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Project Euler 64: Find the continued fractions for N ≤ 10000 have an odd period
Problem Description
All square roots are periodic when written as continued fractions and can be written in the form:
| √N = a0 + |
1
|
||
| a1 + |
1
|
||
| a2 + |
1
|
||
| a3 + … | |||
For example, let us consider √23:
| √23 = 4 + √23 — 4 = 4 + |
1
|
= 4 + |
1
|
|
|
1
√23—4 |
1 + |
√23 – 3
7 |
||
If we continue we would get the following expansion:
| √23 = 4 + |
1
|
|||
| 1 + |
1
|
|||
| 3 + |
1
|
|||
| 1 + |
1
|
|||
| 8 + … | ||||
The process can be summarised as follows:
| a0 = 4, |
1
√23—4 |
= |
√23+4
7 |
= 1 + |
√23—3
7 |
|
| a1 = 1, |
7
√23—3 |
= |
7(√23+3)
14 |
= 3 + |
√23—3
2 |
|
| a2 = 3, |
2
√23—3 |
= |
2(√23+3)
14 |
= 1 + |
√23—4
7 |
|
| a3 = 1, |
7
√23—4 |
= |
7(√23+4)
7 |
= 8 + | √23—4 | |
| a4 = 8, |
1
√23—4 |
= |
√23+4
7 |
= 1 + |
√23—3
7 |
|
| a5 = 1, |
7
√23—3 |
= |
7(√23+3)
14 |
= 3 + |
√23—3
2 |
|
| a6 = 3, |
2
√23—3 |
= |
2(√23+3)
14 |
= 1 + |
√23—4
7 |
|
| a7 = 1, |
7
√23—4 |
= |
7(√23+4)
7 |
= 8 + | √23—4 |
It can be seen that the sequence is repeating. For conciseness, we use the notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.
The first ten continued fraction representations of (irrational) square roots are:
√2=[1;(2)], period=1
√3=[1;(1,2)], period=2
√5=[2;(4)], period=1
√6=[2;(2,4)], period=2
√7=[2;(1,1,1,4)], period=4
√8=[2;(1,4)], period=2
√10=[3;(6)], period=1
√11=[3;(3,6)], period=2
√12= [3;(2,6)], period=2
√13=[3;(1,1,1,1,6)], period=5
Exactly four continued fractions, for N ≤ 13, have an odd period.
How many continued fractions for N ≤ 10000 have an odd period?
Analysis
This one appeared easier than it was. It wasn’t until I read this article on computing square roots using a continued fraction expansion that I could code it for finding a solution.
Project Euler 64 Solution
Runs < 0.150 seconds in Python 2.7.
Use this link to get the Project Euler 64 Solution Python 2.7 source.
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