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a(n) = A292945(A163511(n)).
8

%I #8 Sep 30 2017 16:07:43

%S 0,0,0,0,0,0,0,1,0,0,0,0,0,0,2,2,0,0,0,1,0,0,0,0,0,0,0,1,4,4,4,5,0,0,

%T 0,0,0,0,2,2,0,0,0,1,0,0,0,0,0,0,0,0,0,0,2,3,8,8,8,8,8,8,10,10,0,0,0,

%U 1,0,0,0,0,0,0,0,1,4,4,4,5,0,0,0,0,0,0,2,2,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0

%N a(n) = A292945(A163511(n)).

%C Because A292945(n) = a(A243071(n)), the sequence works as a "masking function" where the 1-bits in a(n) (always a subset of the 1-bits in binary expansion of n) indicate which numbers are of the form 6k+5 in binary tree A163511 (or its mirror image tree A005940) on that trajectory which leads from the root of the tree to the node containing A163511(n).

%C The AND - XOR formulas just restate the fact that J(-3|n) = J(-1|n)*J(3|n), as the Jacobi-symbol is multiplicative (also) with respect to its upper argument.

%H Antti Karttunen, <a href="/A292946/b292946.txt">Table of n, a(n) for n = 0..8191</a>

%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>

%F a(n) = A292945(A163511(n)).

%F a(n) = A292264(n) AND (A292256(n) XOR A292274(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987).

%F a(n) = A292264(n) AND (A292254(n) XOR A292271(n)). [See comments.]

%F For all n >= 0, A292942(n) + A292944(n) + a(n) = n.

%o (Scheme) (define (A292946 n) (A292945 (A163511 n)))

%Y Cf. A005940, A163511, A292945.

%Y Cf. also A292247, A292248, A292254, A292256, A292264, A292271, A292274, A292592, A292593, A292942, A292944 (for similarly constructed sequences).

%K nonn

%O 0,15

%A _Antti Karttunen_, Sep 28 2017