Summation of GCD of all the pairs up to n

Given a number n, find sum of all GCDs that can be formed by selecting all the pairs from 1 to n.

Examples:

Input : n = 4
Output : 7
Explanation: Numbers from 1 to 4 are: 1, 2, 3, 4
Result = gcd(1,2) + gcd(1,3) + gcd(1,4) + gcd(2,3) + gcd(2,4) + gcd(3,4)
= 1 + 1 + 1 + 1 + 2 + 1
= 7

Input : n = 12
Output : 105

Input : n = 1
Output : 0

Input : n = 2
Output : 1

[Naive Approach] Using Nested Loops - O(n² * log(n)) Time

A Naive approach is to run two loops one inside the other. Select all pairs one by one, find GCD of every pair and then find sum of these GCDs.

[Efficient Approach] Using Euler's Totient Function - O(MAX*log(log MAX)) Time and O(MAX) Space

The idea is to use Euler's Totient function Φ(n) for an input n is count of numbers in {1, 2, 3, ..., n} that are relatively prime to n, i.e., the numbers whose GCD (Greatest Common Divisor) with n is 1. For example, ?(4) = 2, ?(3) = 2 and ?(5) = 4. There are 2 numbers smaller or equal to 4 that are relatively prime to 4, 2 numbers smaller or equal to 3 that are relatively prime to 3. And 4 numbers smaller than or equal to 5 that are relatively prime to 5.

The idea is to convert given problem into sum of Euler Totient Functions.

Sum of all GCDs where j is a part of pair is and j is greater element in pair:

Our final result is

The above equation can be written as :

For every possible GCD 'g' of j. Here count(g) represents count of pairs having GCD equals to g. For every such pair(i, j), we can write : gcd(i/g, j/g) = 1

We can re-write our previous equation as
Sum_j = ∑d * ϕ(j/d)
For every divisor d of j and phi[] is Euler Totient number

Example : j = 12 and d = 3 is one of divisor of j so in order to calculate the sum of count of all pairs having 3 as gcd we can simple write it as
=> 3*ϕ[12/3]
=> 3*ϕ[4]
=> 3*2
=> 6

Therefore sum of GCDs of all pairs where 12 is greater part of pair and 3 is GCD.
GCD(3, 12) + GCD(9, 12) = 6.

Complete Example :
N = 4
Sum₁ = 0
Sum₂ = 1 [GCD(1, 2)]
Sum₃ = 2 [GCD(1, 3) + GCD(2, 3)]
Sum₄ = 4 [GCD(1, 4) + GCD(3, 4) + GCD(2, 4)]

Result = Sum₁ + Sum₂ + Sum₃ + Sum₄
= 0 + 1 + 2 + 4
= 7

Below is the implementation of above idea. We pre-compute Euler Totient Functions and result for all numbers till a maximum value. The idea used in implementation is based this post.

Summation of 4 = 7
Summation of 12 = 105
Summation of 5000 = 61567426

Time complexity: O(MAX*log(log MAX))
Auxiliary space: O(MAX)