factors = map(lambda x: int( (-1)**(x-1) ) , k)
maxN = 100
p_n = [0 for i in range(1+maxN)]
p_n[0] = 1
p_n[1] = 1
for n in range(2, 1+maxN):
i = 0
while pents[i] <= n:
p_n[n] = p_n[n] + factors[i] * p_n[ n - pents[i] ]
i = i + 1
N = 100
# Exclude the single partition of size n (question asks for 'sums' only)
print p_n[N] - 1
# Attempt number 3, about the same speed as 2.
abcdCombinatorialUtils.nPartitions.cache = {}
print abcdCombinatorialUtils.nPartitions(N) - 1
# Let p(n) represent the number of different ways in which n coins can
# be separated into piles. For example, five coins can separated into
# piles in exactly seven different ways, so p(5)=7.
# OOOOO
# OOOO O
# OOO OO
# OOO O O
# OO OO O
# OO O O O
# O O O O O
# Find the least value of n for which p(n) is divisible by one
# million.
from abcdCombinatorialUtils import nPartitions
nPartitions.cache = {}
n = 1
while nPartitions(n) % 1000000 > 0:
n += 1
print n, nPartitions(n)
# 55374 and a very big number
