[seqfan] Re: Code

israel at math.ubc.ca israel at math.ubc.ca
Thu Jun 22 08:12:39 CEST 2017


All the polynomial equations with integer coefficients relating F_n 
and F_{n+1} are generated by 
  G(F_n, F_{n+1}) = F_n^2 + F_n F_{n+1} - F_{n+1}^2 + (-1)^n = 0
This particular identity is just 
   G(F_n, F_{n+1))(F_n^3+2 F_n^2 F_{n+1}-F_{n+1}^3-(-1)^n (F_n + F_{n+1})) 
= 0

Cheers,
Robert
  

On Jun 21 2017, Peter Lawrence wrote:Here's one way of generating 
recurrences of this type. Suppose you have a polynomial expression P(F_n, 
F_{n+1}) = 0. With phi = (1+sqrt(5))/2, F_n = (phi^n - (-1)^n 
phi^(-n))/sqrt(5) and F_{n+1} = (phi^(n+1) + (-1)^n phi^(-n-1))/sqrt(5)


>Hans,
>         As I am not familiar with Mathematica (I'm just a C/C++ 
> programmer), would it be too much to ask for an explanation of this 
> formula, in other words not a proof of its correctness, but rather the 
> details of how it computes, what algorithm is being specified here ? Are 
> F(n) and F(n+1) inputs, and if so then what is the output, F(n+2) ?
>
>Thanks,
>Peter Lawrence.
>
>
>> On Jun 21, 2017, at 3:51 PM, seqfan-request at list.seqfan.eu wrote:
>> 
>> Message: 17 Date: Wed, 21 Jun 2017 15:32:01 -0400 From: Hans Havermann 
>> <gladhobo at bell.net <mailto:gladhobo at bell.net>> To: Sequence Fanatics 
>> Discussion list <seqfan at list.seqfan.eu <mailto:seqfan at list.seqfan.eu>> 
>> Subject: [seqfan] Code Message-ID: 
>> <C6CC6D76-FE75-4251-ACCC-61CCD0F6C652 at bell.net 
>> <mailto:C6CC6D76-FE75-4251-ACCC-61CCD0F6C652 at bell.net>> Content-Type: 
>> text/plain; charset=us-ascii
>> 
>> I presume that the inclusion of code in sequences is to be able to 
>> generate/verify/extend that sequence. Simpler and faster is better. 
>> Browsing the OEIS today I chanced upon this Mathematica code for 
>> Fibonacci numbers < https://oeis.org/A000045 <https://oeis.org/A000045> 
>> >:
>> 
>> Table[Fibonacci[n]^5 - Fibonacci[1 + n] + 3 Fibonacci[n]^4 Fibonacci[1 
>> + n] + Fibonacci[n]^3 Fibonacci[1 + n]^2 - 3 Fibonacci[n]^2 Fibonacci[1 
>> + n]^3 - Fibonacci[n] Fibonacci[1 + n]^4 + Fibonacci[1 + n]^5, {n, 1, 
>> 10}]
>> 
>> It's an interesting Fibonacci identity worthy perhaps of mention in the 
>> comments (if it isn't already there) but I'm not sure it ought to be 
>> where it is.
>
>
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>
>



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