## Problem Statement

Continued fractions problem - this problem asks about the continued fraction representation of $\sqrt{2}$. In particular, we are to examine the first 1,000 continued fraction terms, determine what they are, and count how many times the numerator has more digits than the denominator.

## Solution Technique

Also see blog post: Computing square roots using continued fractions: https://charlesreid1.github.io/computing-square-roots-part-2-using-continued-fractions.html

This utilizes a recurrence relation for the numerator and denominator, so we can start with the first few terms and obtain as many terms as we wish.

The recurrence relation is: $a_i = a_{i-1} + 2 b_{i-1}$ $b_i = a_{i-1} + b_{i-1}$

For $\sqrt{2}$, we have $a(1) = 3, b(1) = 2$

## Code

    import java.math.BigInteger;
/**
* Find the number of terms in the first 1,000 iterations
* of the continued fraction of sqrt(2) whose denominator
* has more digits than its numerator.
*
* This utilizes the recurrence relation for the nth iteration,
* a(1) = 3,  b(1) = 2
*
* a_i = a_{i-1} + 2 b_{i-1}
*
* b_i = a_{i-1} + b_{i-1}
*
* I actually implemented this in Excel, to begin with, just cuz,
* but these numbers get REALLY huge, REALLY fast.
*/
public class ContinuedFraction {
public static final BigInteger TWO = new BigInteger("2");
public static void main(String[] args) {
BigInteger aim1 = new BigInteger("3");
BigInteger bim1 = new BigInteger("2");
BigInteger ai, bi;

int nterms = 1000;
int noverflows = 0;
for(int i=0; i<nterms; i++) {