汉明数，即任意整数 n = 2^i * 3^j * 5^k  ijk为非负整数

fun(1) = (0, 0, 0) = 1
fun(2) = (1, 0, 0) = 2
fun(3) = (0, 1, 0) = 3
fun(4) = (2, 0, 0) = 4
fun(5) = (0, 0, 1) = 5
fun(6) = (1, 1, 0) = 6
fun(7) = (3, 0, 0) = 8
fun(8) = (0, 2, 0) = 9
fun(9) = (1, 0, 1) = 10
fun(10) = (2, 1, 0) = 12

i=0, j=0, k=0, a[0]=1;
2*a[0]、3*a[0]、5*a[0] 中 2*a[0] 最小，所以 a[1]=2*a[0]=2，i=1;
2*a[1]、3*a[0]、5*a[0] 中 3*a[0] 最小，所以 a[2]=3*a[0]=3, j=1;
2*a[1]、3*a[1]、5*a[0] 中 2*a[1] 最小，所以 a[3]=2*a[1]=4, i=2;
……
求 2*a[i]、3*a[j]、5*a[k] 中哪个最小，那就就是下一个fun，相应的ijk加1

def hamming(n):
    s = 1
    import collections
    q2 = collections.deque()
    q3 = collections.deque()
    q5 = collections.deque()
    for x in range(n - 1):
        q2.append(s * 2)
        q3.append(s * 3)
        q5.append(s * 5)
        s = min(q2[0], q3[0], q5[0])
        if q2[0] == s: q2.popleft()
        if q3[0] == s: q3.popleft()
        if q5[0] == s: q5.popleft()
    return s

def hamming(n):
    bases = [2, 3, 5]
    expos = [0, 0, 0]
    hamms = [1]
    for _ in range(1, n):
        next_hamms = [bases[i] * hamms[expos[i]] for i in range(3)]
        next_hamm = min(next_hamms)
        hamms.append(next_hamm)
        for i in range(3):
            expos[i] += int(next_hamms[i] == next_hamm)
    return hamms[-1]

数学底子差了，看不懂。