#
# Given these facts, we know that if n_k is the product of the k smallest
# primes, then for all n < n_k, n/φ(n) < n_k/φ(n_k).
#
# Finally, then, the solution is simply the largest product of consecutive
# primes starting from 2 that fits within the given maximum.
def prime_product():
product = 1
for prime in primes:
product *= prime
yield product
prime_products = MonatonicIncreasingSequence(prime_product())
# Based in part on the tail recipe in the Python itertools documentation.
def last(iterable):
'''Return the final element of an iterator'''
return deque(iterable, maxlen=1)[0]
if __name__ == '__main__':
try:
maximum = int(argv[1])
except IndexError:
maximum = 1000000
print(last(prime_products.range(maximum)))
from sequence import MonatonicIncreasingSequence
def fibonacci_gen(a=1, b=1):
yield a
while True:
yield b
a, b = b, a + b
# Strictly this is a lie, as the sequence is monatonic non-decreasing with
# a=b=1, but not monatonic increasing. Thankfully it being monatonic increasing
# after the first two elements is sufficient for everything to work.
fibonacci = MonatonicIncreasingSequence(fibonacci_gen())
from itertools import count from sequence import MonatonicIncreasingSequence triangles = MonatonicIncreasingSequence((i * (i + 1)) // 2 for i in count(1)) squares = MonatonicIncreasingSequence(i**2 for i in count(1)) pentagons = MonatonicIncreasingSequence(n * (3 * n - 1) // 2 for n in count(1)) hexagons = MonatonicIncreasingSequence(n * (2 * n - 1) for n in count(1)) heptagons = MonatonicIncreasingSequence(n * (5 * n - 3) // 2 for n in count(1)) octagons = MonatonicIncreasingSequence(n * (3 * n - 2) for n in count(1))
Triangle, pentagonal, and hexagonal numbers are generated by the following
formulae:
Triangle T(n) = n(n+1)/2 1, 3, 6, 10, 15, ...
Pentagonal P(n) = n(3n−1)/2 1, 5, 12, 22, 35, ...
Hexagonal H(n) = n(2n−1) 1, 6, 15, 28, 45, ...
It can be verified that T(285) = P(165) = H(143) = 40755.
Find the next triangle number that is also pentagonal and hexagonal.
'''
from itertools import count
from sequence import MonatonicIncreasingSequence
triangles = MonatonicIncreasingSequence(n * (n + 1) // 2 for n in count(1))
pentagons = MonatonicIncreasingSequence(n * (3 * n - 1) // 2 for n in count(1))
hexagons = MonatonicIncreasingSequence(n * (2 * n - 1) for n in count(1))
if __name__ == '__main__':
# Want the first number over 40755 that is in triangles, pentagons and
# hexagons. It makes little difference which we iterate over, since we'll
# need to generate numbers in each sequence up to the number under test in
# any case, so let's just go with the wording in the question.
for number in triangles.range(40756, None):
if number in pentagons and number in hexagons:
print(number)
break
#!/usr/bin/env python3
'''
Multiples of 3 and 5
If we list all the natural numbers below 10 that are multiples of 3 or 5, we
get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
'''
from itertools import takewhile, count
from sequence import MonatonicIncreasingSequence
multiples = MonatonicIncreasingSequence(n for n in count()
if n % 3 == 0 or n % 5 == 0)
if __name__ == '__main__':
print(sum(multiples.range(999)))
#!/usr/bin/env python3
'''
Multiples of 3 and 5
If we list all the natural numbers below 10 that are multiples of 3 or 5, we
get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
'''
from itertools import count
from sequence import MonatonicIncreasingSequence
multiples = MonatonicIncreasingSequence(n for n in count()
if n % 3 == 0 or n % 5 == 0)
if __name__ == '__main__':
print(sum(multiples.range(999)))
