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%I #36 Feb 09 2021 01:57:08
%S 1,15,24,42,60,64,69,78,90,100,114,123,133,147,153,177,186,198,222,
%T 231,240,258,259,270,276,288,289,306,339,360,366,393,402,403,414,429,
%U 438,447,459,474,477,492,495,501,507,511,522,582,588,594,600
%N Numbers k such that sigma(k) == k (mod 9).
%C That is, numbers k that satisfy the following:
%C A010878(k) = A105852(k) or A010878(k) = A010878(A000203(k)).
%C A010888(k) = A190998(k) or A010888(k) = A010888(A000203(k)).
%H Ivan N. Ianakiev, <a href="/A240597/b240597.txt">Table of n, a(n) for n = 1..10000</a>
%F A010888(a(n)) = A010888(A000203(a(n))).
%F A010888(a(n)) = A190998(a(n)).
%e sigma(15) = 24. 24 == 15 (mod 9), therefore 15 is in the sequence.
%t Select[Range[1000],Mod[#,9]==Mod[DivisorSigma[1,#],9]&]
%Y Cf. A000203, A190998, A010878, A105852, A010888.
%K easy,nonn
%O 1,2
%A _Ivan N. Ianakiev_, Sep 13 2014