def is_sub_prime(n): tf = True for i in range(1, len(str(n))): n_sub = int(str(n)[:-i]) tf = isp.is_prime(n_sub) and tf if(tf): tf2 = True for j in range(1, len(str(n))): tf2 = isp.is_prime(int(str(n)[j:])) and tf2 if(tf and tf2): return n else: return 0
def prime_number_of_divisors(n):
count = 0
for i in range(1, n+1):
if n % i == 0:
count += 1
return is_prime(count)
def prime_palindrome(num):
"""
判断一个数是不是回文素数
:param num: 正整数
:return: 是回文素数返回True，不是回文素数返回False
"""
# 请在此处添加代码 #
# *************begin************#
if (is_palindrome(num) and is_prime(num)):
return True
return False
def prime_number_of_divisors(n):
list_of_divisors = [n]
for i in range(1, n):
if n % i == 0:
list_of_divisors.append(i)
#print(list_of_divisors)
#print(len(list_of_divisors))
if is_prime(len(list_of_divisors)):
return True
else:
return False
def prime_number_of_divisors(n):
list_of_divisors = [n]
for i in range(1,n):
if n % i == 0:
list_of_divisors.append(i)
#print(list_of_divisors)
#print(len(list_of_divisors))
if is_prime(len(list_of_divisors)):
return True
else:
return False
def generate_primes_aux(k_min, k_max, p_max, npick):
assert npick > 0
for k in range(k_min, k_max + 1):
c_start = p_max >> k
if c_start % 2 == 0:
c_start -= 1
while c_start % 3 != 0:
c_start -= 2
nleft = npick
for c in range(c_start, -3, -6):
p = (c << k) + 1
if is_prime(p):
yield p
nleft -= 1
if not nleft:
break
__author__ = "Ajay Rangarajan, Jeyashree Krishnan" __email__ = "rangarajan, [email protected]" import numpy as np import isprime as isp max_a = 0 max_b = 0 max_n = 0 for a in range(-999,1000): for b in range(-1001,1001): n=0 p = 5 while ((p>0) and isp.is_prime(p)==True): n = n + 1 p = n**2 + a*n + b if(n>max_n): max_n = n max_a = a max_b = b print max_a * max_b """ Euler discovered the remarkable quadratic formula: n2+n+41n2+n+41 It turns out that the formula will produce 40 primes for the consecutive integer values 0≤n≤390≤n≤39. However, when n=40,402+40+41=40(40+1)+41n=40,402+40+41=40(40+1)+41 is divisible by 41, and certainly when n=41,412+41+41n=41,412+41+41 is clearly divisible by 41. The incredible formula n2−79n+1601n2−79n+1601 was discovered, which produces 80 primes for the consecutive values 0≤n≤790≤n≤79. The product of the coefficients, −79 and 1601, is −126479.
__author__ = "Ajay Rangarajan, Jeyashree Krishnan" __email__ = "rangarajan, [email protected]" import numpy as np import isprime as isp ## find if a given number is pandigital def is_pandigital(n): d = len(str(n)) sort = "".join(sorted(str(n))) if (d > 9): print 'error' return 0 else: val = '' for i in range(1, d + 1): val = val + str(i) if (sort == val): return n else: return 0 maxval = 90000000 for i in np.arange(maxval, 0, -1): # print 'Entered here!' if (is_pandigital(i)): if (isp.is_prime(i)): print i break
# -*- coding: utf-8 -*-
"""
Created on Thu Sep 24 15:42:33 2015
@author: mbreyt
"""
import isprime
total = 0
for i in range(1, 1001):
if isprime.is_prime(i):
total += i
print(total)
from decimal import * import itertools import isprime as isp def is_sub_prime(n): tf = True for i in range(1, len(str(n))): n_sub = int(str(n)[:-i]) tf = isp.is_prime(n_sub) and tf if(tf): tf2 = True for j in range(1, len(str(n))): tf2 = isp.is_prime(int(str(n)[j:])) and tf2 if(tf and tf2): return n else: return 0 k = 0 n = 10 ans = 0 while(k<12): if(isp.is_prime(n)): if(is_sub_prime(n)): k = k+1 print is_sub_prime(n) ans = ans + is_sub_prime(n) n = n + 1 print ans
string = ''
for j in p:
string = string + j
out.append(string)
return out
def digit(n):
digit_str = []
string = str(n)
for val in string:
digit_str.append(val)
return digit_str
count = 0
for n in range(1, 1000001):
str_digit = digit(n)
nums = longestWord(str_digit)
tf = True
for i in range(0, len(nums)):
tf = isp.is_prime(int(nums[i])) and tf
if (tf == True):
count = count + 1
print nums, ' is circular prime'
print count
# a = list(itertools.permutations(str_digit, len(str_digit)))
# for i in range(0, len(a)):
# print a[1][1]
def is_perfect_sqaure(n):
if (np.floor(n) == n):
return True
else:
return False
# initialize counter for non-convergence of Goldbach's Conjecture
k = 1
search = True
while (search == True):
# initialize starting odd number
odd_no = 2 * k + 1
# Check if the generated odd number is composite. If not increment 'k'
if (isp.is_prime(odd_no) == False):
# 'odd_no' is composite
tf = []
# Generate all odd numbers below the given odd composite number
for i in range(1, k):
prime = 2 * i + 1
# selected_prime = 0
# selected_square = 0
# Check if these odd numbers are prime
if (isp.is_prime(prime)):
# calculate difference between the odd number and any prime number less that itself
diff = odd_no - prime
sq_no = np.sqrt(diff / 2.0)
tf.append(is_perfect_sqaure(sq_no))
if (is_perfect_sqaure(sq_no)):
selected_prime = prime # Select the number for which we have a solution to Goldbach's conjectures
from isprime import is_prime x = int(input()) print(is_prime(x))
def test_is_prime(result, expected):
_ip = isprime.is_prime(result)
assert _ip == expected
