# Python Program for Find sum of odd factors of a number

• Last Updated : 25 Jun, 2022

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

Input : n = 30
Output : 24
Odd dividers sum 1 + 3 + 5 + 15 = 24

Input : 18
Output : 13
Odd dividers sum 1 + 3 + 9 = 13

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 = (p1a1)*(p2a2)* … (pkak).

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

To find sum of odd factors, we simply need to ignore even factors and their powers. For example, consider n = 18. It can be written as 2132 and sun of all factors is (1)*(1 + 2)*(1 + 3 + 32). Sum of odd factors (1)*(1+3+32) = 13. To remove all even factors, we repeatedly divide n while it is divisible by 2. After this step, we only get odd factors. Note that 2 is the only even prime.

## python3

 # Formula based Python3 program# to find sum of all divisors# of n.import math # Returns sum of all factors# of n.def sumofoddFactors( n ):         # Traversing through all    # prime factors.    res = 1         # ignore even factors by    # of 2    while n % 2 == 0:        n = n // 2         for i in range(3, int(math.sqrt(n) + 1)):                 # While i divides n, print        # i and divide n        count = 0        curr_sum = 1        curr_term = 1        while n % i == 0:            count+=1                         n = n // i            curr_term *= i            curr_sum += curr_term                 res *= curr_sum         # This condition is to    # handle the case when    # n is a prime number.    if n >= 2:        res *= (1 + n)         return res # Driver coden = 30print(sumofoddFactors(n)) # This code is contributed by "Sharad_Bhardwaj".

Output:

24

Time complexity: O(sqrt(n))

Auxiliary Space: O(1)

Please refer complete article on Find sum of odd factors of a number for more details!

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