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title	id	challengeType	forumTopicId
Amicable pairs	5949b579404977fbaefcd737	5	302225

Description

Two integers $N$ and $M$ are said to be amicable pairs if $N \neq M$ and the sum of the proper divisors of $N$ ($\mathrm{sum}(\mathrm{propDivs}(N))$) $= M$ as well as $\mathrm{sum}(\mathrm{propDivs}(M)) = N$. Example: 1184 and 1210 are an amicable pair, with proper divisors:

1, 2, 4, 8, 16, 32, 37, 74, 148, 296, 592 and
1, 2, 5, 10, 11, 22, 55, 110, 121, 242, 605 respectively.

Instructions

Calculate and show here the Amicable pairs below 20,000 (there are eight).

Tests

tests:
  - text: <code>amicablePairsUpTo</code> should be a function.
    testString: assert(typeof amicablePairsUpTo === 'function');
  - text: <code>amicablePairsUpTo(300)</code> should return <code>[[220,284]]</code>.
    testString: assert.deepEqual(amicablePairsUpTo(300), answer300);
  - text: <code>amicablePairsUpTo(3000)</code> should return <code>[[220,284],[1184,1210],[2620,2924]]</code>.
    testString: assert.deepEqual(amicablePairsUpTo(3000), answer3000);
  - text: <code>amicablePairsUpTo(20000)</code> should return <code>[[220,284],[1184,1210],[2620,2924],[5020,5564],[6232,6368],[10744,10856],[12285,14595],[17296,18416]]</code>.
    testString: assert.deepEqual(amicablePairsUpTo(20000), answer20000);

Challenge Seed

function amicablePairsUpTo(maxNum) {

  return true;
}

After Test

const answer300 = [[220, 284]];
const answer3000 = [
  [220, 284],
  [1184, 1210],
  [2620, 2924]
];
const answer20000 = [
  [220, 284],
  [1184, 1210],
  [2620, 2924],
  [5020, 5564],
  [6232, 6368],
  [10744, 10856],
  [12285, 14595],
  [17296, 18416]
];

Solution

// amicablePairsUpTo :: Int -> [(Int, Int)]
function amicablePairsUpTo(maxNum) {
  return range(1, maxNum)
    .map(x => properDivisors(x)
      .reduce((a, b) => a + b, 0))
    .reduce((a, m, i, lst) => {
      const n = i + 1;

      return (m > n) && lst[m - 1] === n ?
        a.concat([
          [n, m]
        ]) : a;
    }, []);
}

// properDivisors :: Int -> [Int]
function properDivisors(n) {
  if (n < 2) return [];

  const rRoot = Math.sqrt(n);
  const intRoot = Math.floor(rRoot);
  const blnPerfectSquare = rRoot === intRoot;
  const lows = range(1, intRoot)
  .filter(x => (n % x) === 0);

  return lows.concat(lows.slice(1)
    .map(x => n / x)
    .reverse()
    .slice(blnPerfectSquare | 0));
}

// Int -> Int -> Maybe Int -> [Int]
function range(m, n, step) {
  const d = (step || 1) * (n >= m ? 1 : -1);

  return Array.from({
    length: Math.floor((n - m) / d) + 1
  }, (_, i) => m + (i * d));
}

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