[seqfan] Re: Fibonacci sequences sharing no terms

[Jack Brennen]

No, they won't ever share any terms (assuming you don't "work backwards" 
from 4,6 to get 2).

Note that the second sequence:

   4,6,10,16,26,42,68,110,...

is simply equal to the sum of these two sequences:

   3,5,8,13,21,34,55,89,...
   1,1,2, 3, 5, 8,13,21,...

Think about that.  :)




On 9/24/2013 8:14 AM, Eric Angelini wrote:
> Hello Seqfans,
> please forgive me if this question is naïve.
> Consider Fib(1,2) = 1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,...
> Now will Fib(4,6) = 4,6,10,16,26,42,68,110,178,288,466,... share at some point
> one term (or more) with Fib(1,2)?
> If yes or no, do we know why? Are there constraints on 'a' and 'b' leading to
> a Fib(a,b) which will never share any term with Fib(1,2)?
> Many thanks,
> É.
> [I've never subscribed to Fibonacci Quarterly...]
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>
>
>