Express 2^N in exactly N+1 powers of 2

Given a positive integer N, the task is to express 2^N in the sum of powers of 2 and in exactly N+1 terms. Print those N+1 terms.

Example:

Input: N = 4
Output: 1 1 2 4 8
Explanation: 2^4 = 2^0 + 2^0 + 2^1 + 2^2 + 2^3 = 1 + 1 + 2 + 4 + 8

Input: N = 5
Output: 1 1 2 4 8 16  

Approach: As we know:

2^0 + 2^1 +…. 2^(N-1) = 2^N -1

Therefore, adding 1 at the beginning of the series of powers of 2 will result in 2^N in exactly N+1 terms.

1 + 2^0 + 2^1 +…. 2^(N-1) = 2^N

Below is the implementation of the above approach

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find exactly N + 1 terms
// of powers of 2 series,  whose sum is 2^N
void powerOf2(long long N)
{
    cout << 1 << ' ';
    for (int i = 0; i < N; ++i) {
        cout << (long long)pow(2, i) << ' ';
    }
}

// Driver Code
int main()
{
    long long N = 5;

    powerOf2(N);
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
// Function to find exactly N + 1 terms
// of powers of 2 series,  whose sum is 2^N
static void powerOf2(long N)
{
    System.out.print(1);
    for (int i = 0; i < N; ++i) {
        System.out.print(" " + (long)Math.pow(2, i));
    }
}

// Driver Code
public static void main(String args[])
{
    long N = 5;
    powerOf2(N);
}
}
// This code is contributed by Samim Hossain Mondal

C#

// C# program for the above approach
using System;
class GFG
{
// Function to find exactly N + 1 terms
// of powers of 2 series,  whose sum is 2^N
static void powerOf2(long N)
{
    Console.Write(1);
    for (int i = 0; i < N; ++i) {
        Console.Write(" " + (long)Math.Pow(2, i));
    }
}

// Driver Code
public static void Main()
{
    long N = 5;
    powerOf2(N);
}
}
// This code is contributed by Samim Hossain Mondal

Python3

# Python3 program for the above approach
import math

# Function to find exactly N + 1 terms
# of powers of 2 series,  whose sum is 2^N
def powerOf2(N) :

    print(1,end= ' ');
    for i in range(N) :
        print(int(math.pow(2, i)),end = ' ');

# Driver Code
if __name__ == "__main__" :

    N = 5;

    powerOf2(N);

    # This code is contributed by AnkThon

Javascript

<script>
// JavaScript program for the above approach

// Function to find exactly N + 1 terms
// of powers of 2 series,  whose sum is 2^N
function powerOf2(N)
{
    document.write(1 + ' ');
    for (var i = 0; i < N; ++i) {
        document.write(Math.pow(2, i) + ' ');
    }
}

// Driver Code
N = 5;
powerOf2(N);

// This code is contributed by AnkThon
</script>

Output

1 1 2 4 8 16 

Time Complexity: O(N*log(N))
Auxiliary Space: O(1)