WAP to Find Number of Divisor & sum of All Proper Devisor of Number Efficiently

Given a natural number n (1 <= n <= 500000), please output the summation of all its proper divisors.
Definition: A proper divisor of a natural number is the divisor that is strictly less than the number.
e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.

Input
An integer stating the number of test cases (equal to about 200000), and that many lines follow, each containing one integer between 1 and 500000 inclusive.
Output
One integer each line: the divisor summation of the integer given respectively.
Example
Sample Input:
3
2
10
20
Sample Output:
1
8
22

Writing the O(N) code is not the big deal for this(traverse till N/2)..question but writing quality code O(sqrt(n)) is efficient way to solve it.Thats why great companies used to ask such question because the wants quality code.

public static int getSumProperDivisors(int number)
{
    int n=number;
    int sum=1;
    
        int numOfDiv=0;
        int loop=(int)Math.sqrt(number);

        for (int i=2;i<=loop;i++)
        {
            if(n%i == 0)
            {
                numOfDiv += 2;
                sum += i + n/i;
            }

        }

        if(loop*loop==n)
        {
            numOfDiv--;
            sum -= loop;
        }

        
    return sum;
}

TC o(sqrt(n))
SC O(1)