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Search: id:A091285
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| A091285 |
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Numbers n such that sigma[3,n] is divisible by square of phi[n]. |
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+0
8
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OFFSET
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1,2
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COMMENT
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First 8 terms are solutions to relations for all j as follows: {sigma[6j+3,x]/phi[x]^2 is integer} for j=1,...300. Full proof is possible in knowledge of divisors corresponding sigma[k,x] and phi[x.]
No more terms through 10^9. - Ryan Propper (rpropper(AT)cs.stanford.edu), Jan 18 2008
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EXAMPLE
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n=14: phi[n]^2=36,sigma[3,n]=3096=36.86
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MATHEMATICA
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Empirical test for very high powers of divisors is: t = {1, 2, 3, 6, 14, 42, 3810, 13243560} Table[{6*j+3, Union[Table[IntegerQ[DivisorSigma[6*j + 3, Part[t, k]]/EulerPhi[Part[t, k]]^2], {k, 1, 8}]]}, {j, 1, 300}]; output={exponent, True}.
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PROGRAM
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(PARI) for(n = 1, 10^9, if(sigma(n, 3) % (eulerphi(n)^2) == 0, print1(n, ", "))) - Ryan Propper (rpropper(AT)cs.stanford.edu), Jan 18 2008
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CROSSREFS
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Cf. A000010, A001158, A015773, A015779.
Sequence in context: A098641 A056569 A094468 this_sequence A109459 A118986 A091138
Adjacent sequences: A091282 A091283 A091284 this_sequence A091286 A091287 A091288
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KEYWORD
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more,nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu), Feb 03 2004
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