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%I #16 Dec 27 2016 08:18:15
%S 2,3,4,5,7,8,9,11,13,17,19,23,25,27,29,31,36,37,41,43,47,49,53,59,61,
%T 67,71,73,79,83,89,97,101,103,107,109,113,121,125,127,131,137,139,149,
%U 151,157,163,167,169,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,289,293
%N Numbers n such that A001158(n) == 1 (mod n).
%C Essentially the same as A087797. - _Ilya Gutkovskiy_, Dec 26 2016
%e a(1) = 2 because sigma_3(2) = 1^3 + 2^3 = 9 and 9 == 1 (mod 2);
%e a(2) = 3 because sigma_3(3) = 1^3 + 3^3 = 28 and 28 == 1 (mod 3);
%e a(3) = 4 because sigma_3(4) = 1^3 + 2^3 + 4^3 = 73 and 73 == 1 (mod 4), etc.
%t Select[Range[300], Mod[DivisorSigma[3, #1], #1] == 1 & ]
%Y Cf. A000040, A000430, A001158, A001248, A030078, A087797.
%K nonn,easy
%O 1,1
%A _Ilya Gutkovskiy_, Oct 04 2016
%E Edited by _Ilya Gutkovskiy_, Dec 26 2016