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

Here's a solution in Python, based on Will Ness answer but taking some shortcuts and using pure integer math to avoid running into log space numerical accuracy errors:

import math

def next_regular(target):
    """
    Find the next regular number greater than or equal to target.
    """
    # Check if it's already a power of 2 (or a non-integer)
    try:
        if not (target & (target-1)):
            return target
    except TypeError:
        # Convert floats/decimals for further processing
        target = int(math.ceil(target))

    if target <= 6:
        return target

    match = float('inf') # Anything found will be smaller
    p5 = 1
    while p5 < target:
        p35 = p5
        while p35 < target:
            # Ceiling integer division, avoiding conversion to float
            # (quotient = ceil(target / p35))
            # From https://stackoverflow.com/a/17511341/125507
            quotient = -(-target // p35)

            # Quickly find next power of 2 >= quotient
            # See https://stackoverflow.com/a/19164783/125507
            try:
                p2 = 2**((quotient - 1).bit_length())
            except AttributeError:
                # Fallback for Python <2.7
                p2 = 2**(len(bin(quotient - 1)) - 2)

            N = p2 * p35
            if N == target:
                return N
            elif N < match:
                match = N
            p35 *= 3
            if p35 == target:
                return p35
        if p35 < match:
            match = p35
        p5 *= 5
        if p5 == target:
            return p5
    if p5 < match:
        match = p5
    return match

In English: iterate through every combination of 5s and 3s, quickly finding the next power of 2 >= target for each pair and keeping the smallest result. (It's a waste of time to iterate through every possible multiple of 2 if only one of them can be correct). It also returns early if it ever finds that the target is already a regular number, though this is not strictly necessary.

I've tested it pretty thoroughly, testing every integer from 0 to 51200000 and comparing to the list on OEIS http://oeis.org/A051037, as well as many large numbers that are ±1 from regular numbers, etc. It's now available in SciPy as fftpack.helper.next_fast_len, to find optimal sizes for FFTs (source code).

I'm not sure if the log method is faster because I couldn't get it to work reliably enough to test it. I think it has a similar number of operations, though? I'm not sure, but this is reasonably fast. Takes <3 seconds (or 0.7 second with gmpy) to calculate that 2¹⁴² × 3⁸⁰ × 5⁴⁴⁴ is the next regular number above 2² × 3⁴⁵⁴ × 5²⁴⁹+1 (the 100,000,000th regular number, which has 392 digits)