def problem58():
primes = 0
for shell in count(1):
# We only need to check the 4 cornes and there is an easy formula to
# generate all of them
primes += sum(1 for i in range(1, 4)
if isPrime((2 * shell + 1) ** 2 - 2 * i * shell))
if primes / (4 * shell + 1) < 0.10:
return 2 * shell + 1
def problem49():
primes = primesUpTo(10000)
for a in dropwhile(lambda x: x <= 1487, primes):
# the bound comes from wanting c = 2b - a ≤ 10000
for b in dropwhile(lambda x: x <= a, primes):
if b >= (10000 + a) / 2:
break
c = b + (b - a)
# Do the same digits first since I think that is the most time
# consuming part
if sameDigits(a, b) and sameDigits(b, c) and isPrime(c):
return int(str(a) + str(b) + str(c))
def problem50(): GOAL = 10**6 listOfPrimes = primesUpTo(GOAL) recordLength = 21 maxPrime = 953 for index in count(): for length, running in enumerate(accumulate(listOfPrimes[index:])): if length + 1 > recordLength and isPrime(running): recordLength = length + 1 maxPrime = running if running > GOAL: break # If there is no hope, break out if sum(listOfPrimes[index: index+recordLength]) > GOAL: break return maxPrime
def problem27():
record = 0
ab = 0
checker = Orderedlist(primesUpTo(1998))
# We know there is a record of 80
# this removes many possibilities for low values of b
# since the most number of primes for f(n) is b-1
for b in checker[80:1000]:
for p1 in checker[0:1000 + b + 1]:
a = p1 - b - 1
n = 2
p2 = a + p1 + (2 * n - 1)
while isPrime(abs(p2)):
n += 1
p2 = a + p2 + (2 * n - 1)
if n > record:
record = n
ab = a * b
return ab
def primeCheck(i,s):
return isPrime(int(m.replace(i, s)))
def remarkable(p,q): return isPrime(int(str(q) + str(p))) and isPrime(int(str(p) + str(q)))