def compositenum(number):
if is_prime(number):
return False
primelist = [i for i in range(start-1,1,-1) if is_prime(i)]
for i in primelist:
for j in range(1,int(number**0.5)+1):
if i + 2*j**2 == number:
return False
return True
def qf(a, b):
ip = True
n = -1
while ip:
n = n + 1
ip = is_prime(n**2 + a*n + b)
return n
def circular_right_prime(num):
res = [x for x in str(num)]
for i in range(len(res)):
if is_prime(convert(res)) is True:
a = res.pop(0)
res.append(a)
else:
return False
return True
from tools import is_prime_old
fl = [1,2,3,4,5,6,7]#,8]#,9]
l = fl[len(fl)-1]
pands = []
def permut(list, string):
val = 0
if string != "":
val = int(string)
pands.append(val)
for i in range(0, len(list)):
tmp = list[i]
del list[i]
string = string + str(tmp)
permut(list, string)
list.insert(i, tmp)
string = string[:-1]
permut(fl, "")
pands.sort()
pands.reverse()
for pan in pands:
if is_prime(pan):
print pan
break
def is_composite_odd(x):
if (is_prime(x) is False) and (x % 2 != 0):
return True
else:
return False
def is_composite_odd(x):
if (is_prime(x) is False) and (x % 2 != 0):
return True
else:
return False
prime_list = []
square_list = [2]
for x in range(2, large_num):
square_list.append(2*x*x)
if is_prime(x) is True:
prime_list.append(x)
if is_composite_odd(x) is True:
found: bool = True
# check if it's a prime + twice the square of any number
for prime in prime_list:
if prime > x:
break
for square in square_list:
if (square + prime == x):
found = False
break
# By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
# What is the 10 001st prime number?
from tools import is_prime
primes = []
n = 3
primes.append(2)
while len(primes) <= 10000:
if is_prime(n):
primes.append(n)
n += 2
print primes[10000]
def check_is_perm(number):
return is_prime(number+3330) and sorted([int(i) for i in str(number)])==sorted([int(i) for i in str(number+3330)])
known_primes = [-1]*max
sum = 0
for i in range(min,max):
s = str(i)
t = []
prime = True
for j in range(0,len(s)):
temp = s[j:].lstrip('0')
if temp != '':
t.append(temp)
for j in range(1,len(s)+1):
temp = s[:j]
if temp != '':
t.append(temp)
for sp in t:
i = int(sp)
if known_primes[i] == 0:
prime = False
break
if known_primes[i] == -1:
if is_prime(i):
known_primes[i] = 1
else:
known_primes[i] = 0
prime = False
break
if prime:
sum += i
print sum
t3=6
c=1
t4=8
d=1
nr = 0
nr_of_primes = 0
while True:
a += t1
t1 += 8
b += t2
t2 += 8
c += t3
t3 += 8
d += t4
t4 += 8
nr += 4
if is_prime(a):
nr_of_primes += 1
if is_prime(b):
nr_of_primes += 1
if is_prime(c):
nr_of_primes += 1
#if is_prime(d):
# nr_of_primes += 1
if nr_of_primes*10 < nr:
break
print b-a-1
import itertools
import tools
four_digit_primes = []
for n in range(1001, 9999, 2):
if tools.is_prime(n):
four_digit_primes.append(n)
print(four_digit_primes)
for prime_1 in four_digit_primes:
permutations_str = list(itertools.permutations([e for e in str(prime_1)]))
permutations = []
for permutation in permutations_str:
permutations.append(int(''.join(e for e in permutation)))
for prime_2 in list((set(four_digit_primes) & set(permutations)) -
set([prime_1])):
for prime_3 in list((set(four_digit_primes) & set(permutations)) -
set([prime_1, prime_2])):
if abs(prime_1 - prime_2) == abs(prime_2 - prime_3) or abs(
prime_1 - prime_2) == abs(prime_1 - prime_3) or abs(
prime_1 - prime_3) == abs(prime_2 - 3):
print(prime_1, prime_2, prime_3)
from tools import is_prime
best = 0
for a in range(-1000, 1000):
for b in range(-1000, 1000):
running = True
n = 0
while running:
num = n * n + a * n + b
if is_prime(num):
n += 1
else:
if n >= best:
print n, a, b, a*b
best = n
running = False
# The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
# Find the sum of all the primes below two million.
from tools import is_prime
primes = [2]
for i in range(3, 2 * 10 ** 6, 2):
if is_prime(i):
primes.append(i)
print sum(primes)
import tools
def f(n, a, b):
return n**2 + a * n + b
primes_under_1000 = tools.sieve_of_eratosthenes(int(1e3))
consecutive_primes = []
a_max = 0
b_max = 0
for b in primes_under_1000:
for a in range(-b - 1, 1000):
if tools.is_prime(1 + a + b):
temp = []
n = 0
while tools.is_prime(f(n, a, b)):
temp.append(f(n, a, b))
n += 1
if len(temp) > len(consecutive_primes):
consecutive_primes = temp
a_max = a
b_max = b
print(a_max, b_max)
print(a_max * b_max)
print(consecutive_primes)
print(len(consecutive_primes))
import tools
upper_limit = 1000000
primes = tools.sieve_of_eratosthenes(upper_limit)
res = 0
count = 0
for i, prime_1 in enumerate(primes):
temp = [prime_1]
for prime_2 in primes[i + 1:]:
if temp[-1] + prime_2 < upper_limit:
temp.append(temp[-1] + prime_2)
else:
break
for j, p in enumerate(reversed(temp)):
if tools.is_prime(p):
temp = temp[:len(temp) - j]
break
if len(temp) > count:
res = temp[-1]
count = len(temp)
print(res)
print(count)
import math
import tools
found = False
n = 35
primes_smaller_than_n = [i for i in range(2, n + 1) if tools.is_prime(i)]
print(primes_smaller_than_n)
while not found:
if tools.is_prime(n):
primes_smaller_than_n.append(n)
n += 2
continue
else:
for prime in primes_smaller_than_n:
temp = (n - prime) / 2
temp_list = [
i for i in range(1, int(math.sqrt(n) + 1)) if i**2 == temp
]
if len(temp_list) > 0:
print(
str(n) + ' = ' + str(prime) + ' + 2x' + str(temp_list[0]) +
'^2')
n += 2
break
elif prime == primes_smaller_than_n[-1] and len(temp_list) == 0:
found = True
print(n)
Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.
37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18 5 4 3 12 29
40 19 6 1 2 11 28
41 20 7 8 9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49
It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 62%.
If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%? """
from tools import is_prime
start = 1
diag = [1]
step = 2
prime = []
while True:
for i in range(4):
start +=step
diag.append(start)
if is_prime(start):
prime.append(start)
step +=2
if len(prime) / float(len(diag)) < 0.1 :
print step -1
break
from tools import is_prime
x,n = 1,0
while n < 10001:
x = x + 1
if is_prime(x):
n = n + 1
print x
import math
from tools import is_prime
x = 600851475143
y = int(math.sqrt(x))+1
while y > 1:
if x % y == 0 and is_prime(y):
print y
break
y = y - 1
#!/usr/bin/python
#-*- coding: utf-8 -*-
"""
Prime permutations
Problem 49
The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.
There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.
What 12-digit number do you form by concatenating the three terms in this sequence?"""
from tools import is_prime
import itertools
primelist = (i for i in range(1000,10000) if is_prime(i))
def check_is_perm(number):
return is_prime(number+3330) and sorted([int(i) for i in str(number)])==sorted([int(i) for i in str(number+3330)])
permprimelist = []
for i in primelist:
if check_is_perm(i):
if check_is_perm(i+3330):
permprimelist.append([i,i+3330,i+6660])
solution = ''
for i in permprimelist[1]:
solution +=str(i)
print solution
