def gen_primes(M=M, odd_only=False):
from mathplus import sieve
sieves = sieve(M)
if odd_only:
primes = [str(i) for i, f in enumerate(sieves) if f and i != 2]
else:
primes = [str(i) for i, f in enumerate(sieves) if f]
f = open(_get_prime_file(M, odd_only), 'w')
f.write(','.join(primes))
f.close()
def gen_primes(M=M, odd_only=False):
from mathplus import sieve
sieves = sieve(M)
if odd_only:
primes = [str(i) for i, f in enumerate(sieves) if f and i != 2]
else:
primes = [str(i) for i, f in enumerate(sieves) if f]
f = open(_get_prime_file(M, odd_only), 'w')
f.write(','.join(primes))
f.close()
#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?
#Answer:
#296962999629
from time import time
t = time()
from mathplus import sieve, permutations
ret = {}
primes = sieve(10000)
p = [i for i, f in enumerate(primes) if f and i >= 1000]
for n in p:
c = []
for i in permutations(
(n // 1000, (n % 1000) // 100, (n % 100) // 10, n % 10)):
if 0 in i: break
m = i[0] * 1000 + i[1] * 100 + i[2] * 10 + i[3]
if primes[m] and m not in c: c.append(m)
if len(c) >= 3:
c = sorted(c)
if c[0] in ret: continue
for i in range(1, len(c) - 1):
for j in range(i):
if c[i] * 2 - c[j] in c:
ret[c[0]] = (c[j], c[i], 2 * c[i] - c[j])
#n² + 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 time import time
t = time()
from mathplus import sieve, product
M = 15000
primes = sieve(M)
# one of (1+a+b, b*b+a*b+b) is not a prime, so n*n+a*n+b has less than range(b) primes
ab = [[a, b, b] for a, b in product(range(
-1000 + 1, 1000, 2), [i for i in range(83, 1000, 2) if primes[i]])]
n = 0
lastone = None
while len(ab) > 1:
lastone = ab[0]
x = 2 * n + 1
for v in ab:
v[2] += x + v[0]
ab = [[a, b, x] for a, b, x in ab if x > 0 and (x >= M or primes[x])]
n += 1
lastone = ab[0]
#!/usr/bin/python
# -*- coding: utf-8 -*-
#The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
#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 time import time; t = time()
from mathplus import sieve
M = 1000000
S = sieve(M)
count = 2
for k in range(3, M, 2):
if not S[k]: continue
s = str(k)
if any((i in s) for i in '024568'): continue
for i in range(len(s)):
if not S[int(s)]: break
s = s[1:]+s[0]
else:
count += 1
print(count)#, time()-t
#It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.
#9 = 7 + 2×12
#15 = 7 + 2×22
#21 = 3 + 2×32
#25 = 7 + 2×32
#27 = 19 + 2×22
#33 = 31 + 2×12
#It turns out that the conjecture was false.
#What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
#Answer:
#5777
from time import time
t = time()
from mathplus import sieve, isqrt
M = 10000
p = sieve(M)
for n in range(35, M, 2):
if p[n]: continue
for i in range(1, isqrt((n - 3) / 2) + 1):
if p[n - 2 * i * i]: break
else:
print(n) #, time()-t
break
#It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.
#9 = 7 + 2×12
#15 = 7 + 2×22
#21 = 3 + 2×32
#25 = 7 + 2×32
#27 = 19 + 2×22
#33 = 31 + 2×12
#It turns out that the conjecture was false.
#What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
#Answer:
#5777
from time import time; t=time()
from mathplus import sieve, isqrt
M = 10000
p = sieve(M)
for n in range(35, M, 2):
if p[n]: continue
for i in range(1, isqrt((n-3)/2)+1):
if p[n-2*i*i]: break
else:
print(n)#, time()-t
break
# -*- coding: utf-8 -*-
#The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
#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 time import time
t = time()
from mathplus import sieve
M = 1000000
S = sieve(M)
count = 2
for k in range(3, M, 2):
if not S[k]: continue
s = str(k)
if any((i in s) for i in '024568'): continue
for i in range(len(s)):
if not S[int(s)]: break
s = s[1:] + s[0]
else:
count += 1
print(count) #, time()-t
#n² + 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 time import time; t=time()
from mathplus import sieve, product
M = 15000
primes = sieve(M)
# one of (1+a+b, b*b+a*b+b) is not a prime, so n*n+a*n+b has less than range(b) primes
ab = [[a, b, b] for a, b in product(
range(-1000+1, 1000, 2),
[i for i in range(83, 1000, 2) if primes[i]]
)]
n = 0
lastone = None
while len(ab) > 1:
lastone = ab[0]
x = 2*n+1
for v in ab:
v[2] += x+v[0]
ab = [[a, b, x] for a, b, x in ab if x > 0 and (x >= M or primes[x])]
n += 1
# -*- coding: utf-8 -*-
#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?
#Answer:
#296962999629
from time import time; t=time()
from mathplus import sieve, permutations
ret = {}
primes = sieve(10000)
p = [i for i, f in enumerate(primes) if f and i >= 1000]
for n in p:
c = []
for i in permutations((n//1000, (n % 1000)//100, (n % 100)//10, n%10)):
if 0 in i: break
m = i[0]*1000+i[1]*100+i[2]*10+i[3]
if primes[m] and m not in c: c.append(m)
if len(c) >= 3:
c = sorted(c)
if c[0] in ret: continue
for i in range(1, len(c)-1):
for j in range(i):
if c[i]*2-c[j] in c:
ret[c[0]] = (c[j], c[i], 2*c[i]-c[j])
assert len(ret) == 2
