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Relation 4

We saw in equation (4) how a fibonacci number F_n could be expressed in terms of four other smaller fibonacci numbers. Notice that as the index k increases from 1 to n − 2, the other index n − k decreases from n − 1 to 2. So we can try to choose some k for which some of these four indices (k,k + 1,n−k − 1,n−k) turn out to be the same.

To make k and n−k as equal, we need k = n∕2, which is possible only if n is even. Say, n = 2m. Then placing n = 2m and k = n∕2 = m in equation (4), we get for all m > 1:

Now, to make k and n − k − 1 as equal, we need k = (n − 1)∕2, which is possible only if n is odd. Say, n = 2m + 1. Then placing n = 2m + 1 and k = (n − 1)∕2 = m in equation (4), we get for all m ≥ 1:

Thus for n being even or odd, we were able to express F_n in terms of two smaller fibonacci numbers which occur almost halfway earlier in the fibonacci sequence.

Example: for n = 11 and so m = 5, F₁₁ = 89 = F₅² + F₆² = 5² + 8²

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