%I #7 Mar 16 2019 21:46:24
%S 0,1,0,2,1,1,0,2,0,3,1,2,0,1,1,2,1,1,0,4,0,2,1,2,2,1,0,2,0,3,1,2,1,2,
%T 1,3,0,1,0,4,1,1,0,3,1,2,1,2,0,5,1,2,0,1,3,2,0,1,1,4,0,2,0,2,1,2,1,4,
%U 1,3,0,3,1,1,2,2,1,2,0,4,0,2,1,3,2,1,0,3,0,3,0,4,1,2,1,2,1,1,1,6,0,2,1,2,1
%N Number of divisors d of n such that A323243(d) == 1 (mod 3).
%H Antti Karttunen, <a href="/A324831/b324831.txt">Table of n, a(n) for n = 1..10000</a> (based on Hans Havermann's factorization of A156552)
%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>
%F a(n) = A000005(n) - (A324830(n) + A324832(n)).
%F For all n >= 1, a(A000040(n)) = A000035(n).
%o (PARI) A324831(n) = sumdiv(n,d,(1==(A323243(d))%3));
%Y Cf. A000005, A000203, A156552, A323243, A324830, A324832.
%K nonn
%O 1,4
%A _Antti Karttunen_, Mar 16 2019
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