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a(n) = A292253(A163511(n)).
8

%I #9 Sep 30 2017 16:06:29

%S 0,1,2,2,4,4,4,4,8,8,8,9,8,8,8,8,16,16,16,16,16,16,18,19,16,16,16,17,

%T 16,16,16,17,32,32,32,33,32,32,32,32,32,32,32,32,36,36,38,39,32,32,32,

%U 32,32,32,34,34,32,32,32,32,32,32,34,35,64,64,64,64,64,64,66,67,64,64,64,65,64,64,64,65,64,64,64,65,64,64,64,64,72,72,72,72,76,76

%N a(n) = A292253(A163511(n)).

%C Because A292253(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 the numbers that are either of the form 12k+1 or of the form 12k+11 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 formula just restates 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="/A292254/b292254.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) = A292253(A163511(n)).

%F a(n) = A292264(n) AND (A292274(n) XOR A292942(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987). [See comments.]

%F For all n >= 0, a(n) + A292944(n) + A292256(n) = n.

%o (Scheme) (define (A292254 n) (A292253 (A163511 n)))

%Y Cf. A005940, A163511, A292253.

%Y Cf also A292247, A292248, A292256, A292264, A292271, A292274, A292592, A292593, A292942, A292944, A292946 (for similarly constructed sequences).

%K nonn

%O 0,3

%A _Antti Karttunen_, Sep 28 2017