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A099496
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a(n) = (-1)^n * Fibonacci(2*n+1).
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6
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1, -2, 5, -13, 34, -89, 233, -610, 1597, -4181, 10946, -28657, 75025, -196418, 514229, -1346269, 3524578, -9227465, 24157817, -63245986, 165580141, -433494437, 1134903170, -2971215073, 7778742049, -20365011074, 53316291173, -139583862445, 365435296162, -956722026041
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OFFSET
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0,2
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COMMENTS
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With interpolated zeros, a Chebyshev transform of A056594, which has g.f. 1/(1+x^2). The image of G(x) under the Chebyshev transform is (1/(1+x^2))G(x/(1+x^2)).
a(n) is the ceiling of the inverse fractional error in approximating phi, the golden section, by the ratio of two successive terms in the Fibonacci series. - Adam Helman (helman(AT)san.rr.com), May 09 2010
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LINKS
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FORMULA
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G.f.: (1+x)/(1+3x+x^2).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)(-1)^k*cos((n-2k)*Pi/2) (with interpolated zeros);
a(n) = Fibonacci(n+1)(-1)^(n/2)(1 + (-1)^n)/2 (with interpolated zeros).
From Adam Helman (helman(AT)san.rr.com), May 09 2010: (Start)
a(n) = ceiling( phi / (Fibonacci(n+1)/Fibonacci(n) - phi) ).
An exact expression for the inverse fractional error is phi / (Fibonacci(n+1)/Fibonacci(n) - phi) = (phi/sqrt(5)) * ((-1)^n *(phi^2n) - 1). (End)
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EXAMPLE
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a(3) = (-1)^3 * Fibonacci(2 * 3 + 1) = -Fibonacci(7) = -13. - Indranil Ghosh, Feb 04 2017
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MAPLE
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seq((-1)^n*combinat:-fibonacci(2*n+1), n=0 .. 100); # Robert Israel, Jul 02 2015
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MATHEMATICA
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Table[(-1)^n Fibonacci[2 n + 1], {n, 0, 30}] (* Harvey P. Dale, Aug 22 2016 *)
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PROG
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CROSSREFS
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KEYWORD
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easy,sign
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AUTHOR
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STATUS
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approved
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