**Problem 57:**

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?

**Idea:**

This problem was deceptively simple: just continue the expansion 1000 times, counting how many times there are more digits in the numerator than the denominator. After many attempts, I couldn’t figure out an iterative way, but then I realized this is a powerful example of recursion, and the “deceptively” part vanished and became simple.

I wrote a base “root2ConvergenceFraction” that would give me the numerator and denominator (BigInteger form, just to be safe) after n number of expansions (the recursion). Did that from 1 to 1000 and counted the number of instances when the numberator had more digits than the denominator.

int answer = 0; BigInteger[] tmp; for(int i = 1; i < = 1000; i++) { tmp = EulerUtils.root2ConvergenceFraction(i); if(tmp[0].toString().length() > tmp[1].toString().length()) { answer++; } }