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A136541
Numbers n such that sum of the proper divisors of n is equal to (3/4)*phi(n).
0
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
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
Sequence in context: A263908 A358808 A111922 * A263105 A284072 A281444
KEYWORD
more,nonn
AUTHOR
Farideh Firoozbakht, Jan 08 2008
STATUS
approved