def euler026(maxd):
''' A unit fraction contains 1 in the numerator.
The decimal representation of the unit fractions with denominators 2 to 10 are given
1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
>>> euler026(10)
7
>>> euler026(1000)
983
'''
b = Biggest()
for p in primes(maxd):
b.set(find_repeat(p), p)
return b.data
def euler027(limit):
'''Quadratic primes
Euler discovered the remarkable quadratic formula:
n**2 + n + 41
It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39.
However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when
n = 41, 41n**2 + 41 + 41 is clearly divisible by 41.
The incredible formula n**2 - 79n + 1601 was discovered, which produces 80 primes for the
consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is -126479.
Considering quadratics of the form:
n**2 + an + b, where |a| < 1000 and |b| < 1000
where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |-4| = 4
Find the product of the coefficients, a and b, for the quadratic expression that produces
the maximum number of primes for consecutive values of n, starting with n = 0.
'''
P = Primes()
B = Biggest()
for a, b in product(range(-limit+1, limit), primes(limit)):
B.set(len(list(takewhile(lambda x: quadratic(x, a, b) in P, count()))), a * b)
return B.data