Problem

Given a number $N$, find the sum of all numbers less than or equal to $N$ that are co-prime with $N$.

In case you forgot, a number $x$ is co-prime with $N$ if $gcd(x,N) = 1$.

For example, if $N = 10$, then the following numbers are co-prime with it: $[1, 3, 7, 9]$. Therefore, sum of co-prime numbers will be $1 + 3 + 7 + 9 = 20$.

Solution

Let us define a function $f(n)$, which gives us sum of all numbers less than or equal to $n$ that are co-prime to $n$. Then we can calculate the value of $f(n)$ with the following formula:

$$\bbox[yellow,5px]{
f(n) = \frac{\phi(n)}{2}n
}$$

where $\phi(n)$ is Euler Phi Function.

For example, for $n = 10$, then we get the sum of co-prime as:

$$\begin{align}
f(10) & = \frac{\phi(10)}{2} \times 10 \\
& = \frac{4}{2} \times 10 \\
& = 20
\end{align}
$$

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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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