Nth Even Fibonacci Number
Given a value n, find the n’th even Fibonacci Number.
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
Time complexity: O(2^n)
Auxiliary Space: O(1)
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).
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