Talk:Lucas number

message: Comments should be on Talk:Lucas sequence, at the proposed target article which the merge box links to on "Discuss". Duplicating here is not necessary. PrimeHunter 18:12, 21 January 2007 (UTC)
replies: text: Against merger This article is about a specific important sequence. It is simple and easy to understand. The other article is very abstract and difficult to understand and is about many sequences, some of which may not be so important as this one. However, we should have a link to it as a generalization. In other words: Keep It Simple, Stupid! JRSpriggs 13:32, 21 January 2007 (UTC)
thread_title: Comments on proposed merger

message: In the Wiki article on the Pascal Triangle there is a subsection showing how the Fibonacci numbers can be generated by summing samplings taken across every other diagonal. I've seen, online (but can't remember URL), an analogue to the Pascal triangle with one of the 1's sides replaced by 2's. Repeating the aforementioned procedure on this modified Pascal system gives the Lucas numbers in one direction, and the Fibonacci in the other. In fact any Fibonacci-like sequence, leading to the Golden Ratio, can be created by adjusting the sides of the triangle, and the 'seeds' of these sequences are in fact the numbers on the sides. So for the classical Pascal Triangle, with 1,1 on the sides, gives (1,1),2,3,5,8... and you can see that ..(1,2),3,5,8.. leads to a similar, but offset Fib. In the other direction (2,1),3,4,7,11... you get Lucas. This may seem like original research (it was for me) but I have to believe that mathematicians have known about this for ages, and there must be published sources one could hang the donkey tail onto so as to be able to include these facts into the main articles. Anyone know of any? Thanks.  — Preceding unsigned comment added by 96.234.78.93 (talk) 11:28, 13 October 2011 (UTC)
thread_title:  Lucas numbers generated by Pascal Triangle analogue 

message: The link "A Tutorial on Generalized Lucas Numbers" is no longer valid. Anybody knows a new location?  — Preceding unsigned comment added by 217.5.143.100 (talk) 07:55, 11 June 2014 (UTC)
replies: text: I haven't found a live version. It's archived at https://web.archive.org/web/20130825030815/http://nakedprogrammer.com/LucasNumbers.aspx and http://www.archive.today/LBOG1 PrimeHunter (talk) 09:09, 11 June 2014 (UTC)
thread_title:  Links 

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thread_title:  External links modified 

message: Hereby I want to send some generalisation formulas for two Items in the chapter Relationship to Fibonacci numbers. 
The item \,L_n = F_{n-1}+F_{n+1}=F_n+2F_{n-1} = F_{n+2}-F_{n-2} can be more generalise by the following equation:
: 
  L_n :=
  \begin{cases}
    {F_{n-m}+F_{n+m} \over F_{m}}           & \text{if m is uneven and } m => 1;\\
    {F_{n-m}-F_{n+m} \over F_{m}}           & \text{if m is even and } m => 2\\
   \end{cases}
 
where m is de position away from the Lucas number to find.
and item \,F_n = {L_{n-1}+L_{n+1} \over 5} = {L_{n-3}+L_{n+3} \over 10}  can be generalised with the following equations
: 
  F_n :=
  \begin{cases}
    {L_{n+m} + L_{n-m} \over {5F_{m}}}           & \text{if m is uneven and } m => 1;\\
    {L_{n+m} - L_{n-m} \over {5F_{m}}}           & \text{if m is even and } m => 2\\
   \end{cases}
 
which we can rewrite as
: 
  F_nF_m :=
  \begin{cases}
    {L_{n+m} + L_{n-m} \over 5}           & \text{if m is uneven and } m => 1;\\
    {L_{n+m} - L_{n-m} \over 5}           & \text{if m is even and } m => 2\\
   \end{cases}
 
or only written as Lucas numbers
: 
  F_n :=
  \begin{cases}
    {L_{n+m} + L_{n-m} \over {L_{m+1} + L_{m-1}}}           & \text{if m is uneven and } m => 1;\\
    {L_{n+m} - L_{n-m} \over {L_{m+1} + L_{m-1}}}           & \text{if m is even and } m => 2\\
   \end{cases}
 
— Preceding unsigned comment added by 86.84.102.126 (talk) 06:31, 22 September 2018 (UTC)
thread_title:  Relationship to Fibonacci numbers addition 

message: Next identities are very usefull if you need to simplify some complex fibs expressions.
Besides, they are elegant. 
* F_{n+k} + (-1)^k F_{n-k} = L_k F_n, \quad F_{n+k} - (-1)^k F_{n-k} = L_n F_k
* L_{n+k} + (-1)^k L_{n-k} = L_k L_n, \quad L_{n+k} - (-1)^k L_{n-k} = 5 F_n F_k
--Ivigan (talk) 06:33, 5 July 2019 (UTC)
replies: text: I have edited your comment for readability.  Three of these identities are in the article already (the second is the same as the first).  The other one does not involve Fibonacci numbers, but you placed it in the section titled "Relationship to Fibonacci numbers".  It can also easily be derived from formulas that are in the article.  --JBL (talk) 10:14, 5 July 2019 (UTC)
thread_title: Useful identities