Problem 57
Square root convergents
It is possible to show that the square root of two can be expressed as an infinite continued fraction.
$$\sqrt 2 =1+ \frac 1 {2+ \frac 1 {2 +\frac 1 {2+ \dots}}}$$
By expanding this for the first four iterations, we get:
$1 + \frac 1 2 = \frac 32 = 1.5$
$1 + \frac 1 {2 + \frac 1 2} = \frac 7 5 = 1.4$
$1 + \frac 1 {2 + \frac 1 {2+\frac 1 2}} = \frac {17}{12} = 1.41666 \dots$
$1 + \frac 1 {2 + \frac 1 {2+\frac 1 {2+\frac 1 2}}} = \frac {41}{29} = 1.41379 \dots$
The next three expansions are $\frac{99}{70}$, $\frac{239}{169}$, and $\frac{577}{408}$, but the eighth expansion, $\frac{1393}{985}$, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?
平方根逼近
$2$的平方根可以表示为如下无限连分数：
$$\sqrt 2 =1+ \frac 1 {2+ \frac 1 {2 +\frac 1 {2+ \dots}}}$$
将这个连分数进行四次迭代展开，分别可以得到：
$1 + \frac 1 2 = \frac 32 = 1.5$
$1 + \frac 1 {2 + \frac 1 2} = \frac 7 5 = 1.4$
$1 + \frac 1 {2 + \frac 1 {2+\frac 1 2}} = \frac {17}{12} = 1.41666 \dots$
$1 + \frac 1 {2 + \frac 1 {2+\frac 1 {2+\frac 1 2}}} = \frac {41}{29} = 1.41379 \dots$
接下来的三次迭代展开得到的分别是$\frac{99}{70}$、$\frac{239}{169}$和$\frac{577}{408}$，直到第八次迭代展开$\frac{1393}{985}$，分子的位数第一次超过分母的位数。
在前一千次迭代展开中，有多少个分数满足分子的位数多于分母的位数？