--description--

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

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 n expansions, how many fractions contain a numerator with more digits than denominator?

--hints--

squareRootConvergents(10) should return a number.

assert(typeof squareRootConvergents(10) === 'number');

squareRootConvergents(10) should return 1.

assert.strictEqual(squareRootConvergents(10), 1);

squareRootConvergents(100) should return 15.

assert.strictEqual(squareRootConvergents(100), 15);

squareRootConvergents(1000) should return 153.

assert.strictEqual(squareRootConvergents(1000), 153);

--seed--

--seed-contents--

function squareRootConvergents(n) {

  return true;
}

squareRootConvergents(1000);

--solutions--

function squareRootConvergents(n) {
  function countDigits(number) {
    let counter = 0;
    while (number > 0) {
      counter++;
      number = number / 10n;
    }
    return counter;
  }

  // Use BigInt as integer won't handle all cases
  let numerator = 3n;
  let denominator = 2n;
  let moreDigitsInNumerator = 0;

  for (let i = 2; i <= n; i++) {
    [numerator, denominator] = [
      numerator + 2n * denominator,
      denominator + numerator
    ];

    if (countDigits(numerator) > countDigits(denominator)) {
      moreDigitsInNumerator++;
    }
  }
  return moreDigitsInNumerator;
}