$f_a(x)=a^x(a>0,a\ne 1)$，我们要求$\sum _{a=2}^n\left(a\sum _{b=a}^n\lfloor f_a^{-1}(b)\rfloor \lceil f_b^{-1}(a)\rceil \right)$。
$$\begin{gathered} \sum _{a=2}^n\left(a\sum _{b=a}^n\lfloor f_a^{-1}(b)\rfloor \lceil f_b^{-1}(a)\rceil \right) \\ \sum _{a=2}^n\left(a\sum _{b=a}^n\lfloor {\log }_{a}b\rfloor \lceil {\log }_{b}a\rceil \right) \\ b\ge a，则有\lceil {\log }_{b}a\rceil =1 \\ \sum _{a=2}^n\left(a\sum _{b=a}^n\lfloor {\log }_{a}b\rfloor \right) \end{gathered}$$

且容易发现，当$a\times a>n$时，有$\lfloor {\log }_{a}b\rfloor$恒为$1$，所以可以单独用公式计算。

#include <bits/stdc++.h>using namespace std;const int mod = 998244353, inv2 = mod + 1 >> 1, inv6 = (mod + 1) / 6;typedef long long ll;ll calc1(ll l, ll r) {return (l + r) % mod * ((r - l + 1) % mod) % mod * inv2 % mod;
}ll calc2(ll n) {n %= mod;return n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);ll a, n, ans = 0;cin >> n;for (a = 2; a * a <= n; a++) {for (ll l = a, r, k = 1; l <= n; l = r + 1, k++) {r = min(l * a - 1, n);int tot = (r - l + 1) % mod;ans = (ans + a * tot % mod * k % mod) % mod;}}// for (; a <= n; a++) {//原本我们是这样算的，当时这里可以变成，自然幂次求和的形式，所以可以快速算出来。//   ans = (ans + a * (n - a + 1) % mod) % mod;// }ans = (ans + (n + 1) % mod * calc1(a, n) % mod) % mod;ans = ((ans - calc2(n) + calc2(a - 1)) % mod + mod) % mod;cout << ans << "\n";return 0;
}