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A214657
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Floor of the moduli of the zeros of the complex Fibonacci function.
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2
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0, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 108, 110, 112, 114, 116, 118, 120, 122, 124
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OFFSET
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0,3
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COMMENTS
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For the complex Fibonacci function F(z) and its zeros see the T. Koshy reference given in A214315. There the formula for the real and imaginary parts of the zeros is also given.
F: C -> C, z -> F(z) with F(z) := (exp(log(phi)*z) - exp(i*Pi*z)*exp(-log(phi)*z))/(2*phi-1), with phi = (1+sqrt(5))/2 and the imaginary unit i.
The zeros in the complex plane lie on a straight line with angle Phi = -arctan(2*log(phi)/Pi). They are equally spaced with distance tau defined below. Phi is approximately -0.2972713044, corresponding to about -17.03 degrees. The moduli are |z_0(k)| = tau*k, with tau = 2*Pi/sqrt(Pi^2 + (2*log(phi))^2), which is approximately 1.912278633.
a(n) = floor(tau*n) is a Beatty sequence with the complementary sequence b(n) = floor(sigma*n), with sigma:= tau/(tau-1), approximately 2.096156332.
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REFERENCES
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Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.
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LINKS
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FORMULA
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a(n) = floor(n*tau), n>=0, with tau = |z_0(1)| = 2*Pi/sqrt(Pi^2 + (2*log(phi))^2).
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EXAMPLE
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The complementary Beatty sequences a(n) and b(n) start:
n: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ...
a(n): 0 1 3 5 7 9 11 13 15 17 19 21 22 24 26 28 30 32 34 ...
b(n): (0)2 4 6 8 10 12 14 16 18 20 23 25 27 29 31 33 35 37 ...
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MATHEMATICA
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Table[Floor[2*n*Pi/Sqrt[Pi^2 + (2*Log[GoldenRatio])^2]], {n, 0, 100}] (* G. C. Greubel, Mar 09 2024 *)
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PROG
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(Magma) R:= RealField(100); [Floor(2*n*Pi(R)/Sqrt(Pi(R)^2 + (2*Log((1+Sqrt(5))/2))^2)) : n in [0..100]]; // G. C. Greubel, Mar 09 2024
(SageMath) [floor(2*n*pi/sqrt(pi^2 +4*(log(golden_ratio))^2)) for n in range(101)] # G. C. Greubel, Mar 09 2024
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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