|
| |
|
|
A136541
|
|
Numbers n such that sum of the proper divisors of n is equal to (3/4)*phi(n).
|
|
0
| | |
|
|
|
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 5-th 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?
|
|
|
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: A099370 A118641 A111922 * A183551 A114071 A057981
Adjacent sequences: A136538 A136539 A136540 * A136542 A136543 A136544
|
|
|
KEYWORD
| more,nonn
|
|
|
AUTHOR
| Farideh Firoozbakht (mymontain(AT)yahoo.com), Jan 08 2008
|
| |
|
|
