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# Sum of first N natural numbers with all powers of 2 added twice

• Last Updated : 21 Jul, 2022

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 value ` `double` `power(``int` `N, ``int` `P) ` `{ ` `    ``return` `pow``(N, P); ` `} `   `// Function to calculate the ` `// log base 2 of an integer ` `int` `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 sum ` `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 ` `int` `main() ` `{ ` `    ``int` `n = 4; `   `    ``cout << (specialSum(n)) << endl; `   `    ``return` `0; ` `} `   `// This code is contributed by divyeshrabadiya07 `

## Java

 `// Java program to implement ` `// the above approach ` `import` `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 approach` `import` `math`   `# Function to raise N to the` `# power P and return the value` `def` `power(N, P):` `    `  `    ``return` `math.``pow``(N, P)`   `# Function to calculate the` `# log base 2 of an integer` `def` `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 sum` `def` `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 approach ` `using` `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),  since no extra space has been taken.

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