Esempio n. 1
0
def S(n):
	primes = sieve(n)
	print 'primes done'
	total = 0
	for i in range(len(primes)):
		if primes[i] * primes[i + 1] > n:
			break
		print i
		p = primes[i]
		for j in range(i + 1, len(primes)):
			q = primes[j]
			if p * q > n:
				break
			total += M(p, q, n)

	return total
Esempio n. 2
0
def S(n):
    primes = sieve(n)
    print 'primes done'
    total = 0
    for i in range(len(primes)):
        if primes[i] * primes[i + 1] > n:
            break
        print i
        p = primes[i]
        for j in range(i + 1, len(primes)):
            q = primes[j]
            if p * q > n:
                break
            total += M(p, q, n)

    return total
Esempio n. 3
0
def euler_phi(n):
	primes = sieve(1000)

	answer = 1
	for prime in primes:
		if n == 1:
			break
		power = 0
		while n % prime == 0:
			power += 1
			n /= prime 
		if power > 0:
			answer *= (prime - 1) * prime**(power - 1)

	assert n == 1 or isPrimeMR(n)
	if n > 1:
		answer *= n - 1
	return answer
Esempio n. 4
0
def euler_phi(n):
    primes = sieve(1000)

    answer = 1
    for prime in primes:
        if n == 1:
            break
        power = 0
        while n % prime == 0:
            power += 1
            n /= prime
        if power > 0:
            answer *= (prime - 1) * prime**(power - 1)

    assert n == 1 or isPrimeMR(n)
    if n > 1:
        answer *= n - 1
    return answer
Esempio n. 5
0
def lcm(n):
    primes = sieve(n + 1)
    l = range(1, n + 1)
    r = 1
    for i in range(n):
        while r % l[i] != 0:
            for p in primes:
                if l[i] % p == 0:
                    r *= p
        if r % l[i] == 0:
            continue
        j = 0
        while j < len(primes):
            p = primes[j]
            if l[i] % p == 0:
                l[i] /= p
                r *= p
            else:
                j += 1
    return r
Esempio n. 6
0
def lcm(n):
    primes = sieve(n+1)
    l = range(1,n+1)
    r = 1
    for i in range(n):
        while r % l[i] != 0:
            for p in primes:
                if l[i] % p == 0:
                    r *= p
        if r % l[i] == 0:
            continue
        j = 0
        while j < len(primes):
            p = primes[j]
            if l[i] % p == 0:
                l[i] /= p
                r *= p
            else:
                j += 1
    return r
Esempio n. 7
0
from helpers import prime_factorizations, crt, sieve
from itertools import product
'''
153651073760956
[Finished in 2605.0s]

'''


limit = 2 * 10**7
# N = 2 * 10 ** 7

pfs = prime_factorizations(limit / 20)
primes = sieve(limit)
setprimes = set(primes)

def get_pf(n):
	if n < len(pfs):
		return pfs[n]
	else:
		if n in setprimes:
			return {n : 1}
		ans = {}
		for p in primes:

			if n < len(pfs) or n in setprimes:
				break
			power = 0
			while n % p == 0:
				power += 1
				n /= p 
Esempio n. 8
0
# A sieve could be better at this, but requires a good estimate on the upper bound of the nth prime which I didn't know off hand
primes = [2,3,5,7,11,13,17,19,23,29]
sprimes = [2,3,5] # primes up to the square root of the biggest prime in primes

n = primes[-1]
while len(primes) < 10001:
    n += 2
        if n ** 0.5 > sprimes[-1]:
            sprimes.append(primes[len(sprimes)])
    if all([n % p != 0 for p in sprimes]):
        primes.append(n)

print primes[-1]

# I later found that nlogn + nloglogn is a good upperbound approximation:
from math import log
from helpers import sieve
n = 10001
print sieve(int(n*log(n) + n*log(log(n))))[10000]
Esempio n. 9
0
from fractions import Fraction
from helpers import sieve



primes = set(sieve(500))

memo = {}

def left_hop(n):
	if n > 1:
		return n - 1
	assert n == 1
	return 2

def right_hop(n):
	if n < 500:
		return n + 1
	assert n == 500
	return 499

def probability(n, string):
	if (n,string) in memo:
		return memo[(n, string)]
	elif len(string) == 1:
		if n in primes and string == 'P':
			memo[(n,string)] = Fraction(2, 3)
			return Fraction(2, 3)
		if n in primes and string == 'N':
			memo[(n,string)] = Fraction(1, 3)
			return Fraction(1, 3)
Esempio n. 10
0
from helpers import sieve
from itertools import combinations
from fractions import Fraction

primes = sieve(25)

n = reduce(lambda a, b: a * b, primes)
phin = reduce(lambda a, b: a * b, [i - 1 for i in primes])
'''
Just have to get i as small as possible where the first printed value is less than the second.

Answer is the third.

'''
i = 4
print float(i * phin) / (i * n - 1)
print float(15499) / 94744

print i * n
for (p1, p2) in combinations(primes, 2):
    potential_n = n * p1
    phi_potential_n = p2 * phin * p1 / (p1 - 1)
    if float(phi_potential_n) / (potential_n - 1) < float(15499) / 94744:
        print 'wtf'
Esempio n. 11
0
#Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.

from helpers import sieve

p = sieve(1000)[3:]

def cyclelen(n):
    k = 1
    while True:
        if 10**k % n == 1:
            return (k, n)
        k += 1

print max(map(cyclelen, p))
Esempio n. 12
0
from helpers import sieve, prime_factorizations, isPrimeMR

primes = sieve(10**8)
setprimes = set(primes)
pfs = prime_factorizations(10**6)


def get_divisors(pf):
	if len(pf) == 0:
		yield 1
		return
	(prime, power) = pf.popitem()
	for p in range(power + 1):
		for divisor in get_divisors(pf):
			yield prime**p * divisor 
	pf[prime] = power

def get_pf(n):
	if n < len(pfs):
		return pfs[n]
	else:
		if isPrimeMR(n):
			return {n : 1}
		ans = {}
		for p in primes:

			if n < len(pfs) or n in setprimes:
				break
			power = 0
			while n % p == 0:
Esempio n. 13
0
from helpers import get_prime_mask, sieve
from itertools import permutations

limit = 1000000

primes = sieve(limit)
is_prime = get_prime_mask(primes, limit)

def rotate_string(s):
  for i in range(len(s)):
    s = s[1:] + s[0]
    yield s

cycle_primes = []
for p in primes:
  all_prime = True
  for perm in rotate_string(str(p)):
    i = int(''.join(perm))
    all_prime = all_prime & is_prime[i]
  if all_prime:
    cycle_primes.append(p)


print len(cycle_primes)
Esempio n. 14
0
# This is really dumb and takes forever to run, I know
# A smarter way would just check random combos matching ([3,7,9][1,3,7,9]*[3,7,9]) until you find 11
from helpers import sieve, is_prime

print sum([
    int(p) for p in map(str,
                        sieve(1000000)[4:]) if all([
                            is_prime(int(p[i:])) and is_prime(int(p[:i + 1]))
                            for i in range(len(p))
                        ])
])
Esempio n. 15
0
from helpers import sieve

print sum(sieve(2000000))
Esempio n. 16
0
from helpers import sieve, rotations

primes = set(map(str, sieve(1000000)))
print len([p for p in primes if all([r in primes for r in rotations(p)])])
Esempio n. 17
0
from helpers import sieve, is_prime

p = sieve(10000)

total = 0
start = 0
length = 0
maxlen = 0
while True:
    if total + p[start+length] < 1000000:
        total += p[start+length]
        length += 1
        if length > maxlen and is_prime(total):
            if 1000000 - total < p[start+length]:
                print total
                break
            maxlen = length
    else:
        total -= (p[start] + p[start+length-1])
        length -= 2
        start += 1
Esempio n. 18
0
import copy
from helpers import sieve
primes = sieve(30000)
#there's always a solution given for k by 2k = (k,2,1,1,1,1,...,1) so we only need to look in range(1,k+1)
table_of_vals = [30000 for i in range(12001)]
table_of_vals[0] = 0
table_of_vals[1] = 0

sum_list = [[] for i in range(24001)]

def multiply_set(n):
    if(n in primes):
        return([[n]])
    l = []
    for j in range(2,int(n**0.5)+1):
        if(n % j == 0):
            if(sum_list[int(n/j)] != []):
                a = copy.deepcopy(sum_list[int(n/j)])
            else:
                a = copy.deepcopy(multiply_set(int(n/j)))
            for i in a:
                i.append(j)
                i.sort()
                if(l.count(i) == 0):
                    l.append(i)
    l.append([n])
    sum_list[n] = l
    return(l)

def multiply_and_sum_set(n):
    x = copy.deepcopy(multiply_set(n))
Esempio n. 19
0
def int_split(n, d):
  left = n / 10**d
  right = n % 10**d
  if int_concat(left, right) == n:
    return (left, right)

def is_family(f):
  for i in range(len(f)):
    for j in range(i + 1, len(f)):
      left = int_concat(f[i], f[j])
      right = int_concat(f[j], f[i])
      if not (is_prime(left) and is_prime(right)):
        return False
  return True

sieve(100000000)
max_prime = 10000
fam_size = 3

primes = list(prime_gen(max_size = max_prime))
def fams():
  for i in range(1, len(primes)):
    for j in range(i+1, len(primes)):
      if not is_family((primes[i], primes[j])):
        continue
      print (primes[i], primes[j])
      for k in range(j+1, len(primes)):
        if not is_family((primes[i], primes[j], primes[k])):
          continue
        for l in range(k + 1, len(primes)):
          if not is_family((primes[i], primes[j], primes[k], primes[l])) or k > l:
Esempio n. 20
0
"""
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that
the 6th prime is 13.

What is the 10,001st prime number?
"""

from helpers import sieve, primeUpperBound

print sieve(primeUpperBound(10001))[10000]
Esempio n. 21
0
from helpers import sieve, rotations
primes = set(map(str, sieve(1000000)))
print len([p for p in primes if all([r in primes for r in rotations(p)])])
Esempio n. 22
0
Observation: for a given fraction, the number of steps before reducing is the number of steps before we hit the next smallest factor.

We always reduce to 1 / another number

and so we always eliminate all of the smallest factors.

So f(n) = largest_prime_factor(n + 1) - 1.

'''

from helpers import prime_factorizations, isPrimeMR, sieve

limit = 2 * 10**6

primes = sieve(limit + 100)
pfs = prime_factorizations(limit + 100)


def get_pf(n):
    if n < len(pfs):
        return pfs[n]
    else:
        if isPrimeMR(n):
            return {n: 1}
        ans = {}
        for p in primes:

            if n < len(pfs) or isPrimeMR(n):
                break
            power = 0
Esempio n. 23
0
from helpers import sieve, prime_factorizations, isPrimeMR

primes = sieve(10**8)
setprimes = set(primes)
pfs = prime_factorizations(10**6)


def get_divisors(pf):
    if len(pf) == 0:
        yield 1
        return
    (prime, power) = pf.popitem()
    for p in range(power + 1):
        for divisor in get_divisors(pf):
            yield prime**p * divisor
    pf[prime] = power


def get_pf(n):
    if n < len(pfs):
        return pfs[n]
    else:
        if isPrimeMR(n):
            return {n: 1}
        ans = {}
        for p in primes:

            if n < len(pfs) or n in setprimes:
                break
            power = 0
            while n % p == 0:
Esempio n. 24
0
import copy
from helpers import sieve
primes = sieve(30000)
#there's always a solution given for k by 2k = (k,2,1,1,1,1,...,1) so we only need to look in range(1,k+1)
table_of_vals = [30000 for i in range(12001)]
table_of_vals[0] = 0
table_of_vals[1] = 0

sum_list = [[] for i in range(24001)]


def multiply_set(n):
    if (n in primes):
        return ([[n]])
    l = []
    for j in range(2, int(n**0.5) + 1):
        if (n % j == 0):
            if (sum_list[int(n / j)] != []):
                a = copy.deepcopy(sum_list[int(n / j)])
            else:
                a = copy.deepcopy(multiply_set(int(n / j)))
            for i in a:
                i.append(j)
                i.sort()
                if (l.count(i) == 0):
                    l.append(i)
    l.append([n])
    sum_list[n] = l
    return (l)

Esempio n. 25
0
    n = [i for i in str(n)]
    m = [i for i in str(m)]
    while len(n) > len(m):
        answer += num_transitions[int(n[0])]
        n = n[1:]
    for i in range(len(n)):
        num1 = int(n[i])
        num2 = int(m[i])
        answer += num_transition_paired[(num1, num2)]
    return answer


def max_transitions_required(n):
    answer = 0
    answer += sum([num_transitions[int(i)] for i in str(n)])
    while digit_sum(n) != n:
        answer += smart_transitions_required(n, digit_sum(n))
        n = digit_sum(n)
    answer += sum([num_transitions[int(i)] for i in str(n)])
    return answer


primes = [p for p in sieve(2 * 10**7) if p > 10**7]
print 'primes found'
answer = 0
for p in primes:
    answer += sam_transitions_required(p) - max_transitions_required(p)

print answer
# print sam_transitions_required(137)
# print max_transitions_required(137)
Esempio n. 26
0
    if i < p:
        return 0
    a = num_fac_powers(p, i / p)
    if len(memo) > 100000:
        memo.clear()
    memo[(p, i)] = i/p + a
    return i/p + a

def get_smallest_possible(p, i):
    min_possible_power = c * num_fac_powers(p, i)
    potential_n = p + p * (((p - 1) * min_possible_power) / p)
    while num_fac_powers(p, potential_n) < min_possible_power:
        potential_n += p
    return potential_n

primes = sieve(N)
setprimes = set(primes)

print len(primes)
last_chosen_p = 1


last_chosen_p = 1
answer = 0
for i in range(N, 9, -1):
    if i % 100 == 0:
        print i
    if (i+1)/last_chosen_p == i/last_chosen_p and i < N:
        answer += get_smallest_possible(last_chosen_p, i)
        continue
    
Esempio n. 27
0
from helpers import sieve
#from itertools import takewhile
#from helpers import get_primes

print sum(sieve(2000000))
#total = 0
#for p in get_primes():
#    if p < 2E6:
#        total += p
#    else:
#        break
#print total
Esempio n. 28
0
from helpers import sieve, is_prime

p = sieve(10000)

total = 0
start = 0
length = 0
maxlen = 0
while True:
    if total + p[start + length] < 1000000:
        total += p[start + length]
        length += 1
        if length > maxlen and is_prime(total):
            if 1000000 - total < p[start + length]:
                print total
                break
            maxlen = length
    else:
        total -= (p[start] + p[start + length - 1])
        length -= 2
        start += 1
Esempio n. 29
0
Observation: for a given fraction, the number of steps before reducing is the number of steps before we hit the next smallest factor.

We always reduce to 1 / another number

and so we always eliminate all of the smallest factors.

So f(n) = largest_prime_factor(n + 1) - 1.

'''

from helpers import prime_factorizations, isPrimeMR, sieve

limit = 2 * 10**6


primes = sieve(limit + 100)
pfs = prime_factorizations(limit + 100)

def get_pf(n):
	if n < len(pfs):
		return pfs[n]
	else:
		if isPrimeMR(n):
			return {n : 1}
		ans = {}
		for p in primes:

			if n < len(pfs) or isPrimeMR(n):
				break
			power = 0
			while n % p == 0:
Esempio n. 30
0
from helpers import sieve
lim = 50000000

primes = sieve(int(lim**0.5)+1)
l = []

for p1 in primes:
    for p2 in primes:
        if(p1**2+p2**3 > lim):
            break
        for p3 in primes:
            if(p1**2+p2**3+p3**4 > lim):
                break
            l.append(p1**2+p2**3+p3**4)
l = list(set(l))
print(len(l))
Esempio n. 31
0
# I could probably improve this to not recalculate n(v) over and over but instead build it all at once
# But I had basically already written n(v) for solving the earlier coin problem, and it was easy to reuse

from helpers import sieve
primes = sieve(100)


def n(v):
    ways = [1] + [0] * v
    for prime in primes:
        for i in range(prime, v + 1):
            ways[i] += ways[i - prime]
    return ways[v]


v = 2
while n(v) < 5000:
    v += 1
print v, n(v)
Esempio n. 32
0
from helpers import sieve
from itertools import combinations
from fractions import Fraction

primes = sieve(25)

n = reduce(lambda a, b : a * b, primes)
phin = reduce(lambda a,b : a * b, [i - 1 for i in primes])
'''
Just have to get i as small as possible where the first printed value is less than the second.

Answer is the third.

'''
i = 4
print float(i * phin) / (i * n - 1)
print float(15499) / 94744


print i * n
for (p1, p2) in combinations(primes, 2):
	potential_n = n * p1
	phi_potential_n = p2 * phin * p1 / (p1 - 1)
	if float(phi_potential_n) / (potential_n - 1) < float(15499) / 94744:
		print 'wtf'


Esempio n. 33
0
# I could probably improve this to not recalculate n(v) over and over but instead build it all at once
# But I had basically already written n(v) for solving the earlier coin problem, and it was easy to reuse

from helpers import sieve
primes = sieve(100)

def n(v):
    ways = [1]+[0]*v
    for prime in primes:
        for i in range(prime, v+1):
            ways[i] += ways[i-prime]
    return ways[v]

v = 2
while n(v) < 5000:
    v += 1
print v, n(v)
Esempio n. 34
0
	assert n > m
	answer = 0
	n = [i for i in str(n)]
	m = [i for i in str(m)]
	while len(n) > len(m):
		answer += num_transitions[int(n[0])]
		n = n[1:]
	for i in range(len(n)):
		num1 = int(n[i])
		num2 = int(m[i])
		answer += num_transition_paired[(num1, num2)]
	return answer

def max_transitions_required(n):
	answer = 0
	answer += sum([num_transitions[int(i)] for i in str(n)])
	while digit_sum(n) != n:
		answer += smart_transitions_required(n, digit_sum(n))
		n = digit_sum(n)
	answer += sum([num_transitions[int(i)] for i in str(n)])
	return answer

primes = [p for p in sieve(2 * 10 ** 7) if p > 10 ** 7]
print 'primes found'
answer = 0
for p in primes:
	answer += sam_transitions_required(p) - max_transitions_required(p)

print answer
# print sam_transitions_required(137)
# print max_transitions_required(137)
Esempio n. 35
0
from helpers import prime_factorizations, sieve
from operator import mul

from sys import setrecursionlimit

# bold call but you gotta do what you gotta do to not rewrite your shitty recursive solution into one using a stack :^)^)^)^)^)^)^)^)^)
setrecursionlimit(10**4)

N = 120000

prime_factorizations = prime_factorizations(N)
primes = sieve(N)

# print prime_factorizations


def rad(n):
    return reduce(mul, prime_factorizations[n].keys())


def abc_hit(a, b, c):
    rad_factors = set(prime_factorizations[c].keys())
    b_factors = set(prime_factorizations[b].keys())
    a_factors = set(prime_factorizations[a].keys())
    rad_factors = rad_factors.union(b_factors)
    rad_factors = rad_factors.union(a_factors)
    return reduce(
        mul, rad_factors) < c and a_factors.intersection(b_factors) == set()


def possible_nums(prime_candidates, max_num, min_index):
Esempio n. 36
0
from helpers import sieve

def poly(n, a, b):
  return n**2 + a * n + b

a_max = 1000
b_max = 1000
known_max = 81

largest_possible = poly(known_max, a_max, b_max)

primes = sieve(largest_possible)
prime_mask = [False] * largest_possible
for p in primes:
  prime_mask[p] = True

max = 0
max_a = None
max_b = None

for a in range(-a_max, a_max):
  for b in range(-b_max, b_max):
    n = 0
    while True:
      p = poly(n, a, b)
      if not prime_mask[p]:
        break
      else:
        n += 1
    if n > max:
      print "%s primes for %s, %s" % (n, a, b)
Esempio n. 37
0
# This is really dumb and takes forever to run, I know
# A smarter way would just check random combos matching ([3,7,9][1,3,7,9]*[3,7,9]) until you find 11
from helpers import sieve, is_prime
print sum([int(p) for p in map(str,sieve(1000000)[4:]) if all([is_prime(int(p[i:])) and is_prime(int(p[:i+1])) for i in range(len(p))])])
Esempio n. 38
0
from helpers import sieve
lim = 50000000

primes = sieve(int(lim**0.5) + 1)
l = []

for p1 in primes:
    for p2 in primes:
        if (p1**2 + p2**3 > lim):
            break
        for p3 in primes:
            if (p1**2 + p2**3 + p3**4 > lim):
                break
            l.append(p1**2 + p2**3 + p3**4)
l = list(set(l))
print(len(l))