def is_trunc(num):
for x in range(1, len(str(num))):
if not is_prime(int(num[x:])) \
or not is_prime(int(num[:x])):
return False
return True
def prime_count(a, b):
count = 0
for n in range(b + 1):
if is_prime(abs((n * n) + (a * n) + b)):
count += 1
else:
break
return count
def ok(a, p):
"""
Args: two (alleged) primes
Returns: whether the numbers (a)(p) (p)(a) (by concatenating a and p)
are both primes or not
"""
if a % 3 != p % 3: # (a)(p) and (p)(a) mod 3 == (a) + (p)
# as a and p primes, a and p !== 0 [3], and we do not want
# the sum to be a multiple of 3.
# problem if a or p == 3 ?!
return False
b = a * 10 ** len(str(p)) + p
c = p * 10 ** len(str(a)) + a
if b > 10 ** 7 and is_prime(b, l_7) or (b < 10 ** 7 and b in l_7_2):
return c > 10 ** 7 and is_prime(c, l_7) or (c < 10 ** 7 and c in l_7_2)
else:
return False
def problem_58():
l_6 = numpy_sieve(10**6)
nb, i = 0, 0
a = 1
while True:
i += 1
ii = i << 1
a += ii << 2
b = a - ii
c = b - ii
d = c - ii
for j in [b, c, d]:
if is_prime(j, l_6):
nb += 1
if nb * 10 <= (ii << 1):
print(ii + 1) # sol = 26241
break
def ok2(n):
return is_prime(n // s_digit(n)) and (n != 1) and (n // s_digit(n) != 1)
def ok(n):
return n % s_digit(n) == 0
L = [i for i in range(1, 10)]
L_bis = [i for i in range(1, 10)]
for count in range(12):
L_ter = []
for a in L_bis:
for i in range(10):
b = a * 10 + i
if ok(b):
L_ter.append(b)
L.append(b)
L_bis = L_ter
def ok2(n):
return is_prime(n // s_digit(n)) and (n != 1) and (n // s_digit(n) != 1)
L = [a for a in L if ok2(a)]
S = 0
for a in L:
for i in [1, 3, 7, 9]:
if is_prime(a * 10 + i):
S += a * 10 + i
print("\n", S)
from Tools import is_prime
from Tools import sieve_of_e
def rotate(num):
chars = list(num)
tmp = ""
for x in range(1, len(chars)):
tmp += chars[x]
tmp += chars[0]
return tmp
count = 0
primes = sieve_of_e(1000000)
for i in primes:
check = i
circular = True
for x in range(len(str(check))):
if not is_prime(int(check)):
circular = False
break
else:
check = rotate(str(check))
if circular:
count += 1
print(count)
for b in l_p:
l_p2.append(b * l[i])
f.append((i + 1, a, l_p2))
else:
s += p if ok_mieux(p, l_p) else 0
print(s)
print(time() - timer)
# %%% Problem 358
from Tools import is_prime
# we search for p prime such that (10**(p-1)-1)/p = 00000000137...56789
b = int(1 / 0.00000000137)
a = int(1 / 0.00000000138)
# a < p < b
k = 1
while (56789 * k) % 10**5 != 99999:
k += 1
# p ends up with k
l = []
for p in range(a, b):
if p % 10**5 == k and is_prime(p):
l.append(p)
for a in l:
print(9 * (a - 1) // 2)
# s is one of the values
# sol = 3284144505
# %% Problem 245
# C(n) can not be an unit fraction if n = p^k * a with k > 1 and p prime
from Tools import is_prime, phi, decompose_tout
def unit_co_resilience(n):
return (n - 1) % (n - phi(n)) == 0
s = 0
for i in range(2, 2 * 10**11 + 1):
if not is_prime(i):
if unit_co_resilience(i):
s += i
l = [b for a, b in decompose_tout(i)]
if l.count(1) != len(l):
print(i)