def count_inverse_squares(number):
prime_list = list(reversed(list(primes(number / 2 + 1))))
factories = dict(zip(count(3), factorize_to(number)[2:]))
data = defaultdict(list)
for n, fact in factories.items():
data[max(fact.keys())].append((Fraction(1, n * n), (n, )))
last_choices = [(0, tuple())]
all_choices = [(0, tuple())]
for p in prime_list:
choices = data[p]
for values, lc in product(power_set(choices, with_empty=False), last_choices):
values = values + [lc]
value = sum(v for v, l in values)
lst = tuple(chain(*[l for v, l in values]))
if p == 2:
if value == Fraction(1, 4):
yield sorted(lst)
print sorted(lst)
else:
if value.denominator % p != 0 and value <= Fraction(1, 4):
all_choices.append((value, lst))
last_choices = all_choices
all_choices = [(0, tuple())]
def bruteIt():
limit = 10000
s = sieve(limit)
# Only odds numbers
for candidate in range(3, limit, 2):
# Only composite
if not s[candidate]:
# Brute : test all the primes smaller than candidate and check the
# property of the substraction
isGoldbach = False
for p in primes(candidate):
delta = candidate - p
if (int(math.sqrt(delta/2)) == math.sqrt(delta/2)):
isGoldbach = True
break
if not isGoldbach:
print(candidate)
break
def main():
ps = primes(10**6)
biggest = 1
for start in range(len(ps)):
acc = 0
nbTerms = 0
if len(ps[start:]) < biggest:
continue
for p in ps[start:]:
nbTerms += 1
acc += p
if acc > 10**6:
continue
if isPrime(acc) and nbTerms > biggest:
biggest = nbTerms
print("prime ", acc, " with ", nbTerms, " terms")
def main():
ps = dropwhile(lambda x: x < 1000, primes(10000))
for prime in ps:
perms = list(filter(isPrime,
map(int,
set(
map("".join,
permutations(str(prime)))))))
# look for a serie
for first in perms:
for second in perms:
if first >= second:
continue
delta = second - first
# looks for a potential third
if second + delta in perms:
print(first, second, second + delta)
return
def main():
ps = primes(10**6)
for p in dropwhile(lambda x: x <= 56003, ps):
if test(p):
return
