Find sum of divisors of all the divisors of a natural number
Given a natural number n, the task is to find sum of divisors of all the divisors of n.
Input : n = 54 Output : 232 Divisors of 54 = 1, 2, 3, 6, 9, 18, 27, 54. Sum of divisors of 1, 2, 3, 6, 9, 18, 27, 54 are 1, 3, 4, 12, 13, 39, 40, 120 respectively. Sum of divisors of all the divisors of 54 = 1 + 3 + 4 + 12 + 13 + 39 + 40 + 120 = 232. Input : n = 10 Output : 28 Divisors of 10 are 1, 2, 5, 10 Sums of divisors of divisors are 1, 3, 6, 18. Overall sum = 1 + 3 + 6 + 18 = 28
Using the fact that any number n can be expressed as product of prime factors, n = p1k1 x p2k2 x … where p1, p2, … are prime numbers.
All the divisors of n can be expressed as p1a x p2b x …, where 0 <= a <= k1 and 0 <= b <= k2.
Now sum of divisors will be sum of all power of p1 – p10, p11,…., p1k1 multiplied by all power of p2 – p20, p21,…., p2k1
Sum of Divisor of n
= (p10 x p20) + (p11 x p20) +…..+ (p1k1 x p20) +….+ (p10 x p21) + (p11 x p21) +…..+ (p1k1 x p21) +……..+
(p10 x p2k2) + (p11 x p2k2) +……+ (p1k1 x p2k2).
= (p10 + p11 +…+ p1k1) x p20 + (p10 + p11 +…+ p1k1) x p21 +…….+ (p10 + p11 +…+ p1k1) x p2k2.
= (p10 + p11 +…+ p1k1) x (p20 + p21 +…+ p2k2).
Now, the divisors of any pa, for p as prime, are p0, p1,……, pa. And sum of diviors will be (p(a+1) – 1)/(p -1), let it define by f(p).
So, sum of divisors of all divisor will be,
= (f(p10) + f(p11) +…+ f(p1k1)) x (f(p20) + f(p21) +…+ f(p2k2)).
So, given a number n, by prime factorization we can find the sum of divisors of all the divisors. But in this problem we are given that n is product of element of array. So, find prime factorization of each element and by using the fact ab x ac = ab+c.
Below is the implementation of this approach:
For the cases when there are multiple inputs for which we need find the value, we can use Sieve of Eratosthenes as discussed in this post.
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