--description--

A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, $\{a_1, a_2, \ldots , a_k\}$ is called a product-sum number: $N = a_1 + a_2 + \cdots + a_k = a_1 × a_2 × \cdots × a_k$.

For example, 6 = 1 + 2 + 3 = 1 × 2 × 3.

For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k = 2, 3, 4, 5, and 6 are as follows.

<var>k</var>=3: 6 = 1 × 2 × 3 = 1 + 2 + 3

<var>k</var>=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4

<var>k</var>=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2

<var>k</var>=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6

Hence for 2 ≤ k ≤ 6, the sum of all the minimal product-sum numbers is 4 + 6 + 8 + 12 = 30; note that 8 is only counted once in the sum.

In fact, as the complete set of minimal product-sum numbers for 2 ≤ k ≤ 12 is $\{4, 6, 8, 12, 15, 16\}$, the sum is 61.

What is the sum of all the minimal product-sum numbers for 2 ≤ k ≤ limit?

--hints--

productSumNumbers(6) should return a number.

assert(typeof productSumNumbers(6) === 'number');

productSumNumbers(6) should return 30.

assert.strictEqual(productSumNumbers(6), 30);

productSumNumbers(12) should return 61.

assert.strictEqual(productSumNumbers(12), 61);

productSumNumbers(300) should return 12686.

assert.strictEqual(productSumNumbers(300), 12686);

productSumNumbers(6000) should return 2125990.

assert.strictEqual(productSumNumbers(6000), 2125990);

productSumNumbers(12000) should return 7587457.

assert.strictEqual(productSumNumbers(12000), 7587457);

--seed--

--seed-contents--

function productSumNumbers(limit) {

  return true;
}

productSumNumbers(6);

--solutions--

function productSumNumbers(limit) {
  function getProductSums(curProduct, curSum, factorsCount, start) {
    const k = curProduct - curSum + factorsCount;
    if (k <= limit) {
      if (curProduct < minimalProductSums[k]) {
        minimalProductSums[k] = curProduct;
      }
      for (let i = start; i < Math.floor((limit / curProduct) * 2) + 1; i++) {
        getProductSums(curProduct * i, curSum + i, factorsCount + 1, i);
      }
    }
  }

  const minimalProductSums = new Array(limit + 1).fill(2 * limit);
  getProductSums(1, 1, 1, 2);

  const uniqueProductSums = [...new Set(minimalProductSums.slice(2))];

  let sum = 0;
  for (let i = 0; i < uniqueProductSums.length; i++) {
    sum += uniqueProductSums[i];
  }

  return sum;
}