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A214673
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Floor of the moduli of the zeros of the complex Lucas function.
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3
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0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 109, 111, 113, 115, 117, 119, 121, 123, 125
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OFFSET
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0,2
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COMMENTS
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For the complex Lucas function L(z) and its zeros see the commens in A214671 and the Koshy reference.
The modulus rho(k) of the zeros is sqrt(x_0(k)^2 + y_0(k)^2), with x_0(k) = (2*k+1)*(alpha/2) and y_0(k) = (2*k+1)*(b/2), where alpha = 2*(Pi^2)/(Pi^2 + (2*log(phi))^2) and b = 4*Pi*log(phi)/(Pi^2 + (2*log(phi))^2) (see the Fibonacci case A214657) and phi =(1+sqrt(5))/2. This leads to rho(k) = (k+1/2)*tau, with tau = 2*Pi/sqrt(Pi^2 + (2*log(phi))^2), known from the Fibonacci case. tau is approximately 1.912278633.
The zeros lie in the complex plane on a straight line with angle Phi = -arctan(2*log(phi)/Pi). They are equally spaced with distance tau given above. Phi is approximately -.2972713044, corresponding to about -17.03 degrees. This is the same line like in the Fibonacci case A214657, and the zeros of the Lucas function are just shifted on this line by tau/2, approximately 0.9561393165.
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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((2*n+1)*tau/2), n>=0, with tau/2 = rho(0) = 2*Pi / sqrt(Pi^2 + (2*log(phi))^2).
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MATHEMATICA
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Table[Floor[(2*n+1)*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+1)*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+1)*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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