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)
{

    // Initial 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 even 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)
    {

        // Initial 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 even 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):


    # Initial 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 even 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)
    {

        // Initial 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 even 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)
{

    // Initial 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 even 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
?>

Javascript

<script>

// javascript 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)
{

    // Initial odd numbers
    let sumo = 0;

    // Initial even numbers
    let sume = 0;

    // First power of 2
    let x = 1;

    // Check for parity
    // for odd/even
    let cur = 0;

    // Counts the sum
    let ans = 0;
    while (num > 0) {

        // Get the minimum
        // out of remaining num
        // or power of 2
        let 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 even 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

    let n = 4;
    document.write( findSum(n));

// This code is contributed by todaysgaurav


</script>

Output:

10

Time Complexity: O(n) because using inbuilt function min

Auxiliary Space: O(1)