sequence?
Notes:
Brute force with my primes generator should be good enough. If we happen
to get some primes which do have permutations (but not three) we can boot
em.
"""
from itertools import permutations
from projecteuler import all_primes_between
if __name__ == '__main__':
primes = all_primes_between(2, 9999)
x = []
for p in primes:
if p < 1000:
pass
else:
for p2 in permutations(str(p)):
p2 = int(''.join(p2))
if (p2 > p) and (p2 in primes):
p3 = 2 * p2 - p
if (p3 < 10000) and (p3 in primes) and (sorted(str(p3)) == sorted(str(p))):
x.append((p, p2, p3))
print x
What 12-digit number do you form by concatenating the three terms in this
sequence?
Notes:
Brute force with my primes generator should be good enough. If we happen
to get some primes which do have permutations (but not three) we can boot
em.
"""
from itertools import permutations
from projecteuler import all_primes_between
if __name__ == '__main__':
primes = all_primes_between(2, 9999)
x = []
for p in primes:
if p < 1000:
pass
else:
for p2 in permutations(str(p)):
p2 = int(''.join(p2))
if (p2 > p) and (p2 in primes):
p3 = 2 * p2 - p
if (p3 < 10000) and (p3 in primes) and (sorted(str(p3))
== sorted(str(p))):
x.append((p, p2, p3))
print x
Simple enough. Generate a list of primes, check the sum of all the sub-sets
of increasing size. To do this, we make a cumulative sum. Then we check the
offsets (csPrimes - csPrimes[i]) for the largest prime.
"""
from projecteuler import all_primes_between
N = 1000000
# N = 100
if __name__ == "__main__":
primes = all_primes_between(2, N)
primeSums = [0]
for i in range(len(primes)):
primeSums.append(primeSums[-1] + primes[i])
L = len(primeSums)
numberOfPrimes = 0
pMax = 0
for i in range(len(primeSums)):
for j in range(i - numberOfPrimes - 1, -1, -1):
dp = primeSums[i] - primeSums[j]
if dp > N:
break
if dp in primes:
numberOfPrimes = i - j
s = set()
sp = str(p)
L = len(sp)
s = {int(''.join([sp[(offset + j) % L] for j in range(L)]))
for offset in range(L)}
return s
def any_evens(p):
return any(int(el) % 2 == 0 for el in str(p)) or p == 2
if __name__ == '__main__':
n = all_primes_between(2, 1000000)
unchecked = set(n)
circularPrimes = set()
i = 0
for p in n:
if any_evens(p):
unchecked.discard(p)
else:
if p / 1000 > i:
print p
i = p / 1000
if p in unchecked:
ps = perms(p)
isCircular = all(el in n for el in ps)
L = len(sp)
s = {
int(''.join([sp[(offset + j) % L] for j in range(L)]))
for offset in range(L)
}
return s
def any_evens(p):
return any(int(el) % 2 == 0 for el in str(p)) or p == 2
if __name__ == '__main__':
n = all_primes_between(2, 1000000)
unchecked = set(n)
circularPrimes = set()
i = 0
for p in n:
if any_evens(p):
unchecked.discard(p)
else:
if p / 1000 > i:
print p
i = p / 1000
if p in unchecked:
ps = perms(p)
isCircular = all(el in n for el in ps)
Notes:
Simple enough. Generate a list of primes, check the sum of all the sub-sets
of increasing size. To do this, we make a cumulative sum. Then we check the
offsets (csPrimes - csPrimes[i]) for the largest prime.
"""
from projecteuler import all_primes_between
N = 1000000
#N = 100
if __name__ == '__main__':
primes = all_primes_between(2, N)
primeSums = [0]
for i in range(len(primes)):
primeSums.append(primeSums[-1] + primes[i])
L = len(primeSums)
numberOfPrimes = 0
pMax = 0
for i in range(len(primeSums)):
for j in range(i - numberOfPrimes - 1, -1, -1):
dp = primeSums[i] - primeSums[j]
if dp > N:
break
if dp in primes:
numberOfPrimes = i - j
