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