cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A214673 Floor of the moduli of the zeros of the complex Lucas function.

Original entry on oeis.org

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
Offset: 0

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Author

Wolfdieter Lang, Jul 25 2012

Keywords

Comments

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.

References

  • Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.

Crossrefs

Cf. A214657 (Fibonacci case), A214671, A214672.

Programs

  • 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
    
  • Mathematica
    Table[Floor[(2*n+1)*Pi/Sqrt[Pi^2+(2*Log[GoldenRatio])^2]], {n,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

Formula

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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