def prob27():
sieve = prime.prime_table(10**4)
D = {}
for a in range(-999, 999, 2):
for b in range(a + 2, 1000):
if (a*a - 4*b >= 0):
continue
num_primes = 0
for n in range(0, b):
temp = n*n + a*n + b
if (temp < 2):
break
elif (temp == 2):
num_primes += 1
else:
if (temp%2 == 1 and sieve[int((temp-1)/2)]):
num_primes += 1
else:
D[(a,b)] = num_primes
break
max_primes = 0
max_term = (0, 0)
for i,j in D:
if (D[(i,j)] > max_primes):
max_primes = D[(i, j)]
max_term = (i, j)
return [max_term]
def prob37(): sieve = prime_table(10**6) # the set of truncatable primes h = set() n = 11 while len(h) < 11: truncatable = True l = str(n) for i in range (1, len(l) + 1): left_num = int(l[:i]) right_num = int(l[len(l) - i:]) left_index = (left_num - 1) // 2 right_index = (right_num - 1) // 2 # check for prime if not (left_num > 1 and right_num > 1 and sieve[left_index] and sieve[right_index]): truncatable = False break # if n is truncatable, add it to h if truncatable: h.add(n) # increase n by 2 n += 2 if (l[len(l) - 1] == 3): n += 2 return (sum(i for i in h), h)
def prob21(limit):
cache = []
miss = [0, 1, 2]
sieve = prime.prime_table(limit+1)
# 1. Prime numbers are not amicable numbers
for i in range(1, len(sieve)):
if (sieve[i] == True):
miss.append(2*i+1)
for i in range(3, limit+1):
# 2. Find i in cache and miss
if not (i in cache or i in miss):
# 3. Compute sum of divisors for i
p = sum_of_divisors(i)
if (i != p and i == sum_of_divisors(p)):
cache.append(i)
if (p not in cache):
cache.append(p)
print("result: ", i, p)
else:
if (p != i):
miss.append(i)
miss.append(p)
else:
miss.append(i)
return sum(i for i in cache)
def prob23():
limit = 20162
s = 0
abn = set()
prime_sieve = prime.prime_table(limit)
for n in range(1, limit):
if is_abundant(n, prime_sieve):
abn.add(n)
if not any((n-a in abn) for a in abn):
s += n
return s
def prob12(): n = 3 # triangle number = n*(n+1)/2 D[n] = 2 # number of divisors sieve = prime.prime_table(1000) # the prime table MIN_DIVISORS = 500 count = 0 while count <= MIN_DIVISORS: n += 1 n1 = n if (n1%2 ==0): n1 /= 2 D[n1] = 1 for i in range(0, len(sieve)): if sieve[i] == False: continue if i == 0: prime = 2 else: prime = 2 * i + 1 if (prime*prime) > n1: Dn1 *= 2 break exponent = 1 while n1 % (prime) == 0: n1 /= prime exponent += 1 # if a number can be factorized as A = p1^k1 * p2^k2 * p3^k3 ... where # p1, p2, p3.. is prime then it has (k1 + 1)*(k2 + 1)*(k3 + 1) ... divisors if exponent > 1: Dn1 *= exponent if n1 == 1: break count = D[n]*D[n1] D[n] = D[n1] return int(n*(n-1)/2)
def prob35(limit):
# there are 13 circular primes below 1,000,000
num_cprimes = 13
sieve = prime_table(limit)
for n in range(111, limit - 1, 2):
l = str(n)
if not any(d in ("0","2","4","5","6","8") for d in l):
num_digits = len(l)
is_cprime = True
for i in range(0, num_digits):
# shift the string 1 digit to the left
l = l[1:] + l[:1]
# index in the prime table
index = (int(l) - 1) // 2
if (sieve[index] == False):
is_cprime = False
break;
if is_cprime:
# h.add(n)
num_cprimes += 1
return num_cprimes
def extend_primes(self, to, step):
name = 'prime'
display_name = '素数'
_ptable = prime_table(to)
ptable = [n for i, n in enumerate(_ptable) if (i+1)%step==0]
self.append(Sequence(name, display_name, step, ptable, NamedNumber))
