Program to find Nth term in the series 0, 0, 2, 1, 4, 2, 6, 3, 8,…

Given a number N. The task is to write a program to find the N-th term in the below series: 

0, 0, 2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12, 6, 14, 7, 16, 8,…..

Examples: 

Input : N = 10
Output : 4

Input : N = 7
Output : 6

On observing carefully, you will find that the series is a mixture of 2 series:

1. Terms at odd positions in the given series form the series of even numbers in increasing order starting from 0. Like, 0,2,4,6,..
2. Terms at even positions in the given series are derived from the previous term using the formula (previousTerm/2). That is, the terms at even positions are half of their previous term.

Now, it is known that every odd positioned term forms an even series starting from 0 and every even positioned term is the half of the previous odd positioned term.
Therefore first check whether the input number N is even or odd. If it is odd, set N=(N/2) + 1(since there are Two series running parallelly) and find the Nth term by using formula 2*(N-1) ( N-1 because the series starts with 0). 
Similarly, if N is even, set N = N/2, use the previous formula and divide the answer by 2.
Below is the implementation of above approach: 

C++

// CPP program to find N-th term
// in the series
#include <iostream>
#include <math.h>
using namespace std;

// Function to find N-th term
// in the series
void findNthTerm(int n)
{
    // If n is even
    if (n % 2 == 0) {
        n = n / 2;
        n = 2 * (n - 1);
        cout << n / 2 << endl;
    }
    // If n is odd
    else {
        n = (n / 2) + 1;
        n = 2 * (n - 1);
        cout << n << endl;
    }
}

// Driver code
int main()
{
    int X = 10;
    findNthTerm(X);

    X = 7;
    findNthTerm(X);

    return 0;
}

Java

// Java program to find N-th term
// in the series

// Function to find N-th term
// in the series
class GFG
{
static void findNthTerm(int n)
{
    // If n is even
    if (n % 2 == 0)
    {
        n = n / 2;
        n = 2 * (n - 1);
        System.out.println(n / 2);
    }

    // If n is odd
    else
    {
        n = (n / 2) + 1;
        n = 2 * (n - 1);
        System.out.println(n);
    }
}

// Driver code
public static void main(String args[])
{
    int X = 10;
    findNthTerm(X);

    X = 7;
    findNthTerm(X);
}
}

// This code is contributed by Subhadeep

Python 3

# Python 3 program to find N-th term
# in the series

# Function to find N-th term
# in the series
def findNthTerm(n):

    # If n is even
    if (n % 2 == 0):
        n = n // 2
        n = 2 * (n - 1)
        print( n // 2)

    # If n is odd
    else:
        n = (n // 2) + 1
        n = 2 * (n - 1)
        print(n)

# Driver code
if __name__ == "__main__":
    X = 10
    findNthTerm(X);

    X = 7;
    findNthTerm(X)

C#

// C# program to find N-th term
// in the series
using System;

// Function to find N-th term
// in the series
class GFG
{
static void findNthTerm(int n)
{
    // If n is even
    if (n % 2 == 0)
    {
        n = n / 2;
        n = 2 * (n - 1);
        Console.Write(n / 2);
    }

    // If n is odd
    else
    {
        n = (n / 2) + 1;
        n = 2 * (n - 1);
        Console.Write(n);
    }
}

// Driver code
public static void Main()
{
    int X = 10;
    findNthTerm(X);
    Console.Write("\n");
    X = 7;
    findNthTerm(X);
}
}

// This code is contributed
// by Smitha

PHP

<?php
// PHP program to find N-th
// term in the series

// Function to find N-th
// term in the series
function findNthTerm($n)
{
    // If $n is even
    if ($n % 2 == 0)
    {
        $n = $n / 2;
        $n = 2 * ($n - 1);
        echo $n / 2 . "\n";
    }

    // If $n is odd
    else
    {
        $n = (int)($n / 2) + 1;
        $n = 2 * ($n - 1);
        echo $n . "\n";
    }
}

// Driver code
$X = 10;
findNthTerm($X);

$X = 7;
findNthTerm($X);

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>
// JavaScript program to find N-th term
// in the series

// Function to find N-th term
// in the series
function findNthTerm(n)
{
    // If n is even
    if (n % 2 == 0) {
        n = Math.floor(n / 2);
        n = 2 * (n - 1);
        document.write(Math.floor(n / 2) + "<br>");
    }
    // If n is odd
    else {
        n = Math.floor(n / 2) + 1;
        n = 2 * (n - 1);
        document.write(n + "<br>");
    }
}

// Driver code

    let X = 10;
    findNthTerm(X);

    X = 7;
    findNthTerm(X);


// This code is contributed by Surbhi Tyagi

</script>

Output:

4
6

Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.