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

    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;

/// Precalculating factorials under prime modulo
int fact[SQRT_LIM + 10]; /// fact[n] = n!
int invs[SQRT_LIM + 10]; /// invs[n] = n^(-1)
int tcaf[SQRT_LIM + 10]; /// tcaf[n] = (n!)^(-1)
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;
    }
}

/// Calculating binomial coefficient queries
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;
}

/// Linear Sieve
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 || You can add something more
    memset(lpf, 0, sizeof(lpf[0]) * (n + 1));
    fill_n(mu, n + 1, 1);
    /// Main Sieve || Without this, you barely able to achive linear complexity
    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) /// For each number
    {
        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];
        }
    }
}

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

/// Solving for n, k
long long solve(int n, int k)
{
    /// We only care for d that 1 <= d <= sqrt(n)
    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) /// For each fixed p^2
    {
        long long sum = 0;
        for (int p : divisors[d]) /// For each (p | d)
            sum += mu[d / p] * nck(p - 1, k - 1);

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

    res %= MOD;
    return res;
}

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

    /// Assumming constant p = 10^9 + 7
    int n, k;
    cin >> n >> k;
    cout << solve(n, k);
    return 0;
}

vector<int> prime;           /// prime list              = A000040
bool isPrime[SQRT_LIM + 10]; /// characteristic function = A010051
int lpf[SQRT_LIM + 10];      /// lowest prime factor     = A020639
void linear_sieve(int n)
{
    if (n < 1) return ;
    prime.clear();
    prime.reserve(n / log(n - 1));
    memset(lpf, 0, sizeof(lpf[0]) * (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]
        {
            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;
        }
    }
}

long long res[SQRT_LIM + 10];
long long solve(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
        res[d] = nck(d - 1, k - 1);

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 0;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

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

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

We can use factorial formula for binomial coefficients as $$$1 \leq k \leq n < p$$$

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 $$$O(t \log t)$$$

The complexity depends on sieve, calculating result and calculating binomial coefficients

The sieve and binomial coefficients is obviously $$$O(t)$$$ complexity for $$$t = \sqrt{n}$$$

The calculating part is just simply related to the sum of inversion of primes

Therefore we achived $$$O(t \log t)$$$ complexity for $$$t = \sqrt{n}$$$

Literally what we do up to now just optimize the counting part. The counting, which is the core is as the same as original.

The counting part is only about the number of valid sequences bounded by $$$n$$$, or you can say, the number of possible $$$p^2$$$ you can choose when you fix squarefree number $$$u$$$.

And those stuffs above $$$\mu(n)$$$, $$$\phi(n)$$$, sieving or anything else are just optimizations.

$$$\binom{n + k - 1}{k}$$$

and when you fix one number as a similar idea:

$$$s_k(n) = \binom{n + k - 2}{k - 1}$$$

It is standard stars and bars problem.

Instead of thinking the number of way to select $$$a_i$$$.

You can think of the number of value $$$a_i$$$ you have used.

Let $$$x_v$$$ is the number of value $$$v$$$ you used, or the number of $$$a_i = v$$$ in this sequence.

Each sequence can be represented uniquely as a sequence of $$$x_1, x_2, \dots, x_n$$$.

Since we selected $$$k$$$ in total we have $$$x_1 + x_2 + \dots + x_n = k$$$.

Let $$$x_v$$$ is the number of value $$$v$$$ you used, or the number of $$$a_i = v$$$ in this sequence.

For each box $$$v = 1 \dots n$$$, draw $$$x_v$$$ stars ⭐.

You can make $$$k - 1$$$ cut between these box (not cut into the box), and split into $$$k$$$ intervals.

The sum of each interval is the value of selected element.

Since there are $$$n + k - 1$$$ boxes, and you use $$$k - 1$$$ cuts, there are the total of $$$\binom{n + k - 1}{k}$$$ different ways.

long long brute(int n)
{
    set<int> S;
    for (int i = 1; i <= n; ++i)
        S.insert(i * i);

    long long res = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = i; j <= n; ++j)
            if (S.count(i * j))
                for (int k = j; k <= n; ++k)
                    if (S.count(i * k)) 
                        if (S.count(j * k))
                            ++res;

    return res;
}

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;
}

bool is_squarefree[LIM];
int solve(int n, int k)
{
    memset(is_squarefree, true, sizeof(is_squarefree[0]) * (n + 1));
    precal_nck(2 * 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(k + j - 2, k);
    }

    res %= MOD;
    return res;
}

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
void linear_sieve(int n)
{
    if (n < 1) return ;
    prime.clear();
    prime.reserve(n / log(n - 1));
    memset(lpf, 0, sizeof(lpf[0]) * (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]
        {
            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;
        }
    }
}

long long res[SQRT_LIM + 10];
long long solve(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
        res[d] = nck(d + k - 2, k - 1);

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 0;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

vector<int> factor;
int factorize(int n) /// Calculating phi(n) while factorizing (n) in O(sqrt n)
{
    factor.clear();
    int phi = n;

    if (!(n & 1))
    {
        n >>= __builtin_ctz(n);
        factor.push_back(2);
        phi -= phi / 2;
    }

    for (int x = 3; x * x <= n; x += 2)
    {
        if (n % x == 0)
        {
            do n /= x; while (n % x == 0);
            factor.push_back(x);
            phi -= phi / x;
        }
    }

    if (n > 1)
    {
        factor.push_back(n);
        phi -= phi / n;
    }

    return phi;
}

int f[LIM];    /// f[x] = nCk(n, x)
int fact[LIM]; /// n! 
int tcaf[LIM]; /// n!^(-1)
int divp[LIM]; /// x but ignore all factors p[i]
int cntp[LIM][LOG_LIM]; /// cntp[x][i] = Number of time factor p[i] appear in 1..x
void precal(int MOD) /// Calculate f[x] for all x = 1 -> n in O(n log mod + sqrt mod)
{
    int PHIMOD = factorize(MOD);
    for (int x = 1; x <= n; ++x) /// For each part x in n!
    {
        int &t = divp[x] = x;
        for (int i = 0; i < factor.size(); ++i) /// Ignore all factor p[i] of p
        {
            cntp[x][i] = cntp[x - 1][i];
            for (; t % factor[i] == 0; t /= factor[i]) /// Count how many times p[i] appears in 1..n
                ++cntp[x][i];
        }
    }

    fact[0] = fact[1] = 1;
    tcaf[0] = tcaf[1] = 1;
    for (int x = 2; x <= n; ++x) /// Finding n! and n!^(-1)
    {
        fact[x] = (1LL * fact[x - 1] * divp[x]) % MOD;
        tcaf[x] = powMOD(fact[x], PHIMOD - 1, MOD);
    }

    memset(f, 0, sizeof(f[0]) * k);
    for (int x = k; x <= n; ++x)
    {
        /// Calculate nCk % p normally
        f[x] = fact[x];
        mulMOD(f[x], tcaf[k], MOD);
        mulMOD(f[x], tcaf[x - k], MOD);
        for (int i = 0; i < factor.size(); ++i) /// Bringing those factors back
        {
            int p = cntp[x][i] - cntp[k][i] - cntp[x - k][i];
            f[x] = 1LL * f[x] * powMOD(factor[i], p, MOD) % MOD;
        }
    }
}

The sequence contain positive integers only.

The sequence contain negative integers only.

The sequence contain a zero, and the rest are positive integers.

The sequence contain a zero, and the rest are negative integers.

You can flip all the sign to positive, thus negative integers sequence are now the same as positive sequences.

Without zero, the problem has no difference from the original task.

With zero, any product with it is already a perfect square.

Hence zeros dont affect anything other than the remaining number you have not selected.

The sequence contain positive integers only:

• This is the original task as positive integers only means $$$1 \leq a_i \leq n$$$.

The sequence contain negative integers only:

• Flip all the sign doesnt change the product for each pairs.
• The problem now is the same as the sequence contain positive integers only.

The sequence contain a zero, and the rest are positive integers:

• The product of a zero with any number is still a perfect square.
• So the zero only reduce the number you have to select.
• Therefore this is same as the original task but now you have to select $$$k - 1$$$ numbers.

The sequence contain a zero, and the rest are negative integers:

• Just also flip the sign, the answer is the same as above.

The sequence contain zeros only.

The sequence contain positive integers only.

The sequence contain negative integers only.

The sequence contain $$$0 < t < k$$$ positive integers, and the rest $$$k - t$$$ are zeros.

The sequence contain $$$0 < t < k$$$ positive integers, and the rest $$$k - t$$$ are zeros.

The sequence contain zeros only.

The sequence contain $$$0 < t \leq k$$$ non-zero integers, and the rest $$$k - t$$$ are zeros.

You might wondering where is the $$$2 \times \dots$$$ come from.

Yet it is the same as unique number cases.

You have the answer for positive numbers are the same as negative number when you flip the sign.

So you have $$$1 \times \dots$$$ for positive, and $$$1 \times \dots$$$ for negative.

Therefore you have $$$2 \times \dots$$$ in total

You might wondering why we dont have a formula like this:

$$$task_E(n, k) = 1 + \overset{k}{\underset{t = 1}{\Large \Sigma}} C^t_k \times task_B(n, t)$$$.

But you know, since the sequence is ordered.

No matter which number you selected as zero there is one and only one way to sort it back.

You see, there is only one case for fully zeros sequence.

Isnt this sum is familiar to some standard math problem related to binomial coefficient ?

This formula is related to $$$\overset{k}{\underset{t = 0}{\Large \Sigma}} \binom{n + t}{t} = \binom{n + k + 1}{k}$$$

$$$s_k(n) = \overset{k}{\underset{t = 1}{\Large \Sigma}} \binom{d + t - 2}{t - 1} \times 2 = \binom{d + k - 1}{k - 1} \times 2$$$

long long res[SQRT_LIM + 10];
long long solve(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
    {
        res[d] += (k >= 1) * nck(d - 1, k - 1) * 2;
        res[d] += (k >= 2) * nck(d - 1, k - 2) * 2;  
    }

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 0;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

long long res[SQRT_LIM + 10];
long long solve(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
        for (int t = 1; t <= k; ++t)
            res[d] += nck(d + t - 2, t - 1) * 2;

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 1;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}

long long res[SQRT_LIM + 10];
long long solve(int n, int k)
{
    int t = ceil(sqrt(n) + 1) + 1;
    linear_sieve(t);
    precal_nck(t);

    memset(res, 0, sizeof(res[0]) * (t + 1));
    for (int d = 1; d * d <= n; ++d) 
        res[d] += nck(d + k - 1, k - 1) * 2;

    for (int p : prime)
        for (int d = t / p; d > 0; --d)
            res[d * p] -= res[d];

    long long ans = 1;
    for (int d = 1; d * d <= n; ++d)
        ans += res[d] * (n / (d * d));

    ans %= MOD;
    return ans;
}