Project Euler 21 - Amicable Numbers - Solved with JavaScript

const getProperDivisors = (n) => {
  const divisors = [];
  const root = Math.floor(Math.sqrt(n));
  for (let i = 1; i <= root; i++) {
    if (n % i === 0) {
      divisors.push(i);
      if (Math.pow(i, 2) !== n) {
        divisors.push(n / i);
      }
    }
  }
  return divisors.sort(function (a, b) {
    return a - b;
  }).filter(x => x !== n);
};

// Let d(n) be defined as the sum of proper divisors of
// n (numbers less than n which divide evenly into n).
const d = (n) => {
  const divisors = getProperDivisors(n);
  return divisors.reduce((a,b) => a + b, 0);
};

// Iterate over all numbers from 0 to 9999.
const result = Array.from({length: 9999}).map((_, n) => {
    const a = d(n);
    const b = d(a);
    // If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable
    // pair and each of a and b are called amicable numbers.
    return n !== a && n === b ? a : 0;
}).reduce((a,b) => a + b);