project_euler.problem_065.sol1
==============================

.. py:module:: project_euler.problem_065.sol1

.. autoapi-nested-parse::

   Project Euler Problem 65: https://projecteuler.net/problem=65

   The square root of 2 can be written as an infinite continued fraction.

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

   The infinite continued fraction can be written, sqrt(2) = [1;(2)], (2)
   indicates that 2 repeats ad infinitum. In a similar way, sqrt(23) =
   [4;(1,3,1,8)].

   It turns out that the sequence of partial values of continued
   fractions for square roots provide the best rational approximations.
   Let us consider the convergents for sqrt(2).

   1 + 1 / 2 = 3/2
   1 + 1 / (2 + 1 / 2) = 7/5
   1 + 1 / (2 + 1 / (2 + 1 / 2)) = 17/12
   1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / 2))) = 41/29

   Hence the sequence of the first ten convergents for sqrt(2) are:
   1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...

   What is most surprising is that the important mathematical constant,
   e = [2;1,2,1,1,4,1,1,6,1,...,1,2k,1,...].

   The first ten terms in the sequence of convergents for e are:
   2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...

   The sum of digits in the numerator of the 10th convergent is
   1 + 4 + 5 + 7 = 17.

   Find the sum of the digits in the numerator of the 100th convergent
   of the continued fraction for e.

   -----

   The solution mostly comes down to finding an equation that will generate
   the numerator of the continued fraction. For the i-th numerator, the
   pattern is:

   n_i = m_i * n_(i-1) + n_(i-2)

   for m_i = the i-th index of the continued fraction representation of e,
   n_0 = 1, and n_1 = 2 as the first 2 numbers of the representation.

   For example:
   n_9 = 6 * 193 + 106 = 1264
   1 + 2 + 6 + 4 = 13

   n_10 = 1 * 193 + 1264 = 1457
   1 + 4 + 5 + 7 = 17



Functions
---------

.. autoapisummary::

   project_euler.problem_065.sol1.solution
   project_euler.problem_065.sol1.sum_digits


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

.. py:function:: solution(max_n: int = 100) -> int

   Returns the sum of the digits in the numerator of the max-th convergent of
   the continued fraction for e.

   >>> solution(9)
   13
   >>> solution(10)
   17
   >>> solution(50)
   91


.. py:function:: sum_digits(num: int) -> int

   Returns the sum of every digit in num.

   >>> sum_digits(1)
   1
   >>> sum_digits(12345)
   15
   >>> sum_digits(999001)
   28