Fermat numbers are non-negative odd numbers which is valid for all values of k>=0. Only the first seven terms of the sequence are known till date. First, five terms of the series are prime but rest of them are not. The kth term of Fermat number is represented as
The sequence:
3, 5, 17, 257, 65537, 4294967297, 18446744073709551617
For a given N, the task is to find the first N Fermat numbers.
Examples:
Input: N = 4
Output: 3, 5, 17, 257Input: N = 7
Output : 3, 5, 17, 257, 65537, 4294967297, 18446744073709551617
Approach :
Using the above-mentioned formula we will find the Nth term of the series.
Below is the implementation of the above approach :
C++
// CPP program to print fermat numbers #include <bits/stdc++.h> #include <boost/multiprecision/cpp_int.hpp> using namespace boost::multiprecision; #define llu int128_t using namespace std; /* Iterative Function to calculate (x^y) in O(log y) */llu power(llu x, llu y) { llu res = 1; // Initialize result while (y > 0) { // If y is odd, multiply x with the result if (y & 1) res = res * x; // n must be even now y = y >> 1; // y = y/2 x = x * x; // Change x to x^2 } return res; } // Function to find nth fermat number llu Fermat(llu i) { // 2 to the power i llu power2_i = power(2, i); // 2 to the power 2^i llu power2_2_i = power(2, power2_i); return power2_2_i + 1; } // Function to find first n Fermat numbers void Fermat_Number(llu n) { for (llu i = 0; i < n; i++) { // Calculate 2^2^i cout << Fermat(i); if(i!=n-1) cout << ", "; } } // Driver code int main() { llu n = 7; // Function call Fermat_Number(n); return 0; } |
Python3
# Python3 program to print fermat numbers # Iterative Function to calculate (x^y) in O(log y) def power(x, y): res = 1 # Initialize result while (y > 0): # If y is odd, # multiply x with the result if (y & 1): res = res * x # n must be even now y = y >> 1 # y = y/2 x = x * x # Change x to x^2 return res # Function to find nth fermat number def Fermat(i): # 2 to the power i power2_i = power(2, i) # 2 to the power 2^i power2_2_i = power(2, power2_i) return power2_2_i + 1 # Function to find first n Fermat numbers def Fermat_Number(n): for i in range(n): # Calculate 2^2^i print(Fermat(i), end = "") if(i != n - 1): print(end = ", ") # Driver code n = 7 # Function call Fermat_Number(n) # This code is contributed by Mohit Kumar |
output:
3, 5, 17, 257, 65537, 4294967297, 18446744073709551617
Reference:https://en.wikipedia.org/wiki/Fermat_number
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