def sumOfConsectutivePrimes(limit=1000000):
p.genPrimesESieve(1000000, True)
i, j = 1, 2
runningSum, maxPrime = p.primes[i], 0
result, longest = runningSum, 1
for i in xrange(1, len(p.primes)):
j = i+1
count = 1
runningSum = p.primes[i]
if runningSum >= 40:
break
curLongest, curMaxPrime = 2, 0
while runningSum+p.primes[j] < limit:
count +=1
runningSum += p.primes[j]
if p.isPrime(runningSum):
curLongest = count
curMaxPrime = runningSum
j += 1
if curLongest >= longest:
result = curMaxPrime
longest = curLongest
print longest, result
return longest, result
def coefficientProducts(a=1000, b=1000):
aMax, bMax, nMax = (0, 0, 0);
for a in xrange(-1000, 1001):
for b in xrange(-1000, 1001):
n = 0
while p.isPrime(abs(n*n + a*n + b)):
n += 1
if (n > nMax):
aMax = a
bMax = b
nMax = n
return aMax*bMax
def coefficientProducts2():
aMax, bMax, nMax = (0, 0, 0)
for a in xrange(-999, 1001, 2):
key = 1
while p.primes[key] <= 1000:
for sign in (1, -1):
n = 0;
offset = -1 if p.primes[key]%2 == 0 else 0 # Making a even if b is even
while p.isPrime(abs(n*n + (a+offset)*n + sign*p.primes[key])):
n += 1
if (n > nMax):
aMax = a;
bMax = p.primes[key];
nMax = n;
key += 1
return aMax*bMax, nMax
def truncatablePrime(n, leftToRight=True):
'''Returns if all numbers of length ≤ 1 are also prime.
leftToRight flag determines in which direction the new number
is built up from.
'''
n = str(n)
if not leftToRight:
n = n[::-1]
newNumber = ''
for i in xrange(len(n)-1):
if leftToRight:
newNumber += n[i]
else:
newNumber = n[i] + newNumber
if not p.isPrime(int(newNumber)):
return False
return True
