def num_triples(N):
count, do = 0, {-1: True, 1: True}
for y, x in islice(pell_solutions(2, -1), 1, None):
cont = False
for t in (t for t in [-1, 1] if do[t]):
g = gcd(x + t, y - x)
if ((y + t) / g) % 2:
s = 2 * (x + t) * (y + t) / (g * g)
if s <= N:
count += N / s
cont = True
else:
do[t] = False
if not cont:
break
return count
def min_balls(N):
a_min = (1 + 2 * N * (N - 1)) ** 0.5
return (1 + dropwhile(lambda (_, a): a <= a_min, pell_solutions(2, -1)).next()[1]) / 2
Consider the infinite polynomial series AG(x) = xG1 + x2G2 + x3G3 + ..., where Gk is the kth term of the second order recurrence relation Gk = Gk1 + Gk2, G1 = 1 and G2 = 4; that is, 1, 4, 5, 9, 14, 23, ... .
For this problem we shall be concerned with values of x for which AG(x) is a positive integer.
The corresponding values of x for the first five natural numbers are shown below.
x AG(x)
(51)/4 1
2/5 2
(222)/6 3
(1375)/14 4
1/2 5
We shall call AG(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 20th golden nugget is 211345365.
Find the sum of the first thirty golden nuggets.
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'''
from problem094 import pell_solutions
import itertools as it
A0 = it.chain([(7, 1)], ((7 * x + 5 * y, 7 * y + x) for x, y in it.islice(pell_solutions(5), 1, None, 2)))
A1 = ((7 * x - 5 * y, 7 * y - x) for x, y in it.islice(pell_solutions(5), 1, None, 2))
B0 = ((15 * z + y, 3 * y + z) for y, z in it.islice(pell_solutions(5, -1), 0, None, 2))
B1 = ((15 * z - y, 3 * y - z) for y, z in it.islice(pell_solutions(5, -1), 1, None, 2))
C0 = ((2 * (4 * x + 5 * y), 2 * (4 * y + x)) for x, y in it.islice(pell_solutions(5), 0, None, 2))
C1 = ((2 * (4 * x - 5 * y), 2 * (4 * y - x)) for x, y in it.islice(pell_solutions(5), 0, None, 2))
if __name__ == "__main__":
print sum(it.imap(lambda (c, _):(c - 7) / 5, it.islice(it.chain.from_iterable(it.izip(A0, B0, C1, C0, B1, A1)), 1, 31)))