Find the smallest regular number that is not less than N

Question

Regular numbers are numbers that evenly divide powers of 60. As an example, 60² = 3600 = 48 × 75, so both 48 and 75 are divisors of a power of 60.

Answer 1

You know what? I'll put money on the proposition that actually, the 'dumb' algorithm is fastest. This is based on the observation that the next regular number does not, in general, seem to be much larger than the given input. So simply start counting up, and after each increment, refactor and see if you've found a regular number. But create one processing thread for each available core you have, and for N cores have each thread examine every Nth number. When each thread has found a number or crossed the power-of-2 threshold, compare the results (keep a running best number) and there you are.

Answer 2

Okay, hopefully third time's a charm here. A recursive, branching algorithm for an initial input of p, where N is the number being 'built' within each thread. NB 3a-c here are launched as separate threads or otherwise done (quasi-)asynchronously.

1. Calculate the next-largest power of 2 after p, call this R. N = p.

2. Is N > R? Quit this thread. Is p composed of only small prime factors? You're done. Otherwise, go to step 3.

3. After any of 3a-c, go to step 4.

   a) Round p up to the nearest multiple of 2. This number can be expressed as m * 2.
   b) Round p up to the nearest multiple of 3. This number can be expressed as m * 3.
   c) Round p up to the nearest multiple of 5. This number can be expressed as m * 5.

4. Go to step 2, with p = m.

I've omitted the bookkeeping to do regarding keeping track of N but that's fairly straightforward I take it.

Edit: Forgot 6, thanks ypercube.

Edit 2: Had this up to 30, (5, 6, 10, 15, 30) realized that was unnecessary, took that out.

Edit 3: (The last one I promise!) Added the power-of-30 check, which helps prevent this algorithm from eating up all your RAM.

Edit 4: Changed power-of-30 to power-of-2, per finnw's observation.