Zeckendorf-Beatty Bridge

Abstract

Identify the least Zeckendorf digit with the shifted golden Beatty letter.

For a canonical Zeckendorf representation, the conjugate-power error lies on opposite sides of phi^(-3) according to whether index 2 is absent or present. This is exactly the existing golden mechanical window test.

Theorem 1.1 (The least digit is the shifted mechanical letter).

$$\mathrm{leastDigitAbsent}\left(i\right)=\mathrm{goldenMechanicalLetterIsOne}\left(i+1\right)$$

Proof. Machine-checked in Lean as D5/S1/Words/ZeckendorfBeattyBridge.zeckendorf_beatty_bridge (✓ std3). ∎

Source. Repository-derived.

Commentary.

For every natural i, index 2 is absent from wdigits i if and only if goldenMechanicalLetter(i+1) equals one. The shift is part of the statement and is not absorbed into either frozen definition.

Theorem 1.2 (The Fibonacci word has an explicit Beatty floor test).

$$\mathrm{fibWord}\left(Q\right)=\mathrm{ofFn}\left(\mathrm{shiftedGoldenBeattyFloorTest}\left(Q\right)\right)$$

Proof. Machine-checked in Lean as D5/S1/Words/ZeckendorfBeattyBridge.fibWord_eq_beatty_floor (✓ std3). ∎

Source. Repository-derived.

Commentary.

At each valid position i, the Boolean letter is true exactly when floor((i+2)/phi)-floor((i+1)/phi)=1. This follows by rewriting the frozen least-Zeckendorf-digit formula through the bridge above.

References

• Truth anchor: D5/S1/Words/ZeckendorfBeattyBridge.fibWord_eq_beatty_floor
• Truth anchor: D5/S1/Words/ZeckendorfBeattyBridge.zeckendorf_beatty_bridge
• Dependency: D5/S0/Tower/GoldenGapZeckendorf
• Dependency: D5/S1/Words/GoldenMechanicalWord