project_euler.problem_057.sol1
==============================

.. py:module:: project_euler.problem_057.sol1

.. autoapi-nested-parse::

   Project Euler Problem 57: https://projecteuler.net/problem=57
   It is possible to show that the square root of two can be expressed as an infinite
   continued fraction.

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

   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 the denominator?



Functions
---------

.. autoapisummary::

   project_euler.problem_057.sol1.solution


Module Contents
---------------

.. py:function:: solution(n: int = 1000) -> int

   returns number of fractions containing a numerator with more digits than
   the denominator in the first n expansions.
   >>> solution(14)
   2
   >>> solution(100)
   15
   >>> solution(10000)
   1508