N

```
def aliquot_sum(input_num: int) -> int:
"""
Finds the aliquot sum of an input integer, where the
aliquot sum of a number n is defined as the sum of all
natural numbers less than n that divide n evenly. For
example, the aliquot sum of 15 is 1 + 3 + 5 = 9. This is
a simple O(n) implementation.
@param input_num: a positive integer whose aliquot sum is to be found
@return: the aliquot sum of input_num, if input_num is positive.
Otherwise, raise a ValueError
Wikipedia Explanation: https://en.wikipedia.org/wiki/Aliquot_sum
>>> aliquot_sum(15)
9
>>> aliquot_sum(6)
6
>>> aliquot_sum(-1)
Traceback (most recent call last):
...
ValueError: Input must be positive
>>> aliquot_sum(0)
Traceback (most recent call last):
...
ValueError: Input must be positive
>>> aliquot_sum(1.6)
Traceback (most recent call last):
...
ValueError: Input must be an integer
>>> aliquot_sum(12)
16
>>> aliquot_sum(1)
0
>>> aliquot_sum(19)
1
"""
if not isinstance(input_num, int):
raise ValueError("Input must be an integer")
if input_num <= 0:
raise ValueError("Input must be positive")
return sum(
divisor for divisor in range(1, input_num // 2 + 1) if input_num % divisor == 0
)
if __name__ == "__main__":
import doctest
doctest.testmod()
```

The aliquot sum $s(n)$ of a positive integer $n$ is the sum of all proper divisors of $n$, that is, all divisors of $n$ other than the number $n$ itself. That is:

$$ s(n) = \sum_{d | n, d \neq n} {d} $$

So, for example, the aliquot sum of the number $15$ is $(1 + 3 + 5) = 9$

Aliquot sum is a very useful property in Number Theory, and can be used for defining:

- Prime Numbers
- Deficient Numbers
- Abundant Numbers
- Perfect Numbers
- Amicable Numbers
- Untouchable Numbers
- Aliquot Sequence of a number
- Quasiperfect & Almost Perfect Numbers
- Sociable Numbers

- 1 is the only number whose aliquot sum is 0
- The aliquot sums of perfect numbers is equal to the numbers itself
- For a
*semiprime*number of the form $pq$, the aliquot sum is $p + q + 1$ - The Aliquot sum function was one of favorite topics of investigation for the world famous Mathematician, Paul Erdős

We loop through all the numbers from $1$ to $[\frac{n} 2]$ and check if they divide $n$, which if they do we add them as a proper divisor.

The reason we take the upper bound as $[\frac{n} 2]$ is that, the largest possible proper divisor of an even number is $\frac{n} 2 $, and if the number is odd, then its largest proper divisor is less than $[\frac{n} 2]$, hence making it a foolproof upper bound which is computationally less intensive than looping from $1$ to $n$.

The sum which we obtain is the aliquot sum of the number