Editing Lucas number (section)

==Relationship to Fibonacci numbers==
[[File:Comparison_Fibonacci_and_Lucas_numbers.svg|thumb|300px|The first identity expressed visually]]
The Lucas numbers are related to the Fibonacci numbers by many [[identity (mathematics)|identities]].  Among these are the following:
* <math>L_n = F_{n-1}+F_{n+1} = 2F_{n+1}-F_n</math>
* <math>L_{m+n} = L_{m+1}F_{n}+L_mF_{n-1}</math>
* <math>F_{2n} = L_n F_n</math>
* <math>F_{n+k} + (-1)^k F_{n-k} = L_k F_n</math>
* <math>2F_{2n+k} = L_{n} F_{n+k} + L_{n+k} F_{n}</math>
* <math>L_{2n} = 5 F_n^2 + 2(-1)^n = L_n^2 - 2(-1)^n</math>, so <math>\lim_{n\to\infty} \frac{L_n}{F_n}=\sqrt{5}</math>.
* <math> \vert L_n - \sqrt{5} F_n \vert = \frac{2}{\varphi^n} \to 0 </math>
* <math>L_{n+k} - (-1)^k L_{n-k} = 5 F_n F_k</math>; in particular, <math>F_n = {L_{n-1}+L_{n+1} \over 5}</math>, so <math>5F_n + L_n = 2L_{n+1}</math>.

Their [[Closed-form expression|closed formula]] is given as:
:<math>L_n = \varphi^n + (1-\varphi)^{n} = \varphi^n + (- \varphi)^{-n}=\left({ 1+ \sqrt{5} \over 2}\right)^n + \left({ 1- \sqrt{5} \over 2}\right)^n\, ,</math>

where <math>\varphi</math> is the [[golden ratio]]. Alternatively, as for <math>n>1</math> the magnitude of the term <math>(-\varphi)^{-n}</math> is less than 1/2, <math>L_n</math> is the closest integer to <math>\varphi^n</math> or, equivalently, the integer part of <math>\varphi^n+1/2</math>, also written as <math>\lfloor \varphi^n+1/2 \rfloor</math>.

Combining the above with [[Binet's formula]],

:<math>F_n = \frac{\varphi^n - (1-\varphi)^{n}}{\sqrt{5}}\, ,</math>

a formula for <math>\varphi^n</math> is obtained:

:<math>\varphi^n = {{L_n + F_n \sqrt{5}} \over 2}\, .</math>

For integers ''n'' ≥ 2, we also get:

:<math> \varphi^n = L_n - (- \varphi)^{-n} = L_n - (-1)^n L_n^{-1} - L_n^{-3} + R </math>

with remainder ''R'' satisfying

:<math> \vert R \vert < 3 L_n^{-5} </math>.