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 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!