Problem Description

It is possible to show that the square root of two can be expressed as an infinite continued fraction.

√ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + … ))) = 1.414213…

By expanding this for the first four iterations, we get:

1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666…
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379…

The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 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?

Analysis

Expand the series as 2d + n for the numerator and d+n for the denominator and count the number of times the length of the numerator exceeds the denominator as strings.

Solution

Runs < 1 seconds in Python.

n, d, c = 3, 2, 0
 
for x in range(2, 1000):
    n, d = n + d * 2, n + d
    if len(str(n)) > len(str(d)): c += 1
 
print "Answer to PE57 = ",c

Comments

We start the first terms of the series by initializing n to 3 and d to 2 as shown in the description.