--description--

The square root of 2 can be written as an infinite continued fraction.

$\sqrt{2} = 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + ...}}}}$

The infinite continued fraction can be written, $\sqrt{2} = [1; (2)]$ indicates that 2 repeats ad infinitum. In a similar way, $\sqrt{23} = [4; (1, 3, 1, 8)]$. It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for $\sqrt{2}$.

$1 + \dfrac{1}{2} = \dfrac{3}{2}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2}} = \dfrac{7}{5}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}} = \dfrac{17}{12}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}}} = \dfrac{41}{29}$

Hence the sequence of the first ten convergents for $\sqrt{2}$ are:

$1, \dfrac{3}{2}, \dfrac{7}{5}, \dfrac{17}{12}, \dfrac{41}{29}, \dfrac{99}{70}, \dfrac{239}{169}, \dfrac{577}{408}, \dfrac{1393}{985}, \dfrac{3363}{2378}, ...$

What is most surprising is that the important mathematical constant, $e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, ... , 1, 2k, 1, ...]$. The first ten terms in the sequence of convergents for e are:

$2, 3, \dfrac{8}{3}, \dfrac{11}{4}, \dfrac{19}{7}, \dfrac{87}{32}, \dfrac{106}{39}, \dfrac{193}{71}, \dfrac{1264}{465}, \dfrac{1457}{536}, ...$

The sum of digits in the numerator of the 10^th convergent is $1 + 4 + 5 + 7 = 17$.

Find the sum of digits in the numerator of the n^th convergent of the continued fraction for e.

--hints--

convergentsOfE(10) should return a number.

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

convergentsOfE(10) should return 17.

assert.strictEqual(convergentsOfE(10), 17);

convergentsOfE(30) should return 53.

assert.strictEqual(convergentsOfE(30), 53);

convergentsOfE(50) should return 91.

assert.strictEqual(convergentsOfE(50), 91);

convergentsOfE(70) should return 169.

assert.strictEqual(convergentsOfE(70), 169);

convergentsOfE(100) should return 272.

assert.strictEqual(convergentsOfE(100), 272);

--seed--

--seed-contents--

function convergentsOfE(n) {

  return true;
}

convergentsOfE(10);

--solutions--

function convergentsOfE(n) {
  function sumDigits(num) {
    let sum = 0n;
    while (num > 0) {
      sum += num % 10n;
      num = num / 10n;
    }
    return parseInt(sum);
  }

  // BigInt is needed for high convergents
  let convergents = [
    [2n, 1n],
    [3n, 1n]
  ];
  const multipliers = [1n, 1n, 2n];
  for (let i = 2; i < n; i++) {
    const [secondLastConvergent, lastConvergent] = convergents;
    const [secondLastNumerator, secondLastDenominator] = secondLastConvergent;
    const [lastNumerator, lastDenominator] = lastConvergent;
    const curMultiplier = multipliers[i % 3];

    const numerator = secondLastNumerator + curMultiplier * lastNumerator;
    const denominator = secondLastDenominator + curMultiplier * lastDenominator;

    convergents = [lastConvergent, [numerator, denominator]]
    if (i % 3 === 2) {
      multipliers[2] += 2n;
    }
  }
  return sumDigits(convergents[1][0]);
}