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 another possibility I just thought of:

If N is X bits long, then the smallest regular number R ≥ N will be in the range
[2X-1, 2X]

e.g. if N = 257 (binary 100000001) then we know R is 1xxxxxxxx unless R is exactly equal to the next power of 2 (512)

To generate all the regular numbers in this range, we can generate the odd regular numbers (i.e. multiples of powers of 3 and 5) first, then take each value and multiply by 2 (by bit-shifting) as many times as necessary to bring it into this range.

In Python:

from itertools import ifilter, takewhile
from Queue import PriorityQueue

def nextPowerOf2(n):
    p = max(1, n)
    while p != (p & -p):
        p += p & -p
    return p

# Generate multiples of powers of 3, 5
def oddRegulars():
    q = PriorityQueue()
    q.put(1)
    prev = None
    while not q.empty():
        n = q.get()
        if n != prev:
            prev = n
            yield n
            if n % 3 == 0:
                q.put(n // 3 * 5)
            q.put(n * 3)

# Generate regular numbers with the same number of bits as n
def regularsCloseTo(n):
    p = nextPowerOf2(n)
    numBits = len(bin(n))
    for i in takewhile(lambda x: x <= p, oddRegulars()):
        yield i << max(0, numBits - len(bin(i)))

def nextRegular(n):
    bigEnough = ifilter(lambda x: x >= n, regularsCloseTo(n))
    return min(bigEnough)