A132971 a(2*n) = a(n), a(4*n+1) = -a(n), a(4*n+3) = 0, with a(0) = 1.

1, -1, -1, 0, -1, 1, 0, 0, -1, 1, 1, 0, 0, 0, 0, 0, -1, 1, 1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 1, 0, 1, -1, 0, 0, 1, -1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 1, 0, 1, -1, 0, 0, 1, -1, -1, 0, 0, 0, 0, 0, 1, -1

Offset: 0

Author

Michael Somos, Sep 17 2007, Sep 19 2007

Keywords

• sign

Comments

If binary(n) has adjacent 1 bits then a(n) = 0 else a(n) = (-1)^A000120(n).

Fibbinary numbers (A003714) gives the numbers n for which a(n) = A106400(n). - Antti Karttunen, May 30 2017

Examples

			G.f. = 1 - x - x^2 - x^4 + x^5 - x^8 + x^9 + x^10 - x^16 + x^17 + x^18 + ...
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..10922
• Paul Tarau, Emulating Primality with Multiset Representations of Natural Numbers, in Theoretical Aspects of Computing, ICTAC 2011, Lecture Notes in Computer Science, 2011, Volume 6916/2011, 218-238, DOI: 10.1007/978-3-642-23283-1_15.
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A000120, A000621, A003714, A005252, A005940, A008683, A024490, A027935, A106400.

Cf. A085357 (gives the absolute values: -1 -> 1), A286576 (when reduced modulo 3: -1 -> 2).

Programs

Mathematica

m = 100; A[_] = 1;
Do[A[x_] = A[x^2] - x A[x^4] + O[x]^m // Normal, {m}];
CoefficientList[A[x], x] (* Jean-François Alcover, Nov 16 2019 *)

PARI

{a(n) = if( n<1, n==0, if( n%2, if( n%4 > 1, 0, -a((n-1)/4) ), a(n/2) ) )};

PARI

{a(n) = my(A, m); if( n<0, 0, m = 1; A = 1 + O(x); while( m<=n, m *= 2; A = subst(A, x, x^2) - x * subst(A, x, x^4) ); polcoeff(A, n)) };

Python

from sympy import mobius, prime, log
import math
def A(n): return n - 2**int(math.floor(log(n, 2)))
def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
def a(n): return mobius(b(n)) # Indranil Ghosh, May 30 2017

Scheme

(define (A132971 n) (cond ((zero? n) 1) ((even? n) (A132971 (/ n 2))) ((= 1 (modulo n 4)) (- (A132971 (/ (- n 1) 4)))) (else 0))) ;; Antti Karttunen, May 30 2017

Formula

A024490(n) = number of solutions to 2^n <= k < 2^(n+1) and a(k) = 1.

A005252(n) = number of solutions to 2^n <= k < 2^(n+1) and a(k) = -1.

A027935(n-1) = number of solutions to 2^n <= k < 2^(n+1) and a(k) = 0.

G.f. A(x) satisfies A(x) = A(x^2) - x * A(x^4).

G.f. B(x) of A000621 satisfies B(x) = x * A(x^2) / A(x).

a(n) = A008683(A005940(1+n)). [Analogous to Moebius mu] - Antti Karttunen, May 30 2017