def f1(n):
arr = primeBoolList(n)
pdt = pow(2, trunc(log(n, 2)));
for i in range(3, n+1, 2):
if(arr[i]):
k = trunc(log(n, i))
pdt *= pow(i, k)
return pdt
def f(count, beg, end, sum):
L = primeBoolList(end)
i = beg
while i <= end:
k = trunc(log10(i))
k = pow(10, k)
f = i / k
# check the most left digit
if f not in (2,3,5,7):
i += k
continue
# left to right
j = i
while L[j] and k > 0:
j = j % k
k /= 10
if k == 0:
# right to left
j = i / 10
while L[j] and j != 0:
j /= 10
if j == 0:
sum += i
count -= 1
if count == 0:
return sum
# check the most right digit
if i % 10 == 7:
i += 6
else:
i += 2
if N > 0:
return f(count, end+1, end*10, sum)
# Answer: 296962999629
from prime import primeBoolList
def arePermutation(m, n):
c = [0] * 10
for i in range(4):
c[m % 10] += 1
c[n % 10] -= 1
m /= 10
n /= 10
return c == [0] * 10
l = primeBoolList(9999)
i = 1001
while i <= 9999:
if i != 1487 and l[i]:
j = i + 2
k = i + 4
while k <= 9999:
if l[j] and l[k]:
if arePermutation(i, j) and arePermutation(i, k):
print "%d%d%d" % (i, j, k)
j += 2
k += 4
# 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.
#
# Answer: -59231
from prime import primeBoolList
MAX = 1000
pb = primeBoolList(2 * MAX * MAX + MAX)
mn = 0
mm = 0
for a in range(1 - MAX, MAX):
for b in range(1 - MAX, MAX):
n = 0
while True:
c = n * (n + a) + b
if c < 0 or not pb[c]:
break
n += 1
if n > mn:
mn = n
mm = a * b
#
# There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71,
# 73, 79, and 97.
#
# How many circular primes are there below one million?
#
# Answer: 55
from math import trunc
from math import log10
from prime import primeBoolList
from util import pow
N = 1000000
L = primeBoolList(N)
def rotate(n):
k = trunc(log10(n))
if k == 0:
return n
return n / 10 + n % 10 * pow(10, k)
s = 0
for i in range(2, N + 1):
if L[i]:
c = 1
k = rotate(i)
while k != i and L[k]: