[seqfan] Re: Consecutive composite Fibonacci numbers

[Robert Israel]

This is obvious: F_{kn} is divisible by F_k, so whenever integers
j,j+1,...,j+n are composite (and > 4), F_j, ..., F_{j+n} are 
also composite.

Robert Israel                                israel at math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada

On Thu, 18 Nov 2010, Alonso Del Arte wrote:

> Who proved that there is always a run of n consecutive composite Fibonacci
> numbers?
>
> (I'm sure it's either in Fib. Quart. or in Koshy's book, but I have no idea
> what search terms to use to zero in on this particular result).
>
> Any pointers would be appreciated.
>
> Al
>
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