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%I #27 Jun 07 2021 15:27:21
%S 14,15,23,30,33,35,42,46,47,51,54,55,56,60,62,66,69,70,71,77,78,87,92,
%T 94,95,102,105,110,114,115,119,120,123,126,132,135,138,140,141,142,
%U 143,154,155,158,159,161,165,167,168,174,177,182,184,186,187,188
%N Numbers k such that the sum of divisors of k is a multiple of 24.
%H Vincenzo Librandi, <a href="/A207036/b207036.txt">Table of n, a(n) for n = 1..10000</a>
%e 33 is in the sequence because the divisors of 33 are 1, 3, 11, 33 and the sum of divisors of 33 is 1+3+11+33 = A000203(33) = 48 and 48 is a multiple de 24 because 48 divided by 24 gives a positive integer.
%t Select[Range[188], Mod[DivisorSigma[1, #], 24] == 0 &] (* _T. D. Noe_, Feb 27 2012 *)
%Y Cf. A000203, A008606, A183010, A207374.
%K nonn
%O 1,1
%A _Omar E. Pol_, Feb 20 2012