

A091285


Numbers k such that sigma_3(k) is divisible by the square of phi(k).


7



1, 2, 3, 6, 14, 42, 3810, 318990, 13243560, 1108809240, 1719507048, 25330080090, 271984504290
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OFFSET

1,2


COMMENTS

The first 8 terms are solutions to: {sigma_{6j+3}(x)/phi(x)^2 is integer, for j=1,...,300}. A proof is possible with knowledge of respective divisors of sigma_k(x) and phi(x).


LINKS



EXAMPLE

k = 14: phi(k)^2 = 36, sigma_3(k) = 3096 = 36*86.


MATHEMATICA

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}.


PROG

(PARI) for(n = 1, 10^9, if(sigma(n, 3) % (eulerphi(n)^2) == 0, print1(n, ", "))) \\ Ryan Propper, Jan 18 2008


CROSSREFS



KEYWORD

more,nonn


AUTHOR



EXTENSIONS



STATUS

approved



