vector<int> prime; /// prime list
vector<int> lpf;   /// Lowest prime factor, lpf[x] is smallest prime divisor of x
void sieve(int lim = LIM) /// O(n)
{
    prime.assign(1, 2);
    lpf.assign(lim + 1, 2);

    lpf[1] = 1; /// For easier calculation but can cause inf loops
    for (int i = 3; i <= lim; i += 2) {
        if (lpf[i] == 2) prime.push_back(lpf[i] = i);
        for (int j = 0; j < sz(prime) && prime[j] <= lpf[i] && prime[j] * i <= lim; ++j)
            lpf[prime[j] * i] = prime[j];
    }
}

/// mask(x) is smallest positive number that mask(x) * x is a perfect square
int getMask(int x) /// O(log n)
{
    int mask = 1;
    while (x > 1) {
        int p = lpf[x], f = 0;
        do x /= p, f++; while (p == lpf[x]);
        if (f & 1) mask *= p; /// if current power is odd, we mutiple mask with current prime
    }
    return mask;
}

int cnt[LIM];
int magic(int n) /// O(n log max(a))
{
    memset(cnt, 0, sizeof(cnt[0]) * (n + 1));

    ll res = 0;
    for (int a = 1; a <= n; ++a) /// Check all cases of a
        res += cnt[getMask(a)]++;

    res %= MOD;
    return res;
}

int main() /// O(n log max(a))
{
    int n;
    cin >> n;
    sieve(n + 500);
    cout << magic(n);    
    return 0;
}

int solve(int n)
{
    memset(is_squarefree, true, sizeof(is_squarefree[0]) * (n + 1));

    long long res = 0;
    for (int i = 1, j; i <= n; ++i) if (is_squarefree[i]) 
    {
        for (j = 1; i * j * j <= n; ++j)
            is_squarefree[i * j * j] = false;

        res += 1LL * (j - 1) * (j - 2) / 2;
    }

    res %= MOD;
    return res;
}

int solve(int n)
{
    memset(is_squarefree, true, sizeof(is_squarefree[0]) * (n + 1));

    long long res = 0;
    for (int i = 1; i <= n; ++i) if (is_squarefree[i]) 
    {
        for (int j = 1; i * j * j <= n; ++j)
        {
            is_squarefree[i * j * j] = false;
            res += j - 1;
        }
    }

    res %= MOD;
    return res;
}

/// Linear Sieve
vector<bool> isPrime; /// Characteristic function = A010051
vector<int> prime;    /// Prime list              = A000040
vector<int> lpf;      /// Lowest prime factor     = A020639
vector<int> mu;       /// Mobius                  = A008683
vector<int> phi;      /// Euler totient phi       = A000010
void sieve(int n)
{
    if (n < 1) return ;
    /// Extension part
    mu.assign(n + 1, 1);
    phi.assign(n + 1, 1);
    /// Main part
    prime.clear();
    lpf.assign(n + 1, 0);
    isPrime.assign(n + 1, true);
    isPrime[0] = isPrime[1] = false;
    for (int x = 2; x <= n; ++x)
    {
        if (isPrime[x]) /// Func[Prime]
        {
            lpf[x] = x;
             mu[x] = -1;
            phi[x] = x - 1;
            prime.push_back(x);
        }
 
        for (int p : prime) /// Func[Prime * X] <- Func[Prime]
        {
            if (p > lpf[x] || x * p > n) break;
            isPrime[x * p] = 0;
            lpf[x * p] = p;
            if (lpf[x] == p)
            {
                 mu[x * p] = 0;
                phi[x * p] = phi[x] * p;
            }
            else 
            {
                 mu[x * p] = -mu[x];
                phi[x * p] = phi[x] * phi[p];
            }
        }
    }
}

long long solve(int n)
{
    sieve(n + 1);
    long long res = 0;
    for (int x = 1; x <= n; ++x) if (mu[x]) 
    {
        int t = sqrt(n / x);
        res += 1LL * t * (t - 1) / 2;
    }

    res %= MOD;
    return res;
}

It just related to the sum of inversion of primes.

Precalculating smallest prime factor for each $$$n$$$.

Since you known the smallest prime factor for each $$$n$$$, you can factorize any number in $$$O(\log n)$$$.

There are $$$n$$$ number total. Hence the calculating part take $$$O(n \log n)$$$.

The sieve only cost you $$$O(n)$$$ give the total complexity of $$$O(n \log n)$$$.

This just a clickbait. Have a nice day.

Each number will be calculated once.

Though some numbers might be visited many times. It is actually armotized to constant.

Let remind you about something called the sum of inversion of squares.

Just ignore those squarefree. A normal loop through it is still linear.

$$$O\left(\underset{\text{squarefree } p \leq n}{\Large \Sigma} \left \lfloor {\Large \frac{n}{p^2}} \right \rfloor \right) = O\left(\underset{p \leq n}{\Large \Sigma} {\Large \frac{n}{p^2}} \right) = O\left(n \times {\Large \Sigma} {\Large \frac{1}{p^2}} \right) = O\left(n \times {\Large \frac{\pi^2}{6}} \right) = O(n)$$$.

It depends on the sieve as the calculation is already linear.

Each number can be represent as $$$x = lpf[x] \times k$$$ where $$$lpf[x]$$$ is lowest prime factor of $$$x$$$ and $$$k$$$ is an integer

Each $$$x$$$ can be represented uniquely as the form $$$x = lpf[x] \times k$$$.

So you can make a sieve of linear, where each number is visisted once or atmost $$$O(1)$$$

const int LIM = 1e7 + 17;
const int MOD = 1e9 + 7;

int fact[LIM + 1]; /// factorial:         fact[n] = n!
int invs[LIM + 1]; /// inverse modular:   invs[n] = n^(-1)
int tcaf[LIM + 1]; /// inverse factorial: tcaf[n] = (n!)^(-1)
void precal(int n = LIM)
{
    fact[0] = fact[1] = 1;
    invs[0] = invs[1] = 1;
    tcaf[0] = tcaf[1] = 1;
    for (int i = 2; i <= n; ++i)
    {
        fact[i] = (1LL * fact[i - 1] * i) % MOD;
        invs[i] = MOD - 1LL * (MOD / i) * invs[MOD % i] % MOD;
        tcaf[i] = (1LL * tcaf[i - 1] * invs[i]) % MOD;
    }
}

int nCk(int n, int k)
{
    k = min(k, n - k);
    if (k < 0) return 0;

    long long res = fact[n];
    res *= tcaf[k];         res %= MOD;
    res *= tcaf[n - k];     res %= MOD;
    return res;
}

int solve(int n)
{
    memset(is_squarefree, true, sizeof(is_squarefree[0]) * (n + 1));

    long long res = 0;
    for (int i = 1, j; i <= n; ++i) if (is_squarefree[i]) 
    {
        for (j = 1; i * j * j <= n; ++j)
            is_squarefree[i * j * j] = false;

        res += nCk(j - 1, k);
    }

    res %= MOD;
    return res;
}

int solve(int n)
{
    memset(is_squarefree, true, sizeof(is_squarefree[0]) * (n + 1));
    long long res = 0;

    int t = sqrt(n);
    while (t * t < n) ++t;
    while (t * t > n) --t;
    for (int j = t; j > 1; --j)
    {
        for (int i = 1; i * j * j <= n; ++i)
        {
            if (!used[i * j * j])
            {
                used[i * j * j] = true;
                res += j - 1;
            }
        }
    }

    res %= MOD;
    return res;
}

Well, you see, since $$$(a, b, n)$$$ form a three-term geometric progression then you have $$$b^2 = an$$$.

It is not hard to proove that $$$F(N)$$$ will be our answer as we count over all possible squarefree $$$u$$$ for every fixed $$$p^2$$$.

We use $$$g(n)$$$ because its beautiful property, which make the calculation a lot easier.

$$$F(N) = \overset{n}{\underset{p=1}{\Large \Sigma}} f(p)$$$.

and

$$$f(n) = g(n) + 1$$$.

We have

$$$F(n)$$$

$$$= \overset{n}{\underset{p=2}{\Large \Sigma}} g(p) - n$$$

$$$= \overset{n}{\underset{p=2}{\Large \Sigma}} \underset{d^2 | p}{\Large \Sigma} \phi(d) - n$$$

Break the loop, make it simpler.

Reduce a for loop to constant.

$$$F(N)$$$

$$$= \overset{n}{\underset{p=1}{\Large \Sigma}} f(p)$$$

$$$= -n + \overset{n}{\underset{p=2}{\Large \Sigma}} g(p)$$$

$$$= -n + \overset{n}{\underset{p=2}{\Large \Sigma}} \underset{d^2 | p}{\Large \Sigma} \phi(d)$$$

$$$= -n + \overset{\left \lfloor \sqrt{n} \right \rfloor}{\underset{p=1}{\Large \Sigma}} \underset{1 \leq u \times p^2 \leq n}{\Large \Sigma} \phi(p)$$$

$$$= -n + \overset{\left \lfloor \sqrt{n} \right \rfloor}{\underset{p=1}{\Large \Sigma}} \phi(p) \times \left \lfloor \frac{n}{p^2} \right \rfloor$$$

$$$= \overset{\left \lfloor \sqrt{n} \right \rfloor}{\underset{p=2}{\Large \Sigma}} \phi(p) \times \left \lfloor \frac{n}{p^2} \right \rfloor$$$

Well, you can sieve $$$\phi(p)$$$ for all $$$2 \leq p \leq \sqrt{n}$$$ in $$$O\left(\sqrt{n} \log \log \sqrt{n} \right)$$$ or improve it with linear sieve to $$$O(\sqrt{n})$$$.

Therefore you can solve the whole problem in $$$O(\sqrt{n})$$$.

#include <iostream>
#include <cstring>
#include <numeric>
#include <cmath>

using namespace std;

const int MOD = 1e9 + 7;
const int LIM = 1e7 + 17;
const int SQRT_LIM = ceil(sqrt(LIM) + 1) + 1;

int euler[SQRT_LIM];
void sieve_phi(int n)
{
    iota(euler, euler + n + 1, 0);
    for (int x = 2; x <= n; x++) if (euler[x] == x)
        for (int j = x; j <= n; j += x)
            euler[j] -= euler[j] / x;
}

int solve(int n)
{
    sieve_phi(ceil(sqrt(n) + 1) + 1);
    
    long long res = 0;
    for (int p = 2; p * p <= n; ++p)
        res += 1LL * euler[p] * (n / (p * p));

    res %= MOD;
    return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    
    int n;
    cin >> n;
    cout << solve(n);
    return 0;
}

#include <iostream>
#include <cstring>
#include <numeric>
#include <vector>
#include <cmath>

using namespace std;

const int MOD = 1e9 + 7;

vector<int> lpf;
vector<int> prime;
vector<int> euler;
void linear_sieve_phi(int n)
{
    lpf.assign(n + 1, 0);
    euler.assign(n + 1, 1);
    for (int x = 2; x <= n; ++x)
    {
        if (lpf[x] == 0)
        {
            prime.push_back(lpf[x] = x);
            euler[x] = x - 1;                    
        }
        for (int i = 0; i < prime.size() && x * prime[i] <= n; ++i)
        {
            lpf[x * prime[i]] = prime[i];
            if (x % prime[i] == 0) {
                euler[x * prime[i]] = euler[x] * prime[i];    
                break;
            }
            euler[x * prime[i]] = euler[x] * euler[prime[i]];
        }
    }
}

int solve(int n)
{
    linear_sieve_phi(ceil(sqrt(n) + 1) + 1);
    
    long long res = 0;
    for (int p = 2; p * p <= n; ++p)
        res += 1LL * euler[p] * (n / (p * p));

    res %= MOD;
    return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    
    int n;
    cin >> n;
    cout << solve(n);
    return 0;
}

$$$s_k(n) = C^{n-1}_{k-1}$$$

$$$f_k(n) = F_k(n) - F_k(n-1) = s_k(g_k(n)) = \underset{d^2 | n}{\Large \Sigma} (\mu \times s_k)(d) = \underset{d^2 | n}{\Large \Sigma} \underset{p | d} \mu(p) \times s_k\left(\frac{d}{p}\right)$$$

$$$F_k(n) = \overset{n}{\underset{p=1}{\Large \Sigma}} f_k(p)$$$

$$$= F_k(0) + \overset{\left \lfloor \sqrt{n} \right \rfloor}{\underset{d = 1}{\Large \Sigma}}(\mu \times s_k)(d) \times \left \lfloor \frac{n}{d^2} \right \rfloor$$$

$$$= \overset{\left \lfloor \sqrt{n} \right \rfloor}{\underset{d = 1}{\Large \Sigma}} \left \lfloor \frac{n}{d^2} \right \rfloor \times \left ( \underset{p | d}{\Large \Sigma} \mu(p) \times s_k\left(\frac{d}{p}\right) \right)$$$

#include <iostream>
#include <cstring>
#include <numeric>
#include <vector>
#include <cmath>

using namespace std;

const int LIM = 5e6 + 56;
const int SQRT_LIM = ceil(sqrt(LIM) + 1) + 1;
const int MOD = 1e9 + 7;

int fact[SQRT_LIM + 10];
int invs[SQRT_LIM + 10];
int tcaf[SQRT_LIM + 10];
void precal_nck(int n = SQRT_LIM)
{
    fact[0] = fact[1] = 1;
    invs[0] = invs[1] = 1;
    tcaf[0] = tcaf[1] = 1;
    for (int i = 2; i <= n; ++i)
    {
        fact[i] = (1LL * fact[i - 1] * i) % MOD;
        invs[i] = MOD - 1LL * (MOD / i) * invs[MOD % i] % MOD;
        tcaf[i] = (1LL * tcaf[i - 1] * invs[i]) % MOD;
    }
}

int nck(int n, int k)
{
    k = min(k, n - k);
    if (k < 0) return 0;

    long long res = fact[n];
    res *= tcaf[k];         res %= MOD;
    res *= tcaf[n - k];     res %= MOD;
    return res;
}

vector<int> prime;           /// prime list              = A000040
bool isPrime[SQRT_LIM + 10]; /// characteristic function = A010051
int lpf[SQRT_LIM + 10];      /// lowest prime factor     = A020639
int mu[SQRT_LIM + 10];       /// mobius                  = A008683
void linear_sieve(int n)
{
    if (n < 1) return ;
    /// Extension Sieve
    memset(lpf, 0, sizeof(lpf[0]) * (n + 1));
    fill_n(mu, n + 1, 1);
    /// Main Sieve
    prime.clear();
    prime.reserve(n / log(n - 1));
    memset(isPrime, true, sizeof(isPrime[0]) * (n + 1));
    isPrime[0] = isPrime[1] = false;
    for (int x = 2; x <= n; ++x)
    {
        if (isPrime[x]) /// Func[Prime]
        {
            mu[x] = -1;
            lpf[x] = x;
            prime.push_back(x);
        }
        for (int p : prime) /// Func[Prime * X] <- Func[Prime]
        {
            if (p > lpf[x] || x * p > n) break;
            isPrime[x * p] = 0;
            lpf[x * p] = p;
            mu[x * p] = (lpf[x] == p) ? 0 : -mu[x];
        }
    }
}

vector<int> divisors[SQRT_LIM];
void precal_div(int n)
{
    for (int u = n; u >= 1; --u)
    {
        divisors[u].clear();
        for (int v = u; v <= n; v += u)
            divisors[v].push_back(u);
    }
}

long long solve(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);
    precal_div(t);

    long long res = 0;
    for (int d = 1; d * d <= n; ++d)
    {
        long long sum = 0;
        for (int v : divisors[d])
            sum += mu[d / v] * nck(v - 1, k - 1);

        sum %= MOD;
        res += sum * (n / (d * d));
    }

    res %= MOD;
    return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int n, k;
    cin >> n >> k;
    cout << solve(n, k);
    return 0;
}

The complexity depends on Möbius function, and traverse over all divisor of each number

We only care about $$$d$$$ that $$$1 \leq d \leq \sqrt{n}$$$

We can use precalculation, and sieve

As we only use $$$d$$$ that $$$d^2 \leq n$$$, let $$$t = \sqrt{n}$$$ for convenient

We can use linear sieve for $$$\mu{d}$$$ in $$$O(t)$$$

We can use divisor sieve for finding all divisors $$$p$$$ for each $$$v$$$ in $$$O(t \log t)$$$

We can calculate the result in $$$O(t \log t)$$$

Therefore the overall complexity is $$$O(t \log t)$$$