Numbers that have two or more prime factors - primes

Question

def fun(n) :
ans = 0
for i in range(1,n) :
for j in range(i+1,n+1) :
ans += gcd(i,j)
This is the actual function to be calculated when n is given. The upper bound for n will be 1e6 and the time limit will be 2 seconds. Directly using the above function will clearly exceed the Time Limit. So, I've come up with a following solution. It works for small cases, but failing to pass when N>=100.
Here is the c++ code :
# include <cstdio>
# include <bits/stdc++.h>
# include <cstring>
# include <cstdlib>
# include <cctype>
# include <string>
# include <cfloat>
# include <cmath>
# include <iostream>
# include <fstream>
# include <cassert>
# include <iterator>
# include <vector>
# include <queue>
# include <map>
# include <algorithm>
# include <set>
# include <sstream>
# include <stack>
# include <cassert>
# define MEM(a,b,s) memset(a,(b),s)
# define MAX(a,b) ((a) > (b) ? (a) : (b))
# define MIN(a,b) ((a) < (b) ? (a) : (b))
# define MP make_pair
# define pb push_back
# define inf 1000000000
# define MOD 1000000007
# define SZ 1000001
# define lli long long int
# define li long int
# define repeat(i,a) for(lli i=1;i<a;i++)
# define repeat_untill(a) while (a)
# define repeat2(i,j,k) for (j=i+1;j<=k;j++)
# define ch char
# define gi(a) readFast<int>(&a)
# define gl(a) readFast<li>(&a)
# define gll(a) readFast<lli>(&a)
using namespace std;
template <class fast_io>
inline fast_io readFast(fast_io *a)
{
char c=0;
int sign=0;
while(c<33 && c!='-')
c=getchar_unlocked();
*a=0;
if (c=='-')
{
sign=1;
c=getchar_unlocked();
}
while(c>33)
{
*a=*a*10+(c-48);
c=getchar_unlocked();
}
if (sign==1)
return -*a;
else
return *a;
}
int prime[100001];
void sieve()
{
MEM(prime,0,100001);
for (int i=2;i<100001;i++)
{
if (!prime[i])
{
lli j=2;
while (i*j < 100001)
{
prime[i*j] = 1;
j += 1;
}
}
}
}
lli gcd(lli a,lli b)
{
return (b==0?a:gcd(b,a%b));
}
int main(void)
{
lli n;
sieve();
gll(n);
repeat_untill(n)
{
lli i , j , ans =0;
repeat(i,n)
{
lli divs,coprimes=0;
if (n%i == 0)
{
divs = (n/i)-1;
coprimes = (n-i)-divs;
ans += ((divs*i)+coprimes);
}
else
{
lli k = (int)(pow(i,0.5)),m=2,totdiv=0;
if (k*k < i)
k += 1;
while (m<=k)
{
if (!prime[m])
{
divs = (n-i)/m;
totdiv += divs;
ans += ((divs*m));
}
m += 1;
}
ans += (n-i-totdiv);
}
}
cout<<ans<<endl;
gll(n);
}
return (0);
}
The problem here is if there will be n-i such that it has two prime factors, then this algo will fail as for a value i each j greater than i and lesser than n will only be taken into account for gcd calculation once. For example if n-i will be 25 then prime factors that it has will be 2,3,5. Here 15 will be considered twice at 3 and 5. Kindly help how can i eliminate the 15 after performing the action with 3.
Or is there any other approach to calculate the function fun(n) faster below O(n^2)?
Thanks in advance!