Talk:Generalizations of Fibonacci numbers

message: I have reverted  a copyright violation from  per WP:COPYVIO. The text was copied directly and poorly formatted for Wikipedia. I haven't examined reliability of source or notability of content. PrimeHunter 01:22, 28 May 2007 (UTC)
replies: text: It seems notable to me, see the various Fibonacci number generalizations used in Mathematica: .  Well, most of these are probably not "used", but it's still an astonishing variety of generalizations to real and complex numbers.  It's too bad the "author" of the deleted material didn't bother to write it and source it properly, because this is precisely the sort of thing the article needs. Silly rabbit 12:39, 28 May 2007 (UTC)

text: I have limited copyright knowledge but this appears to solve the copyright problem. I don't personally want to reinstate your edit  in that form because it is rather poorly formatted for Wikipedia. Also, I don't know the subject well enough to judge it and I'm not sure your site satisfies WP:RS and WP:V#SELF. Can you format it better and provide an additional source to support it (maybe http://mathworld.wolfram.com/FibonacciNumber.html)? I admit Wikipedia already contains a lot of information that has problems with our policies and guidelines. PrimeHunter 00:05, 30 May 2007 (UTC)

text: I have restored it, with some changes in wording so it's not as much of a glaring potential copyvio.  I actually don't think it's an issue at all, since the page in question is now under the GFDL.  We incorporate stuff from planetmath all the time.  Anyway, I admit it's a rush-job.  The formulas need to be checked for accuracy, and references still need to be provided.  But I think it's a reasonable start.  Silly rabbit 00:22, 30 May 2007 (UTC)
replies: text: Thanks for formatting it. I see you put the source link in a comment. As far as I know (see for example Wikipedia:GNU Free Documentation License resources), we must give the source when it is GFDL-licensed and used so closely, so I have added the link. PrimeHunter 01:11, 30 May 2007 (UTC)
thread_title:  Copyvio reverted 

message: It seems to me that this page is about variations, rather than generalizations, of the fibonacci sequence. Renaming it would be wise. 84.57.159.115 (talk) 14:31, 2 January 2011 (UTC)
thread_title:  Naming issue 

message: One can generate a matrix of Fibonacci-like numbers by continuing to add Fibonacci numbers to terms in vertical columns:
0,  1,  1,  2,  3,  5,  8, 13, 21,...
1,  1,  2,  3,  5,  8, 13, 21, 34,...
2,  1,  3,  4,  7, 11, 18, 29, 47,...
3,  1,  4,  5,  9, 14, 23, 37, 60,...
4,  1,  5,  6, 11, 17, 28, 45, 73,...
The first two horizontal sequences are both Fibonacci, the third is the Lucas sequence, and so on. One can also generate these by taking the sums of numbers through 'shallow' diagonals in modified Pascal Triangles. If the sides of the triangles are defined as (x,y), with the Classical Pascal system being (1,1), then (0,1) still gives Fibonacci, (2,1) gives Fibonacci on one side and Lucas on the other, and so on. What is really interesting is that the terms of equations defining the powers of the Metal Means (see Wiki article on Silver Ratio and bottom of discussion page) have coefficients that are identical to the terms from half the (2,1) triangle, and sum using shallow diagonals to half the Lucas numbers. 67.81.236.32 (talk) 05:26, 21 November 2011 (UTC)
thread_title:  More sequences 

message: According to Binet's formula, 
F_n = \frac{\varphi^n-\psi^n}{\varphi-\psi} = \frac{\varphi^n-\psi^n}{\sqrt 5} 
Since \psi = -\frac{1}{\varphi}, this formula can also be written as
F_n = \frac{\varphi^n-(-\varphi)^{-n}}{\sqrt 5}
Now if you factor the -1 out of the -\varphi, you get
F_n = \frac{\varphi^n-(-1)^{-n}\varphi^{-n}}{\sqrt 5}
And since \frac{1}{-1} = -1, this can be written as 
F_n = \frac{\varphi^n-(-1)^{n}\varphi^{-n}}{\sqrt 5}
And from Euler's identity, -1 = e^{i\pi}, so
F_n = \frac{\varphi^n-e^{i\pi n}\varphi^{-n}}{\sqrt 5}
And from Euler's formula, e^{i\pi n} = \cos (\pi n) + i \sin (\pi n), so
F_n = \frac{\varphi^n-(\cos (\pi n) + i \sin (\pi n))\varphi^{-n}}{\sqrt 5}
If you distribute and separate the real and imaginary parts, you get
F_n = \frac{\varphi^n-\cos (\pi n)\varphi^{-n}}{\sqrt 5} - i \frac{\sin (\pi n)\varphi^{-n}}{\sqrt 5}
In Generalizations of Fibonacci numbers#Extension to all real or complex numbers, It gives the formula:
Fib(x) = \frac{\varphi^x - \cos(x \pi)\varphi^{-x}}{\sqrt{5}}
Notice that this is just the real part of the formula I got with Binet's formula.  Should we include the imaginary part of Binet's formula in the generalization for all real numbers?  What happens when you put complex numbers into Binet's formula?
By the way, when you graph Binet's formula in the negative numbers with the imaginary part, it forms a variation on the golden spiral where it gets wider (or further from its origin) by a factor of \varphi for every half turn it makes.  (The real golden spiral gets wider by a factor of \varphi for every quarter turn it makes.)  — Preceding unsigned comment added by 174.55.185.213 (talk) 18:37, 26 May 2013 (UTC)
thread_title:  Including the imaginary part of Binet's Formula 

message: I don't know if this can get involved with Fibonacci numbers, but maybe we could define:
φ(a,b,c) = (a+sqrt(b))/c
The golden ratio is φ(1,5,2).  — Preceding unsigned comment added by 99.185.0.100 (talk) 01:36, 24 April 2015 (UTC)
thread_title:  Generalizing the golden ratio with a 3-argument function 

message: And then we can have φ2(a,b) as the solution to xa = x+b. The golden ratio is φ(2,1).  — Preceding unsigned comment added by 99.185.0.100 (talk) 02:18, 24 April 2015 (UTC)
thread_title:  And a 2-argument function... 

message: I did not know about this terminology, is it original research? At present, several things seemed not to make sense, so I made some (quite radical) corrections: F_N(n) was defined as product, changed to sum; N was defined with P_r(n-r) as last factor, changed to P_r^(a_r). Thanks in advance if anyone can double-check and/or confirm (or correct otherwise). - — MFH:Talk 15:56, 27 March 2017 (UTC)
thread_title:  N-generated Fibonacci 

message: It seems that there's a mistake at the end of the Similar Integer Sequences section where it's claimed that the probability of no 'n' consecutive tails in 'm' tosses is given as \frac{1}{2^m}F_{m+2}^{(n)}, where F_j^{(i)} would be the j^{th} element of an i^{th} order Fibonacci sequence. I think this formula only holds for m=2 , and the general formula should be \frac{1}{2^m}F_{m+n}^{(n)}.
For example, if m = 4 and n = 3, there are 16 possible sequences, and 3 of them have at least 3 consecutive tails, so there should be 13 sequences without 3 consecutive tails. But F_{m+2}^{(n)} = F_{6}^{(3)} = 7, whereas F_{m+n}^{(n)} = F_{7}^{(3)} = 13. I sent a message to Mathworld/Wolfram about this a few days ago (since it is the linked source of the claim), but haven't heard back yet.
I can provide my work to show the proof, but wasn't aware if that was appropriate for a talk page since it's a few pages long.
Please correct me if I'm mistaken.
Idiot-dogg (talk) 15:09, 16 June 2023 (UTC)
thread_title:  Potential mistake in probability of no 'n' consecutive tails in 'm' tosses