Find sequence by adding 5 or multiplying by 3
This question is a bit related to a previous question. I will give you the same advice I did to the asker of the previous question.
Work from the other direction.
Currently, you're branching out starting from 1 trying to reach 178. Instead, start from 178 and try to reach 1. I'll show you how that will make the problem a lot easier.

We're at 178. Can we divide by 3? 178 mod 3 != 0 so no. Instead we reduce by 5.
178 - 5 == 173. Can we divide by 3? 173 mod 3 != 0 so no, we reduce by 5 instead.
173 - 5 == 168. Can we divide by 3? 168 mod 3 == 0 so yes, we can. Let's do that.
168 / 3 == 56. Wait a minute... this number ends with a 6... Interesting. (56 - 1) mod 5 equals zero, so from here we can just reduce by 5 until we've reached our target of 1.

Instead of branching out in an exponential manner, we've reduced the problem to a linear approach. Simply by flipping things backwards.
Let's say that we don't want to do the end loop of reducing by 5 just because we can (we want the shortest path, right?), then I see no other solution than to branch out recursively. When checking for the division by 3, a lot of branches are removed so it will be possible to find the shortest path by branching.
I'll leave the fun part of implementing this up to you :)
Some comments about your current code
(Which, given the comments above you should remove and completely re-write using the &quot;backwards&quot; approach)
In Java at least, it's recommended to add braces for each if. Also, as you're using return inside each if there's no need for else.
function findSequence(goal) {
  function find(start, history) {
    if (start == goal) {
      return history;
    }
    if (start &gt; goal) {
      return null;
    }
    return find(start + 5, &quot;(&quot; + history + &quot; + 5)&quot;) ||
             find(start * 3, &quot;(&quot; + history + &quot; * 3)&quot;);
  }
  return find(1, &quot;1&quot;);
}

I would also add a parameter to allow switching the start value, it will make it more flexible very easily.
I think it's good that you're hiding the find method inside the other function.