%I #25 Dec 12 2021 19:39:57
%S 1,2,3,6,249,498,118578,99295058,297885174,4005374907
%N Numbers k such that phi(k) divides sigma_12(k).
%C sigma_12(n) = A013960(n) is the sum of the 12th powers of the divisors of n.
%C sigma(24j+12,x)/phi(x) is an integer for j in the range 0, ..., 500 for x = 1, 2, 3, 6, 249, 498, 118578 and supposed to hold for possible larger terms of A015770 and all j. Compare with comments to A015759, A091285, A015762. - _Labos Elemer_, May 27 2004
%C a(11) > 5*10^9. - _Giovanni Resta_, Aug 22 2017
%C All known terms of A015762 (and also of this sequence) are squarefree. In that case, sigma_12(x)/sigma_4(x) = Product_{primes p|x} (p^8 - p^4 + 1) is an integer, so x is also in this sequence. - _M. F. Hasler_, Aug 22 2017
%t Select[Range[1200000],Divisible[DivisorSigma[12,#],EulerPhi[#]]&] (* _Harvey P. Dale_, Dec 04 2015 *)
%Y Cf. A020492, A015759, A015761, A015762, A015763, A015764, A015765, A015766, A015767, A015768, A015769, A015771, A015773, A015774, A094470.
%K nonn,more
%O 1,2
%A _Robert G. Wilson v_
%E Corrected by _Harvey P. Dale_, Dec 04 2015
%E Offset corrected by and a(8)-a(10) from _Giovanni Resta_, Aug 22 2017