"""
A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {a1, a2, ... , ak} is called a product-sum number: N = a1 + a2 + ... + ak = a1 × a2 × ... × ak.

For example, 6 = 1 + 2 + 3 = 1 × 2 × 3.

For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k = 2, 3, 4, 5, and 6 are as follows.

k=2: 4 = 2 × 2 = 2 + 2
k=3: 6 = 1 × 2 × 3 = 1 + 2 + 3
k=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4
k=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2
k=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6

Hence for 2≤k≤6, the sum of all the minimal product-sum numbers is 4+6+8+12 = 30; note that 8 is only counted once in the sum.

In fact, as the complete set of minimal product-sum numbers for 2≤k≤12 is {4, 6, 8, 12, 15, 16}, the sum is 61.

What is the sum of all the minimal product-sum numbers for 2≤k≤12000?
"""


def prodsum(p, s, c, start):
    k = p - s + c  # product - sum + number of factors
    if k < kmax:
        if p < n[k]:
            n[k] = p
        for i in range(start, kmax // p * 2 + 1):
            prodsum(p * i, s + i, c + 1, i)


kmax = int(input("Enter a value for kmax?"))
if kmax > 12:
    kmax += 1
n = [2 * kmax] * kmax
prodsum(1, 1, 1, 2)

print("Project Euler 88 Solution =", sum(set(n[2:])))