解析解

#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
	double n;
	scanf("%lf",&n);
	double a = 1 + sqrt(2), b = 1 - sqrt(2);
	double x;
	x = (9 - 3 * a)*pow(b,(n - 2))* (1 - pow((a / b),(n - 1))) / (1 - a / b) + 3 * pow(a ,(n - 1));
	printf("%.0f",x);
	return 0;
}