Problem 57 : Square root convergents

Problem Statement

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 the denominator?

Solution

import math as mt

n=3
d=2
count =0

for i in range(2,1001):
    n,d = 2*d+n,n+d
    if int(mt.log10(n)) > int(mt.log10(d)): count += 1

print(count)

Output

153