def improved_brute_force(self):
primes = list(sieve_of_eratosthenes(10000))
composites = set(range(3, 10000)) - set(primes)
for composite in filter(lambda c: c % 2, composites):
square = 0
while 2 * (square**2) <= composite:
out = composite - 2 * square**2
if out in primes:
break
square += 1
else:
return composite
def brute_force(self):
primes = list(sieve_of_eratosthenes(10000))
primes.remove(1487)
for start in primes:
sequence = [start]
for i in range(1, 5):
if (s := start + (3330 * i)) in primes:
sequence.append(s)
else:
break
if len(sequence) == 3 and self.all_permutations(sequence):
return ''.join(map(str, sequence))
def solution(self):
# Best number, best cycle length
best_n, best_c = 0, 0
# Only really need to check primes
for i in sieve_of_eratosthenes(1000):
# Infinite loop if i is divisible by 2 or 5
if i % 2 != 0 and i % 5 != 0:
cn = self.cycle_length(i)
# Keep track of max
if best_c < cn:
best_n = i
best_c = cn
return best_n
def brute_force(self):
upper_bound = 1_000_000
primes = list(sieve_of_eratosthenes(4000))
max_length = 0
max_prime = 0
for start in range(len(primes)):
current_length = 0
for end in range(start, len(primes)):
s = sum(primes[start:end])
current_length += 1
if s > upper_bound:
break
elif simple_is_prime(s) and current_length > max_length:
max_length = current_length
max_prime = s
return max_prime
def brute_force(self):
primes = list(sieve_of_eratosthenes(10000))
composites = set(range(3, 10000)) - set(primes)
for composite in filter(lambda c: c % 2, composites):
success = False
for prime in takewhile(lambda p: p < composite, primes):
for square in range(1, int(composite**0.5)):
out = prime + (2 * (square**2))
if composite == out:
success = True
break
if success:
break
if not success:
return composite
def prime(self):
primes = sieve_of_eratosthenes(1000)
n = 3
prime_divisors = 2
c = 0
while c <= 500:
print(c)
n += 1
n1 = n
if n1 % 2 == 0:
n1 /= 2
dn1 = 1
for prime in primes:
if prime * prime > n1:
dn1 *= 2
break
exponent = 1
while n1 % prime == 0:
exponent += 1
n1 /= prime
if exponent > 1:
dn1 *= exponent
if n1 == 1:
break
c = prime_divisors * dn1
prime_divisors = dn1
return n * (n - 1) / 2
def sieve_of_eratosthenes(self, n=2_000_000):
return sum(filter(lambda x: x <= n, sieve_of_eratosthenes(n)))
def eratosthenes_nth_prime(self, n=10001):
# This is a mathmatical function which can approximately predict the limit for the nth prime
# https://www.maa.org/sites/default/files/jaroma03200545640.pdf
limit = round(n * log(n) + n * (log(log(n - 0.9385))))
return nth(sieve_of_eratosthenes(limit), n - 1)
