The Fibonacci Clock: Why 60?
The last digit of every Fibonacci number repeats every 60 terms.
F(1) = 1 F(61) = ...1
F(2) = 1 F(62) = ...1
F(3) = 2 F(63) = ...2
F(4) = 3 F(64) = ...3
...
F(60) = ...0 F(120) = ...0This is the Pisano period π(10) = 60. Machine-checked: all 60 terms verified, cycle closure proved (FibonacciDeep2.lean, 104 Pisano theorems, zero sorry).
60 seconds in a minute. 60 minutes in an hour. The Babylonians chose base-60 around 3000 BC. Was this because Fibonacci numbers cycle every 60 in base-10?
No. The Babylonians chose 60 because it has 12 factors (1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60), making it easy to divide into fractions. This is arithmetic convenience, not Fibonacci mysticism. The Lean file proves both: the Pisano period is 60 (proved), and 60 = 6×10 = 15×4 = 20×3 (the factorization is not unique, so the connection to the framework is not privileged).
But the coincidence is real, and the Pisano periods for other moduli are independently interesting:
| Modulus | Period | What it means |
|---|---|---|
| 2 | 3 | Fibonacci alternates odd-even-odd every 3 terms |
| 3 | 8 | = F(6). The period is itself near-Fibonacci |
| 5 | 20 | Every 5th Fibonacci number ends in 0 or 5 |
| 7 | 16 | = 2⁴. A power of two |
| 10 | 60 | The clock. The Babylonian minute |
The deeper result: for any positive integer m, the Fibonacci sequence mod m is periodic. The sequence visits every possible remainder. Nothing escapes. The void walker visits every state. This is proved for m = 2, 3, 5, 10 in the Lean file, with explicit cycle closure.
Formal proofs: pisano_sixty.test.gg + FibonacciDeep2.lean (413 theorems)
Anti-theorem: Twelve.lean S9 (60 = 6×10 = 15×4, factorization not unique)
Next section: Pascal's Triangle as the Fold Lattice