##Consider the fraction, n/d, where n and d are positive integers. If nd and
##HCF(n,d)=1, it is called a reduced proper fraction.
##
##If we list the set of reduced proper fractions for d  8 in ascending order of
##size, we get:
##
##1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7,
##3/4, 4/5, 5/6, 6/7, 7/8
##It can be seen that there are 3 fractions between 1/3 and 1/2.
##How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper
##fractions for d  12,000?
##
##Note: The upper limit has been changed recently.
def HCF(i, j):
    if j == 0:
        return i
    else:
        return HCF(j, i % j)


def f(n):
    l = [
        (i, j)
        for i in range(4, n + 1)
        for j in range(i // 3 + 1, i // 2 + 1)
        if HCF(i, j) == 1
    ]
    return len(l)


print(f(12000))
