def prependPrimerDigit(num):
result = set()
for i in range(1,10):
candidate = str(i)+str(num)
if myMath.isPrime(int(candidate)):
result.add(int(candidate))
return result
#!/usr/bin/python
import myMath
myMath.initPrimeSieve(1000000)
primes = [x for x in xrange(1000000) if myMath.isPrime(x)]
answer = 0
mostConsecutive = 0
for i in xrange(len(primes)):
j = 1
cursum = primes[i]
while i+j < len(primes) and cursum < 1000000:
cursum+=primes[i+j]
if myMath.isPrime(cursum) and j > mostConsecutive:
answer = cursum
mostConsecutive = j
j+=1
print answer
for m in range(2, top+1): for n in range(m, top//m+1): p = m*n if p<=top: seive[p-2] = 0 primes = [] for i in range(top-1): if seive[i] != 0: primes.append(seive[i]) return primes p_max = 10000 # These have to be high enough for the program to produce the correct answer s_max = 1000 # => 10000, 100 start_tot = time() prime_list = primes(p_max)[1:] # Remove the number 2 can_be_written = [] # Get a large list of composites that we can write in such a way for p in prime_list: for s in range(1,s_max+1): n = p + 2*s**2 if not isPrime(n): can_be_written.append(n) # Get large list of odd composites and check whether each element is in the "can_be_written" list max_comp = p_max+2*s_max**2 for n in [n for n in range(1,max_comp,2) if not isPrime(n)]: # The [...] generates odd composites if n not in can_be_written: print n break print "Time taken:", time()-start_tot
myMath.initPrimeSieve(1000000)
#Euler published the remarkable quadratic formula:
#
# n^2 + n + 41
#
# It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41^2 + 41 + 41 is clearly divisible by 41.
#
# Using computers, the incredible formula n^2 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, 79 and 1601, is 126479.
#
# Considering quadratics of the form:
#
# n^2 + an + b, where |a| 1000 and |b| 1000
#
# where |n| is the modulus/absolute value of n
# e.g. |11| = 11 and |4| = 4
# Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.
maxPrimes=(0,0,0)
for a in range(-999,1000):
for b in range(-999,1000):
n=0
while myMath.isPrime(abs(n*n+a*n+b)):
n+=1
if n > maxPrimes[0]:
maxPrimes = (n,a,b)
print "%d: a=%d b=%d" % maxPrimes
print "%d: a=%d b=%d" % maxPrimes
print "Answer: " + str(maxPrimes[1] * maxPrimes[2])
#!/usr/bin/python
#We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
#
#What is the largest n-digit pandigital prime that exists?
import myMath
poss = [x for x in range(1,8)]
largest = 0
while len(poss) > 2:
perms = myMath.getAllPermutations(poss)
for perm in perms:
if perm > largest and myMath.isPrime(perm):
largest = perm
poss = poss[:-1]
print largest
from myMath import isPrime ndiag = 0 nprime = 0 l = 0 # starting layer n = 1 # starting number is the center ratio = 1. while ratio >= 0.1: if ndiag%4 == 0: l += 1 # move to next layer every four diagonal numbers n += 2*l # move to next diagonal number ndiag += 1 # increase number of diagonals traversed if isPrime(n): nprime += 1 # if prime, increase their count if ndiag%4 == 0: # run only when a loop is complete ratio = nprime*1./(ndiag+1) # update ratio, include the starting "1" in the diagonal count print 'Last number:', n, ' Layer:', l, ' Side length:', 2*l+1, ' Ratio:', ratio
maxN = 0 maxA = 0 maxB = 0 a = 0 b = 0 range = 999 def qSum(n,a,b): return n*n+a*n+b if __name__ == '__main__': for a in xrange(-range,range): for b in xrange(-range,range): n = 0 sum = qSum(n,a,b) while (sum>0) and (myMath.isPrime(sum)): if n > maxN: maxN = n maxA = a maxB = b n += 1 sum = qSum(n,a,b) print a,b,n,sum print maxN, maxA, maxB, maxA*maxB
#27 = 19 + 222
#33 = 31 + 212
#
#It turns out that the conjecture was false.
#
#What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
import myMath
primeSieve = myMath.initPrimeSieve(1000000)
goldbachSieve = [False] * 1000000
goldbachSieve[0] = goldbachSieve[1] = True
#weeding out all prime numbers (since it must be composite) and even numbers (since it must be odd)
for x in xrange(len(goldbachSieve)):
if myMath.isPrime(x) or x % 2 == 0:
goldbachSieve[x] = True
def getNextPrime():
global primeSieve
primeSieveLen = len(primeSieve)
i = 1
while i < primeSieveLen:
if primeSieve[i]:
yield i
i += 1
lastPrime = 1
for prime in getNextPrime():
x = 0
def rightToLeftTest(strnum):
for x in xrange(len(strnum)-1):
if myMath.isPrime(int(strnum[:-x-1])) == False:
return False
return True
import myMath
myMath.initPrimeSieve(10000)
def digits(num):
return [int(dig) for dig in str(num)]
def findArithmeticSequence(numSet):
if len(numSet) >= 3:
numSet = sorted(numSet)
for x in numSet:
numSetCopy = numSet[:]
numSetCopy.remove(x)
for y in numSetCopy:
diff = abs(y-x)
if y + diff in numSetCopy:
return (x,y,y+diff)
return ()
answers = []
for i in xrange(1001,10000):
perms = list(set(myMath.getAllPermutations(digits(i))))
perms = [x for x in perms if myMath.isPrime(x) and x > 1000]
result = findArithmeticSequence(perms)
if result != () and result not in answers:
print result
answers.append(result)
#coding=utf-8
# main
import myMath as mm
num = input("请输入一个整数")
mm.mSqrt(num)
mm.mAbs(num)
mm.isPrime(num)
def valid_pair(x,y):
if isPrime(int(x + y)) and isPrime(int(y + x)): return True
return False
import myMath
myMath.initPrimeSieve(1000000)
resultSieve = bytearray(1000001)
def rotateDigits(num):
carried = num%10
return int(str(carried) + str(int(num/10)))
for i in range(2,len(resultSieve)):
if resultSieve[i] != 0 or any([x == '0' for x in str(i)]):
continue
variations = [i]
success = myMath.isPrime(i)
rotated = i
for x in xrange(len(str(i))-1):
rotated = rotateDigits(rotated)
variations.append(rotated)
success = success and myMath.isPrime(variations[x+1])
for x in variations:
resultSieve[x] = 2 if success else 1
answer = 0
for x in resultSieve:
if x == 2:
answer+=1
