12. - Giovanni Resta, Nov 03 2012

Examples

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

Crossrefs

Cf. A005540, A076373.

Programs

Mathematica

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

Formula

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