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Search: id:A132971
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| A132971 |
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a(2*n) = a(n), a(4*n+1) = -a(n), a(4*n+3) = 0. |
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+0
1
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| 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, -1, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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If binary(n) has adjacent 1 bits then a(n) = 0 else a(n) = (-1)^A000120(n).
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FORMULA
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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).
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EXAMPLE
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1 - x - x^2 - x^4 + x^5 - x^8 + x^9 + x^10 - x^16 + x^17 + x^18 + ...
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PROGRAM
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(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) = local(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 )) }
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CROSSREFS
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Sequence in context: A097806 A167374 A085357 this_sequence A011748 A145361 A130304
Adjacent sequences: A132968 A132969 A132970 this_sequence A132972 A132973 A132974
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KEYWORD
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sign
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AUTHOR
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Michael Somos, Sep 17 2007, Sep 19 2007
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