Is \(F_{100}\) a multiple of \(3\)?
Draft for the Fibonacci number introduction. This week we’re looking at Fibonacci numbers, and other sequences of numbers.
We say that the ‘zeroth’ Fibonacci number is \(0\) and the first Fibonacci number is \(1\). Then, from that point, every Fibonacci number is the sum of the previous two Fibonacci numbers. This means that the sequence begins \(0,1,1,2,3,5,8,13,21,34,55,89,144,...\)
The Fibonacci numbers hide lots of patterns which we’ll explore today. We’ll also look at another couple of sequences.
Here are a couple of ideas for problems to include. Something about the continued fraction of the golden ratio being [1,1,1,...] (probably won’t include), something about Conway’s soldiers (similarly probably won’t include), each new Fibonacci number introduces a new prime factor not seen in previous number, other than \(F_0=0\), \(F_1=1\), \(F_2=1\), \(F_6=8\) and \(F_{12}=144\). (I’m not even sure what the proof of this is).
We have a sequence where the first term (\(x_1\)) is equal to \(2\), and each term is \(1\) minus the reciprocal of the previous term (which we can write as \(x_{n+1}=1-\frac{1}{x_n}\)).
What’s \(x_{57}\)?