Sum of the series 1, 2, 4, 3, 5, 7, 9, 6, 8, 10, 11, 13.. till N-th term

Given a series of numbers 1, 2, 4, 3, 5, 7, 9, 6, 8, 10, 11, 13… The task is to find the sum of all the numbers in series till N-th number.

Examples:

Input: N = 4
Output: 10
1 + 2 + 4 + 3 = 10

Input: N = 10
Output: 55

Approach: The series is basically 2⁰ odd numbers, 2¹ even numbers, 2² even numbers…. The sum of first N odd numbers is N * N and sum of first N even numbers is (N * (N+1)). Calculate the summation for 2ⁱ odd or even numbers and keep adding them to the sum.

Iterate for every power of 2, till the number of iterations exceeds N, and keep adding the respective summation of odd or even numbers according to the parity. For every segment the sum of the segment will be, (current sum of X odd/even numbers – previous sum of Y odd/even numbers), where X is the total sum of odd/even numbers till this segment and Y is the summation of odd/even numbers till the previous when odd/even numbers occurred.

Below is the implementation of the above approach:

C++

// C++ program to implement 
// the above approach 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find the 
// sum of first N odd numbers 
int sumodd(int n) 
{ 
    return (n * n); 
} 
  
// Function to find the 
// sum of first N even numbers 
int sumeven(int n) 
{ 
    return (n * (n + 1)); 
} 
  
// Function to overall 
// find the sum of series 
int findSum(int num) 
{ 
  
    // Initiall odd numbers 
    int sumo = 0; 
  
    // Initial even numbers 
    int sume = 0; 
  
    // First power of 2 
    int x = 1; 
  
    // Check for parity 
    // for odd/even 
    int cur = 0; 
  
    // Counts the sum 
    int ans = 0; 
    while (num > 0) { 
  
        // Get the minimum 
        // out of remaining num 
        // or power of 2 
        int inc = min(x, num); 
  
        // Decrease that much numbers 
        // from num 
        num -= inc; 
  
        // If the segment has odd numbers 
        if (cur == 0) { 
  
            // Summate the odd numbers 
            // By exclusion 
            ans = ans + sumodd(sumo + inc) - sumodd(sumo); 
  
            // Increase number of odd numbers 
            sumo += inc; 
        } 
        // If the segment has evn numbers 
        else { 
  
            // Summate the even numbers 
            // By exclusion 
            ans = ans + sumeven(sume + inc) - sumeven(sume); 
  
            // Increase number of even numbers 
            sume += inc; 
        } 
  
        // Next set of numbers 
        x *= 2; 
  
        // Change parity for odd/even 
        cur ^= 1; 
    } 
  
    return ans; 
} 
  
// Driver code 
int main() 
{ 
    int n = 4; 
    cout << findSum(n); 
  
    return 0; 
} 

Java

// Java program to implement 
// the above approach 
  
class GFG 
{ 
  
    // Function to find the 
    // sum of first N odd numbers 
    static int sumodd(int n) 
    { 
        return (n * n); 
    } 
  
    // Function to find the 
    // sum of first N even numbers 
    static int sumeven(int n) 
    { 
        return (n * (n + 1)); 
    } 
  
    // Function to overall 
    // find the sum of series 
    static int findSum(int num)  
    { 
  
        // Initiall odd numbers 
        int sumo = 0; 
  
        // Initial even numbers 
        int sume = 0; 
  
        // First power of 2 
        int x = 1; 
  
        // Check for parity 
        // for odd/even 
        int cur = 0; 
  
        // Counts the sum 
        int ans = 0; 
        while (num > 0)  
        { 
  
            // Get the minimum 
            // out of remaining num 
            // or power of 2 
            int inc = Math.min(x, num); 
  
            // Decrease that much numbers 
            // from num 
            num -= inc; 
  
            // If the segment has odd numbers 
            if (cur == 0) 
            { 
  
                // Summate the odd numbers 
                // By exclusion 
                ans = ans + sumodd(sumo + inc) - sumodd(sumo); 
  
                // Increase number of odd numbers 
                sumo += inc; 
            }  
              
            // If the segment has evn numbers 
            else
            { 
  
                // Summate the even numbers 
                // By exclusion 
                ans = ans + sumeven(sume + inc) - sumeven(sume); 
  
                // Increase number of even numbers 
                sume += inc; 
            } 
  
            // Next set of numbers 
            x *= 2; 
  
            // Change parity for odd/even 
            cur ^= 1; 
        } 
  
        return ans; 
    } 
  
    // Driver code 
    public static void main(String[] args)  
    { 
        int n = 4; 
        System.out.println(findSum(n)); 
  
    } 
} 
  
// This code contributed by Rajput-Ji 

Python

# Python3 program to implement 
# the above approach 
  
# Function to find the 
# sum of first N odd numbers 
def sumodd(n): 
  
    return (n * n) 
  
# Function to find the 
# sum of first N even numbers 
def sumeven(n): 
  
    return (n * (n + 1)) 
  
  
# Function to overall 
# find the sum of series 
def findSum(num): 
  
  
    # Initiall odd numbers 
    sumo = 0
  
    # Initial even numbers 
    sume = 0
  
    # First power of 2 
    x = 1
  
    # Check for parity 
    # for odd/even 
    cur = 0
  
    # Counts the sum 
    ans = 0
    while (num > 0): 
  
        # Get the minimum 
        # out of remaining num 
        # or power of 2 
        inc = min(x, num) 
  
        # Decrease that much numbers 
        # from num 
        num -= inc 
  
        # If the segment has odd numbers 
        if (cur == 0): 
  
            # Summate the odd numbers 
            # By exclusion 
            ans = ans + sumodd(sumo + inc) - sumodd(sumo) 
  
            # Increase number of odd numbers 
            sumo += inc 
          
        # If the segment has evn numbers 
        else: 
  
            # Summate the even numbers 
            # By exclusion 
            ans = ans + sumeven(sume + inc) - sumeven(sume) 
  
            # Increase number of even numbers 
            sume += inc 
          
  
        # Next set of numbers 
        x *= 2
  
        # Change parity for odd/even 
        cur ^= 1
      
    return ans 
  
# Driver code 
n = 4
print(findSum(n)) 
  
# This code is contributed by mohit kumar 

C#

// C# program to implement 
// the above approach 
using System; 
  
class GFG 
{ 
  
    // Function to find the 
    // sum of first N odd numbers 
    static int sumodd(int n) 
    { 
        return (n * n); 
    } 
  
    // Function to find the 
    // sum of first N even numbers 
    static int sumeven(int n) 
    { 
        return (n * (n + 1)); 
    } 
  
    // Function to overall 
    // find the sum of series 
    static int findSum(int num)  
    { 
  
        // Initiall odd numbers 
        int sumo = 0; 
  
        // Initial even numbers 
        int sume = 0; 
  
        // First power of 2 
        int x = 1; 
  
        // Check for parity 
        // for odd/even 
        int cur = 0; 
  
        // Counts the sum 
        int ans = 0; 
        while (num > 0)  
        { 
  
            // Get the minimum 
            // out of remaining num 
            // or power of 2 
            int inc = Math.Min(x, num); 
  
            // Decrease that much numbers 
            // from num 
            num -= inc; 
  
            // If the segment has odd numbers 
            if (cur == 0) 
            { 
  
                // Summate the odd numbers 
                // By exclusion 
                ans = ans + sumodd(sumo + inc) - sumodd(sumo); 
  
                // Increase number of odd numbers 
                sumo += inc; 
            }  
              
            // If the segment has evn numbers 
            else
            { 
  
                // Summate the even numbers 
                // By exclusion 
                ans = ans + sumeven(sume + inc) - sumeven(sume); 
  
                // Increase number of even numbers 
                sume += inc; 
            } 
  
            // Next set of numbers 
            x *= 2; 
  
            // Change parity for odd/even 
            cur ^= 1; 
        } 
  
        return ans; 
    } 
  
    // Driver code 
    public static void Main(String[] args)  
    { 
        int n = 4; 
        Console.WriteLine(findSum(n)); 
  
    } 
} 
  
// This code has been contributed by 29AjayKumar 

PHP

<?php 
// PHP program to implement 
// the above approach 
  
// Function to find the 
// sum of first N odd numbers 
function sumodd($n) 
{ 
    return ($n * $n); 
} 
  
// Function to find the 
// sum of first N even numbers 
function sumeven($n) 
{ 
    return ($n * ($n + 1)); 
} 
  
// Function to overall 
// find the sum of series 
function findSum( $num) 
{ 
  
    // Initiall odd numbers 
    $sumo = 0; 
  
    // Initial even numbers 
    $sume = 0; 
  
    // First power of 2 
    $x = 1; 
  
    // Check for parity 
    // for odd/even 
    $cur = 0; 
  
    // Counts the sum 
    $ans = 0; 
    while ($num > 0)  
    { 
  
        // Get the minimum 
        // out of remaining num 
        // or power of 2 
        $inc = min($x, $num); 
  
        // Decrease that much numbers 
        // from num 
        $num -= $inc; 
  
        // If the segment has odd numbers 
        if ($cur == 0) 
        { 
  
            // Summate the odd numbers 
            // By exclusion 
            $ans = $ans + sumodd($sumo + $inc) -  
                          sumodd($sumo); 
  
            // Increase number of odd numbers 
            $sumo += $inc; 
        } 
          
        // If the segment has evn numbers 
        else 
        { 
  
            // Summate the even numbers 
            // By exclusion 
            $ans = $ans + sumeven($sume + $inc) -  
                          sumeven($sume); 
  
            // Increase number of even numbers 
            $sume += $inc; 
        } 
  
        // Next set of numbers 
        $x *= 2; 
  
        // Change parity for odd/even 
        $cur ^= 1; 
    } 
  
    return $ans; 
} 
  
// Driver code 
$n = 4; 
echo findSum($n); 
  
// This code contributed by princiraj1992 
?> 

Output:

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