int x = (1 << p) & ~((1 << n) - 1);

R>int x = (1 << p) & ~((1 << n) - 1);
R>

The behavior is undefined if the right operand is negative, or greater than or equal to the length in bits of the promoted left operand.

#include <iostream>
#include <cmath>
#include <cassert>

//целочисленная степень
typedef unsigned long Power;

//Тип для представления огромного числа в виде first * 10^second
typedef std::pair<double, Power> HugeNumber;

//Функция представления числа вида x = 2^(2^n) 0 <= n < 32 
//в виде: x = first * 10^second
HugeNumber huge_from_degree(Power n)
{
  assert(0 <= n && n < 32);
  
  //lg(2) - десятичный логарифм из 2.
  double  log_10_of_2 = std::log(2.0)/std::log(10.0);
  double  p = Power(1) << n;
  double  ex = p * log_10_of_2; //x = 10^ex
  Power   ex_i = Power(ex); //целая часть от ex
  double  ex_f = ex - ex_i;    //дробная часть от ex
  double  v =  std::exp(ex_f * std::log(10.0)); //v = 10^ex_f

  return HugeNumber(v,ex_i);
}

//Оператор вывода в поток для HugeNumber
std::ostream& operator<<(std::ostream& os, const HugeNumber& v)
{
  return os << v.first << " * 10^" << v.second;
}

int main()
{
  std::cout << "2^(2^0) = \t" << huge_from_degree(0) << std::endl;
  std::cout << "2^(2^1) = \t" << huge_from_degree(1) << std::endl;
  std::cout << "2^(2^2) = \t" << huge_from_degree(2) << std::endl;
  std::cout << "2^(2^3) = \t" << huge_from_degree(3) << std::endl;
  std::cout << "2^(2^4) = \t" << huge_from_degree(4) << std::endl;
  std::cout << "2^(2^5) = \t" << huge_from_degree(5) << std::endl;
  std::cout << "2^(2^30) = \t" << huge_from_degree(30) << std::endl;
}

2^(2^0) = 2 * 10^0
2^(2^1) = 4 * 10^0
2^(2^2) = 1.6 * 10^1
2^(2^3) = 2.56 * 10^2
2^(2^4) = 6.5536 * 10^4
2^(2^5) = 4.29497 * 10^9
2^(2^30) = 4.19716 * 10^323228496