A292256 a(n) = A292255(A163511(n)).

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 6, 6, 0, 0, 0, 0, 0, 0, 2, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 8, 8, 8, 9, 12, 12, 12, 12, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 6, 6, 0, 0, 0, 0, 0, 0, 2, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 2, 2, 16, 16, 16, 16, 16, 16

Offset: 0

Author

Antti Karttunen, Sep 28 2017

Keywords

• nonn

Comments

Because A292255(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+5 or of the form 12k+7 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).

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.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8191
• Index entries for sequences related to binary expansion of n
• Index entries for sequences computed from indices in prime factorization

Crossrefs

Cf. A005940, A163511, A292255.

Cf. also A292247, A292248, A292254, A292264, A292271, A292274, A292592, A292593, A292942, A292944, A292946 (for similarly constructed sequences).

Programs

Scheme

(define (A292256 n) (A292255 (A163511 n)))

Formula

a(n) = A292255(A163511(n)).

a(n) = A292264(n) AND (A292274(n) XOR A292946(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987).

a(n) = A292264(n) AND (A292271(n) XOR A292942(n)). [See comments].

For all n >= 0, a(n) + A292944(n) + A292254(n) = n.