def solve(cur_list): print cur_list if len(cur_list) == LIST_LENGTH: print cur_list, sum(cur_list) exit() start = 1 if cur_list == [] else inb(cur_list[-1], p) + 1 for i in p[start:MAX]: recurse = True for j in cur_list: pairing = int(str(j) + str(i)) if pair_dict.get(pairing) != None: recurse = False break if recurse: solve(cur_list + [i])
It can be seen that there are 21 elements in this set.
How many elements would be contained in the set of reduced proper fractions for d <= 1,000,000?
"""
# return int(n * reduce(lambda x,y: x*y, [1-(1./i) for i in pfact]))
from util import inb, pf, primes, totient
from math import sqrt
end = 10 ** 6
ps = primes(end)
ps_sq = primes(int(sqrt(end)))
pfacts = {}
# print ps, inb(2, ps)
for i in xrange(2, end + 1):
if inb(i, ps) != -1:
pfacts[i] = (set([i]), i - 1)
else:
for prime in ps_sq:
if i % prime == 0:
pf, totient = pfacts[i / prime]
if prime in pf:
pfacts[i] = (pf, totient * prime)
else:
pfacts[i] = (pf.union(set([prime])), totient * (prime - 1))
break
print(sum([pfacts[i][1] for i in xrange(2, end + 1)]))
#! /usr/bin/python
from util import primes
from util import inb
p = primes(1000000)
s = []
res = 21
for i in range(len(p)):
s.append(sum(p[i:i+res]))
if s[i] > 1000000:
break
while s != []:
res += 2
for i in range(len(s)):
s[i] += p[i+res-1] + p[i+res-2]
if s[i] > 1000000:
s = s[:i]
break
if inb(s[i],p):
print s[i],res
#! /usr/bin/python
from util import primes
from util import inb
p = primes(1000000)
s = []
res = 21
for i in range(len(p)):
s.append(sum(p[i:i + res]))
if s[i] > 1000000:
break
while s != []:
res += 2
for i in range(len(s)):
s[i] += p[i + res - 1] + p[i + res - 2]
if s[i] > 1000000:
s = s[:i]
break
if inb(s[i], p):
print s[i], res
print "Done\n"
pair_dict = {}
# Use a simple recursive backtracking algorithm
def solve(cur_list):
print cur_list
if len(cur_list) == LIST_LENGTH:
print cur_list, sum(cur_list)
exit()
start = 1 if cur_list == [] else inb(cur_list[-1], p) + 1
for i in p[start:MAX]:
recurse = True
for j in cur_list:
pairing = int(str(j) + str(i))
if pair_dict.get(pairing) != None:
recurse = False
break
if recurse:
solve(cur_list + [i])
# Precompute pairs of primes that when concatenated together produce other primes
for i in range(1,MAX):
print "%d/%d" % (i,MAX)
for j in range(i+1,MAX):
concat = (int(str(p[i]) + str(p[j])), int(str(p[j]) + str(p[i])))
if ((concat[0] < PRIMES_TO) and (len([k for k in concat if inb(k,p) != False]) != 2)) or (len([k for k in concat if is_prime(k,p) != False]) != 2):
pair_dict[concat[0]] = True
solve([])
