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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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified June 25 12:55 EDT 2019. Contains 324352 sequences. (Running on oeis4.)