--description--

Consider the fraction, $\frac{n}{d}$, where n and d are positive integers. If n < d and highest common factor, ${HCF}(n, d) = 1$, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:

$$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}$$

It can be seen that there are 21 elements in this set.

How many elements would be contained in the set of reduced proper fractions for d ≤ limit?

--hints--

countingFractions(8) should return a number.

assert(typeof countingFractions(8) === 'number');

countingFractions(8) should return 21.

assert.strictEqual(countingFractions(8), 21);

countingFractions(20000) should return 121590395.

assert.strictEqual(countingFractions(20000), 121590395);

countingFractions(500000) should return 75991039675.

assert.strictEqual(countingFractions(500000), 75991039675);

countingFractions(1000000) should return 303963552391.

assert.strictEqual(countingFractions(1000000), 303963552391);

--seed--

--seed-contents--

function countingFractions(limit) {

  return true;
}

countingFractions(8);

--solutions--

function countingFractions(limit) {
  const phi = {};
  let count = 0;

  for (let i = 2; i <= limit; i++) {
    if (!phi[i]) {
      phi[i] = i;
    }
    if (phi[i] === i) {
      for (let j = i; j <= limit; j += i) {
        if (!phi[j]) {
          phi[j] = j;
        }
        phi[j] = (phi[j] / i) * (i - 1);
      }
    }
    count += phi[i];
  }

  return count;
}