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"""
Project Euler Problem 25: :math:`1000`-Digit Fibonacci Number
=============================================================
.. module:: solutions.problem25
:synopsis: My solution to problem #25.
The source code for this problem can be
`found here <https://bitbucket.org/nekedome/project-euler/src/master/solutions/problem25.py>`_.
Problem Statement
#################
The Fibonacci sequence is defined by the recurrence relation:
.. math::
F_n = F_{n-1} + F_{n-2}, \\mbox{ where } F_1 = 1 \\mbox{ and } F_2 = 1.
Hence the first :math:`12` terms will be:
F_1 &= 1 \\\\
F_2 &= 1 \\\\
F_3 &= 2 \\\\
F_4 &= 3 \\\\
F_5 &= 5 \\\\
F_6 &= 8 \\\\
F_7 &= 13 \\\\
F_8 &= 21 \\\\
F_9 &= 34 \\\\
F_{10} &= 55 \\\\
F_{11} &= 89 \\\\
F_{12} &= 144
The :math:`12^{th}` term, :math:`F_{12}`, is the first term to contain three digits.
What is the index of the first term in the Fibonacci sequence to contain :math:`1000` digits?
Solution Discussion
###################
Simply iterate over the Fibonacci sequence until an :math:`F_n` is encountered containing :math:`1000` digits.
This check looks for lines that are too long. You can specify the maximum line length.
Solution Implementation
#######################
.. literalinclude:: ../../solutions/problem25.py
:language: python
:lines: 54-
from itertools import dropwhile
from lib.digital import num_digits
from lib.sequence import Fibonaccis
def solve():
""" Compute the answer to Project Euler's problem #25 """
target = 1000
fibs = Fibonaccis()
for n, F_n in dropwhile(lambda elt: num_digits(elt[1]) < target, enumerate(fibs)):
n
(([a-z][a-z0-9_]{2,30})|(_[a-z0-9_]*))$
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F_n
return n
expected_answer = 4782
expected_answer
(([A-Z_][A-Z0-9_]*)|(__.*__))$
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