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Number of divisors d of n such that A323243(d) == 2 (mod 3).
6

%I #6 Mar 16 2019 21:46:32

%S 0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,2,0,1,

%T 0,1,0,0,0,3,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,3,0,0,0,3,0,0,0,1,

%U 0,0,0,2,0,0,0,0,0,0,0,4,0,1,0,1,0,0,0,2,0,2,0,1,0,0,0,2,0,0,0,1,0,1,0,1,0

%N Number of divisors d of n such that A323243(d) == 2 (mod 3).

%H Antti Karttunen, <a href="/A324832/b324832.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) + A324831(n)).

%F a(p) = 0 for all primes p.

%o (PARI) A324832(n) = sumdiv(n,d,(2==(A323243(d))%3));

%Y Cf. A000005, A000203, A156552, A323243, A324830, A324831.

%K nonn

%O 1,32

%A _Antti Karttunen_, Mar 16 2019