def main():
# bounds
k_max = 6 # this also happens to be as large as any coordinate in a sum/product can be
v = 2 * k_max # this is the largest that any of the sums/products can be
b = k_max - 1 - math.ceil(math.log(k_max)) # this is the lower bound on the number of ones in the solution
print(k_max - b)
# solve k_max case directly to get upper bound
min_sum = math.inf
#N_max = [1] * k_max
#while sum(N_max) != reduce(mul, N_max, 1):
min_sums = {}
# get the prime factorizations of all the possible sums below the bound
prime_facs = [euler.prime_factorization(i) for i in range(2, v + 1)]
mults = [Counter(facs).most_common() for facs in prime_facs]
for m in mults:
count_sum = sum([p[1] for p in m])
N = reduce(mul, [p[0] ** p[1] for p in m], 1)
m = N - sum([p[0] * p[1] for p in m])
k = m + count_sum
if k > 1 and N <= 2 ** (k_max - b):
if k not in min_sums or N < min_sums[k]:
min_sums[k] = N
min_sums = set(min_sums.values())
print(prime_facs)
print(mults)
print(min_sums)
print(sum(min_sums))
def problem():
"""
>>> problem()
12375 th is 76576500
"""
limit = 20000
primes = tuple(primes_limit(limit))
amount_prev = 1
for x in range(2, limit):
if x % 2 == 0:
factors = prime_factorization(x + 1, primes)
else:
factors = prime_factorization((x + 1) // 2, primes)
amount = num_divisors(factors)
total = amount_prev * amount
if total > 500:
break
amount_prev = amount
number = x * (x + 1) // 2
print("%d th is %d" % (x, number))
def dpf(n):
return len(set(prime_factorization(n)))
def largest_prime_factor(n):
return max(prime_factorization(n))
