# [seqfan] A000032 Lucas numbers - suggestion

Jeff Anderson peawormsworth at gmail.com
Mon Jun 3 07:41:49 CEST 2019

```Hi,

re: https://oeis.org/A000032

=== quote ===
Let f(n) = phi^n + phi^( - n), then L(2n) = f(2n) and L(2n + 1) = f(2n + 1)
- 2*Sum_{k=0..infinity} C(k)/f(2n + 1)^(2k + 1) where C(n) are Catalan
numbers (A000108). - Gerald McGarvey, Dec 21 2007, modified by Davide
Colazingari, Jul 01 2016
=== end ===

It is known:
L(n) = phi^n + (-phi)^-(n)

and so...
L(2n) = phi^2n + phi^-(2n)
L(2n+1) = phi^(2n+1) - phi^-(2n+1)

Given:
f(n) = phi^n + phi^-(n)

calculate...
f(2n+1) = phi^(2n+1) + phi^-(2n+1)

Assuming this from quote:
L(2n+1) = f(2n + 1) - 2×Sum_{k=0..infinity} C(k)/f(2n + 1)^(2k + 1)

substitute...
phi^(2n+1) - phi^-(2n+1) = phi^(2n+1) + phi^-(2n+1) -2×Sum_{k=0..infinity}
C(k)/f(2n + 1)^(2k + 1)
-2×phi^-(2n+1) = -2×Sum_{k=0..infinity} C(k)/f(2n + 1)^(2k + 1)
phi^-(2n+1) = Sum_{k=0..infinity} C(k)/f(2n + 1)^(2k + 1)

=== conclude ===

The relationship and formula shown is not about Lucas numbers.
Quote may be more appropriate for inclusion on a different series.

Jeff

```