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	id, title, challengeType, forumTopicId, dashedName
| id | title | challengeType | forumTopicId | dashedName | 
|---|---|---|---|---|
| 5949b579404977fbaefcd736 | 9 billion names of God the integer | 5 | 302219 | 9-billion-names-of-god-the-integer | 
--description--
This task is a variation of the [short story by Arthur C. Clarke](https://en.wikipedia.org/wiki/The Nine Billion Names of God#Plot_summary "wp: The Nine Billion Names of God#Plot_summary").
(Solvers should be aware of the consequences of completing this task.)
In detail, to specify what is meant by a "name":
- The integer 1 has 1 name "1".
- The integer 2 has 2 names "1+1" and "2".
- The integer 3 has 3 names "1+1+1", "2+1", and "3".
- The integer 4 has 5 names "1+1+1+1", "2+1+1", "2+2", "3+1", "4".
- The integer 5 has 7 names "1+1+1+1+1", "2+1+1+1", "2+2+1", "3+1+1", "3+2", "4+1", "5".
This can be visualized in the following form:
          1
        1   1
      1   1   1
    1   2   1   1
  1   2   2   1   1
1   3   3   2   1   1
Where row n corresponds to integer n, and each column C in row m from left to right corresponds to the number of names beginning with C.
Optionally note that the sum of the $n$-th row P(n) is the integer partition function.
--instructions--
Implement a function that returns the sum of the $n$-th row.
--hints--
numberOfNames should be function.
assert(typeof numberOfNames === 'function');
numberOfNames(5) should equal 7.
assert.equal(numberOfNames(5), 7);
numberOfNames(12) should equal 77.
assert.equal(numberOfNames(12), 77);
numberOfNames(18) should equal 385.
assert.equal(numberOfNames(18), 385);
numberOfNames(23) should equal 1255.
assert.equal(numberOfNames(23), 1255);
numberOfNames(42) should equal 53174.
assert.equal(numberOfNames(42), 53174);
numberOfNames(123) should equal 2552338241.
assert.equal(numberOfNames(123), 2552338241);
--seed--
--seed-contents--
function numberOfNames(num) {
  return true;
}
--solutions--
function numberOfNames(num) {
  const cache = [
    [1]
  ];
  for (let l = cache.length; l < num + 1; l++) {
    let Aa;
    let Mi;
    const r = [0];
    for (let x = 1; x < l + 1; x++) {
      r.push(r[r.length - 1] + (Aa = cache[l - x < 0 ? cache.length - (l - x) : l - x])[(Mi = Math.min(x, l - x)) < 0 ? Aa.length - Mi : Mi]);
    }
    cache.push(r);
  }
  return cache[num][cache[num].length - 1];
}