This repository has been archived by the owner on Dec 16, 2023. It is now read-only.
/
euler0065.py
59 lines (39 loc) · 1.76 KB
/
euler0065.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
#!/usr/bin/python
r"""Convergents of e
The square root of 2 can be written as an infinite continued fraction.
$$\sqrt{2} = 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 +
\ldots}}}}$$
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 $\sqrt{2}$.
$1 + \dfrac{1}{2} = 3/2$
$1 + \dfrac{1}{2 + \dfrac{1}{2}} = 7/5$
$1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}} = 17/12$
$1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}}} = 41/29$
Hence the sequence of the first ten convergents for $\sqrt{2}$ are:
$$1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378,
\ldots$$
What is most surprising is that the important mathematical constant, $e =
[2; 1,2,1, 1,4,1, 1,6,1, \ldots, 1,2k,1, \ldots]$.
The first ten terms in the sequence of convergents for $e$ are:
$$2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, \ldots$$
The sum of digits in the numerator of the 10^{th} convergent is
$1+4+5+7=17$.
Find the sum of digits in the numerator of the 100^{th} convergent of the
continued fraction for $e$."""
from itertools import count
from eulerlib import convergents, nth
__tags__ = ['e', 'continued fractions', 'sum of digits']
def eCF():
yield 2
for i in count(1):
yield 1
yield 2*i
yield 1
def solve():
return sum(int(c) for c in str(nth(convergents(eCF()), 99)[0]))
if __name__ == '__main__':
print solve()