Project Euler 65

题目

Convergents of \(e\)

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

\(\sqrt{2} = 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dots}}}}\)

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 + = $ \(1 + \dfrac{1}{2 + \dfrac{1}{2}} = \dfrac{7}{5}\) \(1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}} = \dfrac{17}{12}\) \(1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}}} = \dfrac{41}{29}\)

Hence the sequence of the first ten convergents for \(\sqrt{2}\) are:

\(1, \dfrac{3}{2}, \dfrac{7}{5}, \dfrac{17}{12}, \dfrac{41}{29}, \dfrac{99}{70}, \dfrac{239}{169}, \dfrac{577}{408}, \dfrac{1393}{985}, \dfrac{3363}{2378}, ...\)

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

The first ten terms in the sequence of convergents for \(e\) are:

\(2, 3, \dfrac{8}{3}, \dfrac{11}{4}, \dfrac{19}{7}, \dfrac{87}{32}, \dfrac{106}{39}, \dfrac{193}{71}, \dfrac{1264}{465}, \dfrac{1457}{536}, ...\)

The sum of digits in the numerator of the \(10^{\text{th}}\) convergent is \(1 + 4 + 5 + 7 = 17\).

Find the sum of digits in the numerator of the \(100 ^{\text{th}}\) convergent of the continued fraction for \(e\).

解决方案

假设连分数序列为\(a\)，那么根据连分数的产生规则，第\(n\)个逼近值\(b_{n,1}\)如下构造：

\[b_{n,i}= \left \{\begin{aligned} &a_n & & \text{if}\quad i=n \\ &a_i+\dfrac{1}{b_{n,i+1}} & & \text{else} \end{aligned}\right. \]

直接使用分数进行模拟即可。

代码

from fractions import Fraction

N = 100
a = [2]
v = 0
for i in range(1, N + 1):
    a += [1, 2 * i, 1]
t = a[:N][::-1]
w = Fraction(t[0])
for i in range(1, len(t)):
    w = 1 / w + t[i]
ans = sum(int(x) for x in str(w.numerator))
print(ans)