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)
def check_truncation(nNum):
anDigits = []
anToCheck = []
nExponent = 0
nGrowing = 0
while nNum > 0:
anToCheck.append(nNum)
nGrowing = nGrowing + (nNum %10)*(10**nExponent)
anToCheck.append(nGrowing)
#anToCheck.append( sum( map( lambda(x): anDigits.append( nNum%10 ))))
nNum/=10
nExponent += 1
bToReturn = True
for nToCheck in anToCheck:
if not pfp.is_prime(nToCheck) or nToCheck == 1:
bToReturn = False
return bToReturn
def next_odd_composite(nOn):
nOn += 2
while pfp.is_prime(nOn):
nOn +=2
return nOn
import pfp
import enum_pan_digital
#using array of digits over number for ease of setting up next
anDigits = [9,8,7,6,5,4,3,2,1] #digits from least significant to most.
nCount =0 #debugging number of times through loop checking that next works
nLen = 9
while True:
#smallest possible ...
nToCheck = sum(map(lambda (nIndex, nElement): nElement*(10**(len(anDigits) - 1 - nIndex)), enumerate(anDigits)))
#if len(anDigits) != nLen:
# print(nToCheck)
if nToCheck == 2143:
print('should be prime')
if pfp.is_prime(nToCheck):
print(nToCheck)
break
else:
#print(anDigits)
anDigits = enum_pan_digital.enum_pan_digital(anDigits)
#print(anDigits)
#if nCount > 500000:
# break
#nCount += 1