Lagrange Multipliers Can Fail to Determine Extrema
Jeffrey L. Nunemacher · College Mathematics Journal · 2003
where z denotes the greatest integer less than or equal to z. Using the approximation Fn ≈ φn √5 we get d(n) = 1 + log10( φ n √ 5 ) ≈ 1 + n log10 φ − log10 √ 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 = 16, 35, 59, 83, 102, 126, 150, 169, 193, . . . . For example, F16 to F21 are 987, 1597, 2584, 4181, 6765, 10946, which shows a run of only four Fibonacci numbers with four digits.