def amicables():
a = 0
while True:
a += 1
b = sum(divisors(a))
if b > a:
a_hat = sum(divisors(b))
if a == a_hat:
yield (a, b)
def __init__(self, n):
self.abundant = []
self.isAbundant = (n + 1) * [False]
for num in xrange(1, n + 1):
if sum(e.divisors(num)) > num:
self.abundant.append(num)
self.isAbundant[num] = True
def divisor_permutations(n, largest):
if n == 1:
yield []
divisors = euler.divisors(n)
divisors = [d for d in divisors if d > 1 and d <= largest]
for d in divisors:
for rest in divisor_permutations(n / d, d):
yield [d] + rest
def divisor_permutations(n, largest):
if n == 1:
yield []
divisors = euler.divisors(n)
divisors = [ d for d in divisors if d > 1 and d <= largest]
for d in divisors:
for rest in divisor_permutations(n/d, d):
yield [d] + rest
def fraction(n):
total = 0
divisors = euler.divisors(n)
divisors.remove(1)
for i in range(n // 3 + 1, (n + 1) // 2):
for d in divisors:
if i % d == 0:
break
else:
total += 1
return total
def problem_23(s=0):
n = set()
for x in xrange(1, 28123):
if sum(divisors(x)) > x:
n.add(x)
for y in n:
if x-y in n:
break
else:
s += x
return s
def abundants():
n = 1
while True:
if sum(divisors(n)) > n:
yield n
n += 1
def divisors(num):
if num not in DIVISORS_DICT:
DIVISORS_DICT[num] = euler.divisors(num)
return DIVISORS_DICT[num]
#!/usr/bin/python
import euler
tn = 1
c = 1
ld = 1
while ld < 500:
c += 1
tn += c
f = euler.factors(tn)
if (len(f) > 8):
d = euler.divisors(f)
ld = len(d)
print(c, tn, ld)
def d(n):
return sum(divisors(n))
import sys
from math import ceil
from euler import divisors
def triangular():
i = 1
number = 0
while True:
number += i
i += 1
yield number
target = int(sys.argv[1])
for number in triangular():
div = divisors(number)
if len(div) > target:
print(number, sorted(list(div)))
break
"""
A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.
As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
"""
from itertools import combinations_with_replacement
from euler import divisors
abundant_numbers = [x for x in range(1, 28124) if sum(divisors(x)[:-1]) > x]
Ns = set(range(1, 28124))
ns = {x + y for x, y in combinations_with_replacement(abundant_numbers, 2)}
result = sum(Ns - ns)
def d(n): return sum(euler.divisors(n))
import euler
n=1
while euler.divisors(euler.triangle(n)) < 500:
n += 1
print euler.triangle(n)
import sys
import euler
import collections
import math
import itertools
L = 1500000
ks = euler.divisors(L)
ks = list(ks)
ks.sort()
def tri(n, m, k = 1):
nsq = n*n
msq = m*m
a = k * (msq - nsq)
b = k * 2 * m * n
c = k * (msq + nsq)
return (a,b,c)
def perim(n, m, k = 1):
return 2*k*m*(m+n)
def odd(n):
return not even(n)
def even(n):
return n%2==0
def solve(L, k = 1):
for n in xrange(1,L):
for m in xrange(n+1,L):
#t = tri(n,m,k)
p = perim(n,m,k)
if p > L:
"""
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
"""
from itertools import count
from euler import divisors, triangle
for n in count():
if len(divisors(triangle(n))) > 500:
result = triangle(n)
break
from euler import divisors
from itertools import ifilter
# Let d(n) be defined as the sum of proper divisors of n (numbers less
# than n which divide evenly into n).
#
# If d(a) = b and d(b) = a, where a != b, then a and b are an amicable
# pair and each of a and b are called amicable numbers.
#
# For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20,
# 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of
# 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
#
# Evaluate the sum of all the amicable numbers under 10000.
MAX = 10000
d = lambda n: sum(divisors(n, proper=True))
def amicable(t, dnums):
if t[1] > MAX: return False
if t[0] == t[1]: return False
return dnums[t[1]][1] == t[0]
dnums = [ (x, d(x)) for x in xrange(0, MAX+1) ]
anums = ( t[0] for t in dnums if amicable(t, dnums) )
print sum(anums)
def sum_proper_divisors(n):
return sum(divisors(n)) - n
import sys
import euler
import collections
import math
import itertools
L = 1500000
ks = euler.divisors(L)
ks = list(ks)
ks.sort()
def tri(n, m, k=1):
nsq = n * n
msq = m * m
a = k * (msq - nsq)
b = k * 2 * m * n
c = k * (msq + nsq)
return (a, b, c)
def perim(n, m, k=1):
return 2 * k * m * (m + n)
def odd(n):
return not even(n)
def even(n):
return n % 2 == 0
from euler import trianglenums, divisors
for t in trianglenums():
divs = divisors(t)
if len(divs) > 500:
print('{} has {} divisors'.format(t, len(divs)))
print(divs)
break
import euler
n = 1
while euler.divisors(euler.triangle(n)) < 500:
n += 1
print euler.triangle(n)
"""
Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a != b, then a and b are an amicable pair and each of a and b are called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.
"""
from euler import divisors
result = 0
for n in range(1, 10000):
d = sum(divisors(n)[:-1])
if d != n and sum(divisors(d)[:-1]) == n:
result += n
def sum_proper_divisors(n):
return sum(divisors(n)) - n
import sys
from euler import divisors
amicable = set()
for i in range(4, int(sys.argv[1])+1):
if i in amicable: continue
check = sum(sorted(list(divisors(i)))[:-1])
if check == i: continue
if i == sum(sorted(list(divisors(check)))[:-1]):
amicable.add(i)
amicable.add(check)
print("Found "+str(i)+" and "+str(check));
print(sum(amicable))
def testDivisors(self):
self.assert_(euler.divisors(24) == set([1,2,3,4,6,8,12,24]))
from euler import divisors from math import ceil # see https://en.wikipedia.org/wiki/Abundant_number for assertions # see http://oeis.org/A048242/b048242.txt for pre-generated complete list bound = 28123 # all int > can be written as sum of two abundant abundant = set() for i in range(12, bound): if i in abundant: continue check = sum(sorted(list(divisors(i)))[:-1]) if check > i: abundant.add(i) if not check < i: # multiples of perfect and abundant numbers are abundant j = i+i while j <= bound: abundant.add(j) j += i abundant = list(abundant) abundant.sort() summed = set() for i in abundant: for j in abundant: s = i+j if s > bound: break; summed.add(s) print(sum(set(range(1, bound+1)) - summed))
def iter_chain(n):
while True:
yield n
if n == 0:
return
n = sum(euler.divisors(n, primes)) - n