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### szaranczuk's blog

By szaranczuk, history, 18 months ago,

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 $O(n\log n)$.

Implementation

## 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:

• +108