A091285 - OEIS

%I #25 Jan 13 2025 12:20:16

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

%T 271984504290

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

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

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

%t 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}.

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

%Y Cf. A000010, A001158, A015773, A015774.

%K more,nonn

%O 1,2

%A _Labos Elemer_, Feb 03 2004

%E a(10)-a(13) from _Giovanni Resta_, Feb 06 2014

%E Edited by _M. F. Hasler_, Aug 22 2017