由i个点和j个点组成的二分图个数为 $3_{ij}$，减去不联通的部分得到得到由i，j个点组成的联通二分图个数

 $g_{i,j} = 3_{ij} - \sum_{k=1}^i \sum_{l=0}^j g_{k,l} C_{i-1,k-1} C_{j,l} 3_{(i-k)(j-l)}$

然后再dp一遍

#include <bits/stdc++.h>
using namespace std;
#define rep(i, j, k) for (int i = int(j); i <= int(k); ++ i) 
#define dwn(i, j, k) for (int i = int(j); i >= int(k); -- i)
typedef long long LL;
const LL MOD = 175781251;
const int N = 107;
LL fac[N], inv[N], pw[N * N], g[N][N], f[N];
inline LL comb(LL n, LL m) {
    return fac[n] * inv[m] % MOD * inv[n - m] % MOD;
}    
int main() {
    fac[0] = 1; rep(i, 1, N - 1) fac[i] = fac[i - 1] * i % MOD;
    inv[0] = inv[1] = 1;
    rep(i, 2, N - 1) inv[i] = MOD - (MOD / i) * inv[MOD % i] % MOD;
    rep(i, 2, N - 1) (inv[i] *= inv[i - 1]) %= MOD;
    pw[0] = 1;
    rep(i, 1, N * N - 1) pw[i] = pw[i - 1] * 3 % MOD; 
    rep(i, 1, 100) rep(j, 0, 100 - i) {
        g[i][j] = pw[i * j]; // A集合中i 个点标号 1 -i， B集合中j个点标号1-j
        // 枚举A集合中第一个点所在联通二分图
        rep(k, 1, i) rep(l, 0, j) {
            if (k == i && l == j) continue;
            g[i][j] += MOD - g[k][l] * comb(i - 1, k - 1) % MOD * comb(j, l) % MOD * pw[(i - k) * (j - l)] % MOD;
            g[i][j] %= MOD;
        }
    }
    f[0] = 1; 
    rep(i, 1, 100) 
        rep(j, 0, i - 1) 
            rep(k, 0, i - 1 - j) {
                f[i] += g[j + 1][k] * comb(i - 1, j) % MOD * comb(i - 1 - j, k) % MOD * f[i - 1 - j - k] % MOD;
                f[i] %= MOD;
            }

    int n;
    while (scanf("%d", &n), n) {
        cout << f[n] << '\n';
    }
}