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

33, 2889, 235953, 19129689

Offset

1,1

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

Links

Table of n, a(n) for n=1..4.

Formula

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

Example

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

Mathematica

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

Crossrefs

Cf. A005540, A076373.

Sequence in context: A263908 A358808 A111922 * A263105 A284072 A281444

Adjacent sequences: A136538 A136539 A136540 * A136542 A136543 A136544

Keyword

more,nonn

Author

Farideh Firoozbakht, Jan 08 2008

Status

approved