#### Problem

It is possible to show that the square root of two can be expressed as an infinite continued fraction.

By expanding this for the first four iterations, we get:

1 + 1/2 = 3/2 = 1.5

1 + 1/(2 + 1/2) = 7/5 = 1.4

1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666…

1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379…

The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?

#### Solution

// define function to return all the numerator-denominator pairs for the first n expand let expand n = Seq.unfold (fun (num, denom) -> Some((num, denom), (denom*2I+num, denom+num))) (3I, 2I) |> Seq.take n let answer = expand 1000 |> Seq.filter (fun (num, denom) -> num.ToString().Length > denom.ToString().Length) |> Seq.length

If you look at the patterns 3/2, 7/5, 17/12, 41/29, and so on, it’s easy to spot a pattern where the numerator and denominator of iteration n can be derived from the iteration n-1:

Numerator(n) = Numerator(n-1) + 2 * Denominator(n-1)

Denominator(n) = Numerator(n-1) + Denominator(n-1)