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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.

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

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%I A214673 #11 Mar 09 2024 08:16:37
%S A214673 0,2,4,6,8,10,12,14,16,18,20,21,23,25,27,29,31,33,35,37,39,41,43,44,
%T A214673 46,48,50,52,54,56,58,60,62,64,65,67,69,71,73,75,77,79,81,83,85,87,88,
%U A214673 90,92,94,96,98,100,102,104,106,108,109,111,113,115,117,119,121,123,125
%N A214673 Floor of the moduli of the zeros of the complex Lucas function.
%C A214673 For the complex Lucas function L(z) and its zeros see the commens in A214671 and the Koshy reference.
%C A214673 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.
%C A214673 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.
%D A214673 Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.
%H A214673 G. C. Greubel, <a href="/A214673/b214673.txt">Table of n, a(n) for n = 0..10000</a>
%F A214673 a(n) = floor((2*n+1)*tau/2), n>=0, with tau/2 = rho(0) = 2*Pi / sqrt(Pi^2 + (2*log(phi))^2).
%t A214673 Table[Floor[(2*n+1)*Pi/Sqrt[Pi^2+(2*Log[GoldenRatio])^2]], {n,0,100}] (* _G. C. Greubel_, Mar 09 2024 *)
%o A214673 (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
%o A214673 (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
%Y A214673 Cf. A214657 (Fibonacci case), A214671, A214672.
%K A214673 nonn
%O A214673 0,2
%A A214673 _Wolfdieter Lang_, Jul 25 2012