PyPE solution to PE problem 57

    ProjectEuler.net problem 57
    ===========================

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

        2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...

    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
    --------

    Let us first define some value for the kth expansion called phi_k, where

        sqrt(2)_k ~= 1 + 1 / phi_k

    and where

        phi_k = n_k / d_k.

    From the expansion, we can see that

        phi_k+1 = 2 + 1 / phi_k,

    or,

        n_k+1 = 2*n_k + d_k, d_k+1 = n_k.

    Using these terms, our numerator and denominator in the overall kth
    expansion will be

        sqrt(2)_k ~= (n_k + d_k) / n_k.

    We simply iterate over the desired number of expansions and check the length
    of the numerator vs the length of the denominator.

    Answer
    ------

    153