Nth Even Fibonacci Number

07 Jan, 2022
Given a value n, find the n’th even Fibonacci Number.Examples :  
Input  : n  = 3
Output : 34

Input  : n  = 4
Output : 144

Input : n = 7
Output : 10946
The Fibonacci numbers are the numbers in the following integer sequence. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, .... where any number in sequence is given by:  
 Fn = Fn-1 + Fn-2 
      with seed values
      F0 = 0 and F1 = 1.
 
The even number Fibonacci sequence is, 0, 2, 8, 34, 144, 610, 2584.... We need to find n’th number in this sequence.If we take a closer look at Fibonacci sequence, we can notice that every third number in sequence is even and the sequence of even numbers follow following recursive formula.  
Recurrence for Even Fibonacci sequence is:
     EFn = 4EFn-1 + EFn-2
with seed values
     EF0 = 0 and EF1 = 2.

EFn represents n'th term in Even Fibonacci sequence.
How does above formula work? Let us take a look original Fibonacci Formula and write it in the form of Fn-3 and Fn-6 because of the fact that every third Fibonacci number is even. 
Fn = Fn-1 + Fn-2 [Expanding both terms]
   = Fn-2 + Fn-3 + Fn-3 + Fn-4 
   = Fn-2 + 2Fn-3 + Fn-4 [Expanding first term]
   = Fn-3 + Fn-4 + 2Fn-3 + Fn-4
   = 3Fn-3 + 2Fn-4  [Expanding one Fn-4]
   = 3Fn-3 + Fn-4 + Fn-5 + Fn-6 [Combing Fn-4 and Fn-5]
   = 4Fn-3 + Fn-6  

Since every third Fibonacci Number is even, So if Fn is 
even then Fn-3 is even and Fn-6 is also even. Let Fn be
xth even element and mark it as EFx.
If Fn is EFx, then Fn-3 is previous even number i.e. EFx-1
and Fn-6 is previous of EFx-1 i.e. EFx-2
So
Fn = 4Fn-3 + Fn-6
which means,
EFx = 4EFx-1 + EFx-2
 
C++
Java
Python3
C#
PHP
Javascript
// C++ code to find Even Fibonacci//Series using normal Recursion#include<iostream>using namespace std; // Function which return//nth even fibonacci numberlong int evenFib(int n){    if (n < 1)    return n;    if (n == 1)    return 2;     // calculation of    // Fn = 4*(Fn-1) + Fn-2    return ((4 * evenFib(n-1)) +             evenFib(n-2));} // Driver Codeint main (){    int n = 7;    cout << evenFib(n);    return 0;}
// Java code to find Even Fibonacci// Series using normal Recursion class GFG{  // Function which return// nth even fibonacci numberstatic long evenFib(int n){    if (n < 1)    return n;     if (n == 1)    return 2;     // calculation of    // Fn = 4*(Fn-1) + Fn-2    return ((4 * evenFib(n-1)) +                 evenFib(n-2));} // Driver Codepublic static void main (String[] args){    int n = 7;    System.out.println(evenFib(n));}} // This code is contributed by// Smitha Dinesh Semwal
# Python3 code to find Even  Fibonacci# Series using normal Recursion  # Function which return#nth even fibonacci numberdef evenFib(n) :    if (n < 1) :        return n    if (n == 1)  :        return 2      # calculation of    # Fn = 4*(Fn-1) + Fn-2    return ((4 * evenFib(n-1)) + evenFib(n-2))   # Driver Coden = 7print(evenFib(n))  # This code is contributed by Nikita Tiwari.
// C# code to find Even Fibonacci// Series using normal Recursionusing System; class GFG { // Function which return// nth even fibonacci numberstatic long evenFib(int n){    if (n < 1)    return n;     if (n == 1)    return 2;     // calculation of Fn = 4*(Fn-1) + Fn-2    return ((4 * evenFib(n - 1)) +                 evenFib(n - 2));} // Driver codepublic static void Main (){    int n = 7;    Console.Write(evenFib(n));}} // This code is contributed by Nitin Mittal.
<?php// PHP code to find Even Fibonacci// Series using normal Recursion // Function which return// nth even fibonacci numberfunction evenFib($n){    if ($n < 1)        return $n;    if ($n == 1)        return 2;     // calculation of    // Fn = 4*(Fn-1) + Fn-2    return ((4 * evenFib($n-1)) +                 evenFib($n-2));} // Driver Code$n = 7;echo(evenFib($n)); // This code is contributed by Ajit.?>
<script>// Javascript code to find Even Fibonacci// Series using normal Recursion // Function which return// nth even fibonacci numberfunction evenFib(n){    if (n < 1)        return n;    if (n == 1)        return 2;     // calculation of    // Fn = 4*(Fn-1) + Fn-2    return ((4 * evenFib(n-1)) +                 evenFib(n-2));} // Driver Codelet n = 7;document.write(evenFib(n)); // This code is contributed by _saurabh_jaiswal.</script>
Output : 
10946
Time complexity of above implementation is exponential. We can do it in linear time using Dynamic Programming. We can also do it in O(Log n) time using the fact EFn = F3n. Note that we can find n’th Fibonacci number in O(Log n) time (Please see Methods 5 and 6 here). This article is contributed by Shivam Pradhan(anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
nitin mittal
jit_t
_saurabh_jaiswal
sagar0719kumar
sweetyty
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