On today's POI training camp round I have learnt a nice technique that could possibly be useful in some number theory problems. I couldn't find any CF article on it, so I think it's fair enough to share it on my own.

Remark on used notation

In some sums I will use an Iverson notation

Problem: Squarefree Function

Let's define a Squarefree Function $$$f(x)$$$ for any positive integer $$$x$$$ as $$$x$$$ divided by a greatest perfect square, that divides $$$x$$$.

For example: $$$f(1) = 1$$$, $$$f(2) = 2$$$, $$$f(4) = 1$$$, $$$f(6) = 6$$$, $$$f(27) = 3$$$, $$$f(54) = 6$$$, $$$f(800) = 2$$$

Given an array $$$a$$$ of $$$n \leq 10^5$$$ positive integers, where each $$$a_i \leq 10^5$$$ compute sum

\begin{gather} \sum\limits_{1 \leq i,j \leq n}f(a_i\cdot a_j) \mod (10^9 + 7) \end{gather}

Technique: GCD Convolution

You might probably heard about a Sum Convolution. For two arrays $$$b$$$, and $$$c$$$, it is defined as an array $$$d$$$ such that \begin{gather} d_k = \sum\limits_{i + j = k}b_i\cdot c_j \end{gather} If not, it's basically the same thing as a polynomial multiplication. If $$$B(x) = b_0 + b_1x + b_2x^2 + ... + b_nx^n$$$, and $$$C(x) = c_0 + c_1x + c_2x^2 + ... + c_nx^n$$$, then $$$(B \cdot C)(x) = d_0 + d_1x + d_2x^2 + ... + d_{2n}x^{2n}$$$

Let's define GCD Convolution by analogy

Definition

A GCD Convolution of two arrays $$$b$$$, and $$$c$$$, consisting of positive integers, is an array $$$d$$$ such that \begin{gather} d_k = \sum\limits_{gcd(i,j) = k}b_i\cdot c_j \end{gather}

Algorithm to find GCD Convolution

Of course, we can compute it naively by iterating over all pairs of indicies. If $$$b$$$ and $$$c$$$ consists of $$$n$$$ elements then the overall complexity would be $$$O(n^2log(max(b) + max(c)))$$$. But it turns out, that we can do better.

Let's look at the sum of $$$d_k$$$ values, with indicies divisible by some integer $$$g$$$, so that $$$k = gm$$$ is satisfied for some integer m. \begin{gather} \sum\limits_{m=1}^{n/g}d_{gm} = \sum\limits_{m=1}^{n/g}\sum\limits_{gcd(i,j) = gm}b_i\cdot c_j = \sum\limits_{g | gcd(i,j)}b_i\cdot c_j \end{gather}

From the definition of gcd, we know that $$$g | gcd(i,j) \Leftrightarrow g | i \wedge g | j$$$ \begin{gather} \sum\limits_{g | gcd(i,j)}b_i\cdot c_j = \sum\limits_{i,j}b_i\cdot c_j[g \;|\; gcd(i,j)] = \sum\limits_{i,j}b_i\cdot c_j[g \;|\; i][g \;|\; j] = \end{gather} \begin{gather} =\sum\limits_{i,j}\left(b_i[g \;|\; i]\right)\cdot \left(c_j[g \;|\; j]\right) = \left(\sum\limits_{g|i}b_i\right)\left(\sum\limits_{g|j}c_j\right) \end{gather}

We can define $$$B_g = \sum_{m=1}^{n/g}b_{gm}$$$, and $$$C_g = \sum_{m=1}^{n/g}c_{gm}$$$, and $$$D_g = \sum_{m=1}^{n/g}d_{gm}$$$. From above equation one could easily derive $$$D_g = B_g\cdot C_g$$$. Knowing that $$$O(n + \frac{n}{2} + \frac{n}{3} + ...) = O(n\log n)$$$, arrays $$$B$$$ and $$$C$$$ can be computed directly from their definitions in $$$O(n\log n)$$$.

Recovering a $$$d_k$$$ values from D array is simple. All we need is just subtract all the summands of $$$D_i$$$ except for the smallest. So, formally, we have \begin{gather} d_k = D_k - \sum\limits_{m=2}^{n/k}d_{km} \end{gather} Which can be computed using dynamic programming, starting from $$$k = n$$$.

So, the overall complexity of computing a GCD Convolution of two arrays of size $$$n$$$ is $$$O(n\log n)$$$.

template<typename T>
std::vector<T> gcdConvolution(std::vector<T> a, std::vector<T> b, T mod)
{
	int n = a.size();
	std::vector<T> A(n);
	std::vector<T> B(n);
	std::vector<T> D(n);
	for (int i = 0; i < n; i++)
	{
		for (int j = i; j < n; j += i+1)
		{
			A[i] += a[j];
			B[i] += b[j];
			A[i] %= mod;
			B[i] %= mod;
		}
		D[i] = A[i] * B[i];
		D[i] %= mod;
	}
	std::vector<T> d(n);
	for (int i = n-1; i >= 0; i--)
	{
		d[i] = D[i];
		for (int j = 2*i+1; j < n; j += i + 1)
		{
			d[i] -= d[j];
			d[i] = (d[i] + mod) % mod;
		}
	}
	return d;

}

Back to original problem

We can see, that \begin{gather} f(a_i\cdot a_j) = \frac{f(a_i)\cdot f(a_j)}{gcd(f(a_i), f(a_j))^2} \end{gather}

So, having an array $$$w_{f(a_i)} = \sum\limits_if(a_i)$$$ all we need is just to compute a GCD Convolution of $$$w$$$ with itself. Let's denote that convolution by $$$d$$$. Then, by definition \begin{gather} \sum\limits_{i,j :\;gcd(f(a_i), f(a_j)) = k} \frac{f(a_i)\cdot f(a_j)}{gcd(f(a_i), f(a_j))^2} = \frac{d_k}{k^2} \end{gather}

So answer to our problem is just a sum \begin{gather} \sum\limits_{k=1}^{max(f(a_i))}\frac{d_k}{k^2} \end{gather}

Assuming that we have computed $$$f(a_i)$$$ values with sieve, if we denote $$$A = max(a_i)$$$, then overall complexity of this solution is $$$O(n + A\log A)$$$

Practice problems

Actually, I don't have any. I will be glad if you share some problems in comments. All I have is just this:

• GCD Convolution