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N − 1 + y, where 0 ≤ y < 1. Thus, d(n) = N = 1 + log 10 F n − y = 1 + log 10 F n , where z denotes the greatest integer less than or equal to z. Using the approximation F n ≈ φ n √ 5 we get d(n) = 1 + log 10 (φ n √ 5) ≈ 1 + n log 10 φ − log 10 √ 5 ≈ 0.209n + 0.651. Notice that d(n + k) − d(n) ≈ 0.209n + 0.651 + 0.209k − 0.209n + 0.651. Notice that for k = 1, 2, 3 or 4, d(n + k) − d(n) = 0 or 1. But d(n + 5) − d(n) = 1 or 2. When d(n + 5) − d(n) = 1, we get a run of five Fibonnacci numbers with the same number of digits. However, when d(n + 5) − d(n) = 2, we get only four. Using a computer, we found that d(n + 5) − d(n) = 2 at n =

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