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a(n) is the binary length of A324398(n), where A324398(n) = A156552(n) AND (A323243(n) - A156552(n)).
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%I #9 Mar 27 2019 18:57:02

%S 0,0,0,1,0,1,0,1,3,0,0,1,0,1,4,4,0,1,0,1,5,1,0,1,0,1,4,1,0,1,0,1,0,1,

%T 5,4,0,1,7,1,0,1,0,1,3,1,0,1,0,0,2,1,0,1,6,1,9,1,0,1,0,1,3,6,0,1,0,1,

%U 0,1,0,5,0,1,5,1,6,1,0,1,4,1,0,1,8,1,11,1,0,6,7,1,0,1,9,5,0,0,7,5,0,1,0,1,6

%N a(n) is the binary length of A324398(n), where A324398(n) = A156552(n) AND (A323243(n) - A156552(n)).

%H Antti Karttunen, <a href="/A324874/b324874.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 If A324398(n) = 0, a(n) = 0, otherwise a(n) = A070939(A324398(n)) = 1 + A000523(A324398(n)).

%F a(n) = A324868(n) + A324881(n).

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

%o (PARI)

%o A156552(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ From A156552

%o A324398(n) = { my(k=A156552(n)); bitand(k,(A323243(n)-k)); }; \\ Needs also code from A323243.

%o A324874(n) = #binary(A324398(n));

%Y Cf. A000523, A070939, A156552, A323243, A324398, A324862, A324864, A324868, A324881.

%K nonn

%O 1,9

%A _Antti Karttunen_, Mar 27 2019