import pfp
#should probably just pickly the list of the first million primes ... but whatever.
anPrimes = pfp.gen_prime_list(1000000)
print(len(anPrimes))
#can I search the whole space, start at max, back away, contain smallest, while this is too big keep reducing
nMaxLen = len(anPrimes) - 1
nToCheck = sum(anPrimes[:nMaxLen])
while nToCheck > anPrimes[-1]:
#nToCheck = sum(anPrimes[:nMaxLen])
nToCheck -= anPrimes[nMaxLen]
nMaxLen -= 1
print(nMaxLen, 'first check') #
bFound = False
for nLen in range(nMaxLen,22,-1):
#all ranged of lengths starting at max
for nStart in range(0, len(anPrimes) - nLen): #start to end window sized
nToCheck = sum(anPrimes[nStart:nStart+nLen])
#could use same -= trick as above to speed this up, but not important pretty small search spaces.
if nToCheck in anPrimes:
print('new winner ', nToCheck, nLen)
bFound = True
break
if nToCheck > anPrimes[-1]: #no reason to keep searching range once past a million
break
if bFound:
break
def test(a,b,c):
bToReturn = True
if not pfp.is_prime(a):
bToReturn = False
if bToReturn and not pfp.is_prime(b):
bToReturn = False
if bToReturn and not pfp.is_prime(c):
bToReturn = False
if bToReturn and c-b != b-a:
bToReturn = False
comp_digits(a,b,c)
#start by generating list of all primes,
anPrimes = pfp.gen_prime_list(9999)
#print(len(anPrimes), anPrimes)
#find arithmatic
while anPrimes[0] < 1000:
anPrimes.remove(anPrimes[0])
for nIndex, nPrime in enumerate(anPrimes):
#print(nIndex, nPrime)
for nPrimeTwo in anPrimes[nIndex + 1:]:
if (nPrimeTwo + (nPrimeTwo - nPrime)) in anPrimes:
#all three are primes
if comp_digits(nPrime,nPrimeTwo, nPrimeTwo + (nPrimeTwo - nPrime)):
print(nPrime,nPrimeTwo, nPrimeTwo + (nPrimeTwo - nPrime))