Сложение дробей с округлением

n = int(input())
chisl = [i for i in range(n - 1, 0, -1)]
znam = [i ** 2 for i in range(1, n)]
print(sum([int(i / j) for i, j in zip(chisl, znam)]))

n = int(input())
print(sum(int((n - i) / (i * i)) for i in range(1, n)))

print(sum(int((n - i) / (i * i)) for i in range(1, int(n ** 0.5))))

summa = 0
for i in range(n - 1, 1, -1):
 
    ch = i
    zn = (n - i) ** 2
 
    result = ch // zn
 
    if ch < zn:
        break
 
    summa += result
 
print(summa)

for i in range(1, n):
        answ += int((n - i) / (i * i))
    return answ

print(sum(int((n - i) / (i * i)) for i in range(1, int(n ** 0.5) + 1)))

print(sum(int((n - i) / (i * i)) for i in range(2, int(n ** 0.5) + 1)) + n - 1)

sum((n - i) // (i * i) for i in range(1, int(n ** 0.5) + 1))

sum(int((n - i) / (i * i)) for i in range(1, int(n ** 0.5) + 1))

sum((n - i) / i // i for i in range(1, int(n ** 0.5) + 1))

print(6.0 // 2)

sum(i for i in range(1, int(n ** 0.5) + 1))

from math import floor
 
a = int(input())
summa = 0
if a == 2:
    print(1)
elif a == 3:
    print(2)
else:
    for i in range(1 , floor(a**0.5)):
        h = floor((a - i) / (i ** 2))
        summa += h
    print(summa)