KNOTT-5.1 — The Infinite Golden String (101101011011…), Rabbit Tree & Cutting-Line Slope φ

Iterating the Rabbit substitution 1 ↦ 10, 0 ↦ 1 generates the infinite Sturmian Golden String dₙ = ⌊(n + 1)φ⌋ − ⌊nφ⌋, identically realized geometrically as the sequence of vertical (‘1’) and horizontal (‘0’) unit-lattice crossings of the ray y = φ x from the origin.

Curated Source & Catalog Provenance

Catalog ID	KNOTT-5.1
Curated Source	Dr. Ron Knott, The Fibonacci Rabbit Sequence (The Golden String) (#PhiPlot)
Source Attribution	© 1996–2025 Dr. Ron Knott, University of Surrey — Explore Dr. Knott’s original exposition
Exact SymPy Validator	geometor.knott.golden_string.construct_golden_string_cutting_model() (0 sp.Float atoms)
GEOMETOR Concordance	geometor/trees, geometor/divine

Constructive Synthesis & Exact Proof

Dr. Ron Knott’s signature page fibrab.html (KnottGoldenString) unifies three seemingly distinct mathematical worlds—Fibonacci’s monthly rabbit substitution (\(1 \mapsto 10,\; 0 \mapsto 1\)), Sturmian Beatty differences \(d_n = \lfloor (n+1)\phi \rfloor - \lfloor n\phi \rfloor\), and straight lattice-cutting rays—into the single infinite Golden String 1011010110110101101….

• Cutting-Line Realization of the Golden String: Tracing line [OR] (segment OR from origin O to R \((5, 5\phi)\)) of exact slope \(\phi\) across vertical unit grid segments X1XT1, X2XT2, X3XT3 and horizontal grid segments Y1YR1, Y2YR2, Y3YR3: recording 1 at each vertical crossing (V1 – V5 at \((k, k\phi)\)) and 0 at each horizontal crossing (H1, H2, H3 at \((m\Phi, m)\)) spells out 1, 0, 1, 1, 0, 1, 0, 1 in exact abscissa order \(1 < \Phi < 2 < 3 < 2\Phi < 4 < 3\Phi < 5\).

• Palindrome Structure & Recursive Call Tree: Removing the final two characters of every morphism word \(S_n\) (length \(F_n\) with \(F_{n-1}\) ones and \(F_{n-2}\) zeros) leaves a strict palindrome, while naive recursive evaluation of fib(n) performs \(F_{n+1} - 1\) additions.

Machine-Verified Exact Symbolic Invariants

Invariant	Exact Symbolic Value
ray slope	\(- \frac{1}{2} + \frac{\sqrt{5}}{2}\)
H1 x	\(\frac{1}{2} + \frac{\sqrt{5}}{2}\)
H2 x	\(1 + \sqrt{5}\)
H3 x	\(\frac{3}{2} + \frac{3 \sqrt{5}}{2}\)
prefix length S8	\(34\)
ones in S8	\(21\)
zeros in S8	\(13\)
recursive additions F10	\(88\)