Ultraviolet A 10743-blocks on blocks (matrix power)

Link to the question: Ultraviolet A 10743-blocks on blocks

Question: How many types of N-bone cards are there? After images are rotated, they are not in the same group. The grids in a row must be continuous. If the answer is greater than 10000, the last four digits are output.

Solution: After thinking about the recursive formula for one afternoon, I really couldn't stand it. I searched the launched sequence online. Is there a formula for AI = 5? AI? 1? 7? AI? 2 + 4? AI? 3 (I ≥ 5)

PS: if anyone knows how to launch the product, please leave me a message. Thank you ~

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 3;const int mod = 10000;int sign[maxn][maxn] = { {0, 1, 0}, {0, 0, 1}, {4, -7, 5} };struct MAT {    int s[maxn][maxn];    MAT () {}    void set () {        for (int i = 0; i < maxn; i++)            for (int j = 0; j < maxn; j++)                s[i][j] = sign[i][j];    }    MAT operator * (const MAT& a) {        MAT ans;        for (int i = 0; i < maxn; i++) {            for (int j = 0; j < maxn; j++) {                ans.s[i][j] = 0;                for (int k = 0; k < maxn; k++)                    ans.s[i][j] = (ans.s[i][j] + s[i][k] * a.s[k][j] % mod) % mod;            }        }        return ans;    }};MAT pow_mat (int n) {    MAT ans, x;    ans.set();    x.set();    while (n) {        if (n&1)            ans = ans * x;        x = x * x;        n >>= 1;    }    return ans;}int solve (int n) {    if (n < 3)        return n;    else if (n == 3)        return 6;    else if (n == 4)        return 19;    MAT ans = pow_mat(n-5);    return ((ans.s[2][0] * 2 + ans.s[2][1] * 6 + ans.s[2][2] * 19) % mod + mod) % mod;}int main () {    int cas, n;    scanf("%d", &cas);    for (int i = 1; i <= cas; i++) {        scanf("%d", &n);        printf("Case %d: ", i);        int ans = solve(n);        if (n <= 9)            printf("%d\n", ans);        else            printf("%04d\n", ans);    }    return 0;}