A214671 -id:A214671 - cp's OEIS frontend

A214672 Floor of the imaginary parts of the zeros of the complex Lucas function on the left half-plane.

0, 0, 1, 1, 2, 3, 3, 4, 4, 5, 5, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 17, 17, 18, 18, 19, 19, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 31, 31, 32, 32, 33, 33, 34, 35, 35, 36

Offset: 0

Wolfdieter Lang, Jul 25 2012

nonn

For the complex Lucas function L(z) and its zeros see the comments in A214671 and the Koshy reference.

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

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A214656 (Fibonacci case), A214671 (floor of real parts), A214673 (moduli).

R:= RealField(100); [Floor((2*n+1)*(2*Pi(R)*Log((1 + Sqrt(5))/2))/(Pi(R)^2 + (2*Log((1+Sqrt(5))/2))^2)) : n in [0..100]]; // G. C. Greubel, Mar 09 2024

Table[Floor[(2*n+1)*(2*Pi*Log[GoldenRatio])/(Pi^2 + (2*Log[GoldenRatio])^2)], {n, 0, 100}] (* G. C. Greubel, Mar 09 2024 *)

[floor(2*(2*n+1)*pi*log(golden_ratio)/(pi^2 +4*(log(golden_ratio))^2)) for n in range(101)] # G. C. Greubel, Mar 09 2024

a(n) = floor((2*n+1)*b/2), n >= 0, with b/2 = -y_0(0) = 2*Pi*log(phi) / (Pi^2 + (2*log(phi))^2), with phi = (1+sqrt(5))/2. Note that b/2 is approximately 0.2800649542... . The constant b appears in the corresponding Fibonacci case A214656.

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

Wolfdieter Lang, Jul 25 2012

nonn

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.

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

G. C. Greubel, Table of n, a(n) for n = 0..10000

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

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

Table[Floor[(2*n+1)*Pi/Sqrt[Pi^2+(2*Log[GoldenRatio])^2]], {n,0,100}] (* G. C. Greubel, Mar 09 2024 *)

[floor((2*n+1)*pi/sqrt(pi^2 +4*(log(golden_ratio))^2)) for n in range(101)] # G. C. Greubel, Mar 09 2024

a(n) = floor((2*n+1)*tau/2), n>=0, with tau/2 = rho(0) = 2*Pi / sqrt(Pi^2 + (2*log(phi))^2).

cp's OEIS Frontend

A214672 Floor of the imaginary parts of the zeros of the complex Lucas function on the left half-plane.

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