Python Program for Find sum of odd factors of a number

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, ... p_k be prime factors of n. Let a₁, a₂, .. a_k be highest powers of p₁, p₂, .. p_k respectively that divide n, i.e., we can write n as n = (p₁^a₁)*(p₂^a₂)* ... (p_k^a_k).

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 sum 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.

# 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 code
n = 30
print(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!