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id	title	challengeType	forumTopicId	dashedName
5900f3c51000cf542c50fed8	Problem 87: Prime power triples	5	302201	problem-87-prime-power-triples

--description--

The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is 28. In fact, there are exactly four numbers below fifty that can be expressed in such a way:

28 = 2² + 2³ + 2⁴
33 = 3² + 2³ + 2⁴
49 = 5² + 2³ + 2⁴
47 = 2² + 3³ + 2⁴

How many numbers below n can be expressed as the sum of a prime square, prime cube, and prime fourth power?

--hints--

primePowerTriples(50) should return a number.

assert(typeof primePowerTriples(50) === 'number');

primePowerTriples(50) should return 4.

assert.strictEqual(primePowerTriples(50), 4);

primePowerTriples(10035) should return 684.

assert.strictEqual(primePowerTriples(10035), 684);

primePowerTriples(500000) should return 18899.

assert.strictEqual(primePowerTriples(500000), 18899);

primePowerTriples(5000000) should return 138932.

assert.strictEqual(primePowerTriples(5000000), 138932);

primePowerTriples(50000000) should return 1097343.

assert.strictEqual(primePowerTriples(50000000), 1097343);

--seed--

--seed-contents--

function primePowerTriples(n) {

  return true;
}

primePowerTriples(50);

--solutions--

function primePowerTriples(n) {
  function getSievePrimes(max) {
    const primes = [];
    const primesMap = new Array(max).fill(true);
    primesMap[0] = false;
    primesMap[1] = false;

    for (let i = 2; i <= max; i += 2) {
      if (primesMap[i]) {
        primes.push(i);
        for (let j = i * i; j <= max; j = j + i) {
          primesMap[j] = false;
        }
      }
      if (i === 2) {
        i = 1;
      }
    }
    return primes;
  }

  function getPowersSummed(numbers, powers, limit, curSum) {
    if (curSum >= limit) {
      return [];
    } else if (powers.length === 0) {
      return [curSum];
    }

    const powersSummed = [];

    const curPower = powers[0];
    const powersLeft = powers.slice(1);
    for (let i = 0; i < numbers.length; i++) {
      const curNumber = numbers[i];
      const nextSum = curSum + curNumber ** curPower;
      if (nextSum >= limit) {
        return powersSummed;
      }
      const result = getPowersSummed(
        numbers,
        powersLeft,
        limit,
        curSum + curNumber ** curPower
      );
      powersSummed.push(...result);
    }
    return powersSummed;
  }

  const maximumBaseNumber = Math.floor(Math.sqrt(n - 2 ** 3 - 2 ** 4)) + 1;
  const primes = getSievePrimes(maximumBaseNumber);
  const uniqueSums = new Set(getPowersSummed(primes, [2, 3, 4], n, 0));

  return uniqueSums.size;
}

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