Python Program for Find sum of even factors of a number

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Python Program for 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

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) 

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 sun 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 whaich is 1. After this step, we only get even factors. Note that 2 is the only even prime.

# Formula based Python3
# program to find sum 
# of alldivisors of n.
import math
 
# Returns sum of all 
# factors of n.
def sumofFactors(n) :
     
    # If n is odd, then
    # there are no even
    # factors.
    if (n % 2 != 0) :
        return 0
  
    # Traversing through
    # all prime factors.
    res = 1
    for i in range(2, (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= count + 1
  
            n = n // i
  
            # here we remove the
            # 2^0 that is 1. All
            # other factors
            if (i == 2 and count == 1) :
                curr_sum = 0
  
            curr_term = curr_term * i
            curr_sum = curr_sum + curr_term
         
        res = res * curr_sum
         
  
    # This condition is to
    # handle the case when
    # n is a prime number.
    if (n >= 2) :
        res = res * (1 + n)
  
    return res
 
 
# Driver code
n = 18
print(sumofFactors(n))
 
 

Output:

26

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