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Sum of first N natural numbers with all powers of 2 added twice
• Last Updated : 26 Mar, 2021

Given an integer N, the task is to calculate the sum of first N natural numbers adding all powers of 2 twice to the sum.
Examples:

Input: N = 4
Output: 17
Explanation:
Sum = 2+4+3+8 = 17
Since 1, 2 and 4 are 2 0, 2 1 and 2 2 respectively, they are added twice to the sum.
Input: N = 5
Output: 22
Explanation:
The sum is equal to 2+4+3+8+5 = 22,
because 1, 2 and 4 are 2 0, 2 1 and 2 2 respectively.

Naive Approach:
The simplest approach to solve this problem is to iterate upto N, and keep calculating the sum by adding every number once except the powers of 2, which needs to be added twice.
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach:
Follow the steps below to optimize the above approach:

1. Calculate sum of first N natural numbers by the formula (N * (N + 1)) / 2.
2. Now, all powers of 2 needs to be added once more. Sum of all powers of 2 up to N can be calculated as 2 log2(N) + 1 – 1

3. Hence, the required sum is:

(N * (N + 1)) / 2 + 2 log2(N) + 1 – 1

Below is the implementation of the above approach:

## C++

 // C++ program to implement// the above approach#include using namespace std; // Function to raise N to the// power P and return the valuedouble power(int N, int P){    return pow(N, P);} // Function to calculate the// log base 2 of an integerint Log2(int N){     // Calculate log2(N) indirectly    // using log() method    int result = (int)(log(N) / log(2));    return result;} // Function to calculate and// return the required sumdouble specialSum(int n){     // Sum of first N natural    // numbers    double sum = n * (n + 1) / 2;     // Sum of all powers of 2    // up to N    int a = Log2(n);    sum = sum + power(2, a + 1) - 1;     return sum;} // Driver codeint main(){    int n = 4;     cout << (specialSum(n)) << endl;     return 0;} // This code is contributed by divyeshrabadiya07

## Java

 // Java program to implement// the above approachimport java.util.*;import java.lang.Math; class GFG {     // Function to raise N to the    // power P and return the value    static double power(int N, int P)    {        return Math.pow(N, P);    }     // Function to calculate the    // log base 2 of an integer    public static int log2(int N)    {         // Calculate log2(N) indirectly        // using log() method        int result = (int)(Math.log(N)                        / Math.log(2));        return result;    }     // Function to calculate and    // return the required sum    static double specialSum(int n)    {         // Sum of first N natural        // numbers        double sum = n * (n + 1) / 2;         // Sum of all powers of 2        // up to N        int a = log2(n);        sum = sum + power(2, a + 1) - 1;         return sum;    }     // Driver Code    public static void main(String[] args)    {         int n = 4;         System.out.println(specialSum(n));    }}

## Python3

 # Python3 program to implement# the above approachimport math # Function to raise N to the# power P and return the valuedef power(N, P):         return math.pow(N, P) # Function to calculate the# log base 2 of an integerdef Log2(N):         # Calculate log2(N) indirectly    # using log() method    result = (math.log(N) // math.log(2))         return result # Function to calculate and# return the required sumdef specialSum(n):         # Sum of first N natural    # numbers    sum = n * (n + 1) // 2     # Sum of all powers of 2    # up to N    a = Log2(n)    sum = sum + power(2, a + 1) - 1         return sum     # Driver code   if __name__=="__main__":         n = 4    print(specialSum(n)) # This code is contributed by rutvik_56

## C#

 // C# program to implement// the above approachusing System;class GFG{     // Function to raise N to the    // power P and return the value    static double power(int N, int P)    {        return Math.Pow(N, P);    }     // Function to calculate the    // log base 2 of an integer    public static int log2(int N)    {         // Calculate log2(N) indirectly        // using log() method        int result = (int)(Math.Log(N) /                        Math.Log(2));        return result;    }     // Function to calculate and    // return the required sum    static double specialSum(int n)    {         // Sum of first N natural        // numbers        double sum = (double)(n) * (n + 1) / 2;         // Sum of all powers of 2        // up to N        int a = log2(n);        sum = (sum) + power(2, a + 1) - 1;         return sum;    }     // Driver Code    public static void Main(string[] args)    {        int n = 4;         Console.Write(specialSum(n));    }} // This code is contributed by Ritik Bansal

## Javascript


Output:
17.0

Time Complexity: O(log2(N))
Auxiliary Space: O(1)

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