def sp(n, m):
if n == 0: return 1
if n <= 1 or m <= 1: return 0
if n == 2: return 1
s = 0
for i in takewhile(lambda x: x <= m, primes):
ni = n - i
if i <= ni:
s += sp(ni, i)
else:
s += sp(ni, sieves[ni])
return s
def sp(n, m):
if n == 0: return 1
if n <= 1 or m <= 1: return 0
if n == 2: return 1
s = 0
for i in takewhile(lambda x: x <= m, primes):
ni = n - i
if i <= ni:
s += sp(ni, i)
else:
s += sp(ni, sieves[ni])
return s
#!/usr/bin/python # -*- coding: utf-8 -*- #Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder when (pn−1)n + (pn+1)n is divided by pn2. #For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25. #The least value of n for which the remainder first exceeds 109 is 7037. #Find the least value of n for which the remainder first exceeds 1010. #Answer: #21035 #https://github.com/skydark/project-euler/blob/master/123.py from time import time; t=time() from mathplus import get_primes_by_sieve, takewhile M = 10**10 L = 1000000 p, sieves = get_primes_by_sieve(L) for n in takewhile(lambda x: 2*x*p[x-1] <= M, range(7037, L, 2)): pass print(n+2)#, time()-t
def no_primes_larger_than(n, p):
for pp in takewhile(lambda x: x <= p, primes):
while n % pp == 0:
n //= pp
return n == 1
#!/usr/bin/python
# -*- coding: utf-8 -*-
#Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder when (pn−1)n + (pn+1)n is divided by pn2.
#For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25.
#The least value of n for which the remainder first exceeds 109 is 7037.
#Find the least value of n for which the remainder first exceeds 1010.
#Answer:
#21035
#https://github.com/skydark/project-euler/blob/master/123.py
from time import time
t = time()
from mathplus import get_primes_by_sieve, takewhile
M = 10**10
L = 1000000
p, sieves = get_primes_by_sieve(L)
for n in takewhile(lambda x: 2 * x * p[x - 1] <= M, range(7037, L, 2)):
pass
print(n + 2) #, time()-t
def no_primes_larger_than(n, p):
for pp in takewhile(lambda x: x <= p, primes):
while n % pp == 0:
n //= pp
return n == 1
