

A214673


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


2



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

0,2


COMMENTS

For the complex Lucas function L(z) and its zeros see a comment on A214671 where also the Koshy reference is given.
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)*(a/2), where alpha := 2*(Pi^2)/(Pi^2 + (2*log(phi))^2) and a := 4*Pi*log(phi)/(Pi^2 + (2*log(phi))^2) (see the Fibonacci case
A214657). phi:=(1+sqrt(5))/2. This leads to rho(k) = (k+1/2)*tau, with tau:= 2*Pi/sqrt(Pi^2 + (2*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.


LINKS

Table of n, a(n) for n=0..65.


FORMULA

a(n) = floor((2*k+1)*tau/2 ), n>=0, with tau/2 = rho(0) defined in the comment section.


CROSSREFS

Cf. A214671, A214672, A214657 (Fibonacci case).
Sequence in context: A007928 A257219 A092451 * A055962 A246410 A195169
Adjacent sequences: A214670 A214671 A214672 * A214674 A214675 A214676


KEYWORD

nonn


AUTHOR

Wolfdieter Lang, Jul 25 2012


STATUS

approved



