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Search: id:A136541
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| A136541 |
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Numbers n such that sum of the proper divisors of n is equal to (3/4)*phi(n). |
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1
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OFFSET
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1,1
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COMMENT
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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?
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FORMULA
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For n=1,2,3 & 4 a(n)=3^(2n-1)*(4*3^(2n-1)-1).
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EXAMPLE
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sigma(33)-33=48-33=15=(3/4)*20=(3/4)*phi(33).
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MATHEMATICA
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Do[If[DivisorSigma[1, n]-n==3/4*EulerPhi@n, Print[n]], {n, 2000000000}]
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CROSSREFS
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Cf. A005540, A076373.
Sequence in context: A099370 A118641 A111922 this_sequence A114071 A057981 A099828
Adjacent sequences: A136538 A136539 A136540 this_sequence A136542 A136543 A136544
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KEYWORD
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more,nonn
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AUTHOR
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Farideh Firoozbakht (mymontain(AT)yahoo.com), Jan 08 2008
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