def sumFactors(n):
return sum(d for d in range(1, n) if n%d == 0)
# Simple Python 3 program to
# find sum of all divisors of
# a natural number
import math
# Function to calculate sum
# of all divisors of given
# natural number
def divSum(n) :
if(n == 1):
return 1
# Final result of summation
# of divisors
result = 0
# find all divisors which
# divides 'num'
for i in range(2,(int)(math.sqrt(n))+1) :
# if 'i' is divisor of 'n'
if (n % i == 0) :
# if both divisors are same
# then add it only once
# else add both
if (i == (n/i)) :
result = result + i
else :
result = result + (i + n//i)
# Add 1 and n to result as above
# loop considers proper divisors
# greater than 1.
return (result + n + 1)
# Driver program to run the case
n = 30
print(divSum(n))
# This code is contributed by Nikita Tiwari.