Problem

Given a positive integer $N$, find the value of $g(N)$, where $$g(n) = gcd(1,n) + gcd(2,n) + gcd(3,n) + \dots + gcd(n,n) = \sum_{i=1}^n gcd(i,n)$$

For example,
$$ \begin{align}
g(6) & = gcd(1,6) + gcd(2,6) + g(3,6) + gcd(4,6) + gcd(5,6) + gcd(6,6) \\
& = 1 + 2 + 3 + 2 + 1 + 6 \\
& = 15
\end{align}$$

The function $g(n)$ is called GCD Sum Function for obvious reasons.

There is a paper on this topic, “The gcd-sum Function” and in this post, I will attempt to explain it as simply as possible.

GCD Sum Function – $g(n)$

In short, there is a direct formula for calculating the value of $g(n)$.

If the prime factorization of $n$ is $p_{1}^{a_1} \times p_2^{a_2}\times \dots p_k^{a_k}$, then $$g(n) = \prod_{i=0}^k (a_i + 1)p_i^{a_i} – a_ip^{a_i-1}$$

For example, we know that $6 = 2 \times 3$. So using the formula above, we get:
$$ \begin{align}
g(6) & = \big((1+1)2 – 2^{1-1} \big) \big((1+1)3 – 3^{1-1}\big) \\
& = (4-1)(6-1) \\
& = 15
\end{align}$$

Very neat formula. So, as long as we can prime factorize $n$, we can easily calculate gcd-sum for $n$.

For the curious minds who want to know how we got the formula above, please proceed to next section.

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A function $f(n)$ is called multiplicative function if:

1. $f(n)$ is defined for positive integer $n$.
2. $f(1) = 1$
3. $f(mn) = f(m)f(n)$ whenever $gcd(m,n) = 1$.

For example, the following functions are multiplicative:

• Euler Phi – $\phi(n)$
• Number of Divisor – $\sigma_0(n)$
• Sum of Divisor – $\sigma_1(n)$

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