Program to find Nth odd Fibonacci Number

Given an integer N. The task is to find the Nth odd Fibonacci number.

The odd number fibonacci series is as:

1, 1, 3, 5, 13, 21, 55, 89, 233, 377, 987, 1597………….and so on.

Note: In the above series we have omited even terms from the general fibonacci sequence.

Examples:



Input: N = 3
Output: 3

Input: N = 4
Output: 5

Approach:
On observing carefully, it can be deduced that every third Fibonacci number is even, so the Nth odd Fibonacci number is the {(3*N+1)/2}th term in the general Fibonacci sequence.

Below is the implementation of the above approach:

C++

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// C++ program for Nth odd fibonaci number
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find nth odd fibonaci number
int oddFib(int n)
{
    n = (3 * n + 1) / 2;
    int a = -1, b = 1, c, i;
    for (i = 1; i <= n; i++) {
        c = a + b;
        a = b;
        b = c;
    }
    return c;
}
  
// Driver Code
int main()
{
    int n = 4;
      
    cout << oddFib(n);
      
    return 0;
}

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Java














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// Java program for Nth odd fibonaci number  


class GFG  


{ 


      


    // Function to find nth odd fibonaci number  


    static int oddFib(int n)  


    {  


        n = (3 * n + 1) / 2;  


          


        int a = -1, b = 1, c = 0, i;  


          


        for (i = 1; i <= n; i++)  


        {  


            c = a + b;  


            a = b;  


            b = c;  


        }  


        return c;  


    }  


      


    // Driver Code  


    public static void main (String[] args)  


    {  


        int n = 4;  


          


        System.out.println(oddFib(n));  


    }  


} 


  


// This code is contributed by AnkitRai01 



















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Python3














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# Python3 program for Nth odd fibonaci number 


  


# Function to find nth odd fibonaci number 


def oddFib(n): 


    n = (3 * n + 1) // 2


  


    a = -1


    b = 1


    c = 0


    for i in range(1, n + 1): 


        c = a + b 


        a = b 


        b = c 


  


    return c 


  


# Driver Code 


n = 4


  


print(oddFib(n)) 


  


# This code is contributed by mohit kumar 



















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C#














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// C# program for Nth odd fibonaci number  


using System; 


      


class GFG  


{ 


      


    // Function to find nth odd fibonaci number  


    static int oddFib(int n)  


    {  


        n = (3 * n + 1) / 2;  


          


        int a = -1, b = 1, c = 0, i;  


          


        for (i = 1; i <= n; i++)  


        {  


            c = a + b;  


            a = b;  


            b = c;  


        }  


        return c;  


    }  


      


    // Driver Code  


    public static void Main (String[] args)  


    {  


        int n = 4;  


          


        Console.WriteLine(oddFib(n));  


    }  


} 


  


// This code is contributed by 29AjayKumar 



















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Output:


5





Time Complexity: O(N)


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