Find sum of even factors of a number

Given a number n, the task is to find the even factor sum of a number.
Examples: 
 

Input : 30
Output : 48
Even dividers sum 2 + 6 + 10 + 30 = 48

Input : 18
Output : 26
Even dividers sum 2 + 6 + 18 = 26

 

Prerequisite : Sum of factors
As discussed in above mentioned previous post, sum of factors of a number is
Let p1, p2, … pk be prime factors of n. Let a1, a2, .. ak be highest powers of p1, p2, .. pk respectively that divide n, i.e., we can write n as n = (p1^a1)*(p2^a2)* … (pk^ak).
 

Sum of divisors = (1 + p1 + p12 ... p1a1) * 
 (1 + p2 + p22 ... p2a2) *
 ...........................
 (1 + pk + pk2 ... pkak) 

If number is odd, then there are no even factors, so we simply return 0.
If number is even, we use above formula. We only need to ignore 2⁰. All other terms multiply to produce even factor sum. For example, consider n = 18. It can be written as 2¹3² and sum of all factors is (2⁰ + 2¹)*(3⁰ + 3¹ + 3²). if we remove 2⁰ then we get the 
Sum of even factors (2)*(1+3+3²) = 26.
To remove odd number in even factor, we ignore then 2⁰ which is 1. After this step, we only get even factors. Note that 2 is the only even prime. 
Below is the implementation of the above approach. 
 

Output: 
 

26

Time Complexity: O(?n log n) 
Auxiliary Space: O(1)

Please suggest if someone has a better solution which is more efficient in terms of space and time.