A091285 Numbers k such that sigma_3(k) is divisible by the square of phi(k).

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

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

Table of n, a(n) for n=1..13.

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

Cf. A000010, A001158, A015773, A015774.

Sequence in context: A015892 A056569 A094468 * A109459 A118986 A091138

Adjacent sequences: A091282 A091283 A091284 * A091286 A091287 A091288

Keyword

more,nonn

Author

Labos Elemer, Feb 03 2004

Extensions

a(10)-a(13) from Giovanni Resta, Feb 06 2014

Edited by M. F. Hasler, Aug 22 2017

Status

approved