def main(verbose=False):
MAX_n = 10 ** 6 - 1
distinct_solutions = 10
PRIMES = sieve(int(sqrt(MAX_n)) + 1)
factor_hash = {}
count = 0
for n in range(1, MAX_n + 1):
factor_list = factors(n, factor_hash, PRIMES)
if len(num_solutions(factor_list)) == distinct_solutions:
count += 1
return count
def all_values_on_form(form, value):
"""
Returns all lattice points (not necessarily coprime)
that produce the desired value on the form
Given the recurrence for the form, these values
can serve to determine *all* solutions for
the given value due to the repeating nature
of the infinite river
"""
factor_list = factors(value)
valid_factors = [factor for factor in factor_list
if is_power(value/factor, 2)]
roots = all_positive_roots(form)
found = set()
for root in roots:
candidates = seek_up_to_val(root, value)
to_add = [candidate for candidate in candidates
if candidate[1] in valid_factors] + \
[candidate for candidate in root
if candidate[1] in valid_factors]
found.update(to_add)
found = list(found)
# We may get some duplicates from since when we include
# values from the river, we don't check that they come from
# a different iteration of the river
x_mult, y_mult, _ = get_recurrence(form)
checked = found[:]
for candidate in found:
coords, val = candidate
next_x = sum(operator.mul(*pair) for pair in zip(coords, x_mult))
next_y = sum(operator.mul(*pair) for pair in zip(coords, y_mult))
if ((next_x, next_y), val) in found:
checked.remove(((next_x, next_y), val))
# Finally we must scale up factors to account for
# the reduction by a square multiple
result = []
for cell in checked:
(x, y), val = cell
if val < value:
ratio = int(sqrt(value/val))
x *= ratio
y *= ratio
result.append((x,y))
return result
