A049998

[Graeme McRae]

d'Ocagne's identity says F(m)F(n+1)-F(n)F(m+1)=(-1)^n F(m-n), so the 
difference of these products, anyway, is always a Fibonacci number.  If, for 
any given n and m, there is no product of Fibs that lies between F(m)F(n+1) 
and F(n)F(m+1), which is quite likely, since successive Fibs tend to differ 
by almost exactly the golden ratio, then the proof is complete.
--Graeme


----- Original Message ----- 
From: "David Wilson" <davidwwilson at comcast.net>
To: "Sequence Fans" <seqfan at ext.jussieu.fr>
Sent: Wednesday, December 14, 2005 12:49 PM
Subject: Re: A049998


> If A049998 includes only Fibs, that means that the difference of two 
> adjacent numbers that are the product of two Fibs is a Fib.
>
> I will go out on a limb and conjecture that the difference between two 
> adjacent numbers that are the product of n Fibs is the product of n-1 
> Fibs.  I have little evidence, it just seems like it would be a nice thing 
> to be true.
>
> ----- Original Message ----- 
> From: "David Wilson" <davidwwilson at comcast.net>
> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> Sent: Wednesday, December 14, 2005 1:07 PM
> Subject: A049998
>
>
>> Is A049998 composed entirely of Fibonacci numbers?
>>
>> --------------------------------
>> - David Wilson
>
>