KNOTT-6.4 — Fibonacci Arctangent Telescoping Formulae for π

By Cassini’s odd-index relation F₂ₙ₊₁² − 1 = F₂ₙ F₂ₙ₊₂, every even-indexed reciprocal Fibonacci angle splits as arctan(1/F₂ₙ) = arctan(1/F₂ₙ₊₁) + arctan(1/F₂ₙ₊₂), telescoping from arctan(1/F₂) = π/4 into Lehmer’s series π/4 = ∑_{k=1}^∞ arctan(1/F₂ₖ₊₁).

Curated Source & Catalog Provenance

Catalog ID	KNOTT-6.4
Curated Source	Dr. Ron Knott, The Fibonacci Numbers in a Formula for Pi (#piandfib)
Source Attribution	© 1996–2025 Dr. Ron Knott, University of Surrey — Explore Dr. Knott’s original exposition
Exact SymPy Validator	geometor.knott.identities.construct_hayashi_arctan_grid_model() (0 sp.Float atoms)
GEOMETOR Concordance	geometor/wildberger, geometor/fundamentals

Constructive Synthesis & Exact Proof

How do Fibonacci numbers—governed by \(\sqrt{5}\)—generate the circle constant \(\pi\)? On fibpi.html (KnottFibPi), Dr. Ron Knott derives D. H. Lehmer’s (Lehmer1936) reciprocal Fibonacci Euler–Machin series directly from odd-index Cassini's identity:

\[F_{2n+1}^2 - F_{2n} F_{2n+2} = 1 \implies \frac{\frac{1}{F_{2n+1}} + \frac{1}{F_{2n+2}}}{1 - \frac{1}{F_{2n+1} F_{2n+2}}} = \frac{F_{2n+2} + F_{2n+1}}{F_{2n} F_{2n+2}} = \frac{F_{2n+3}}{F_{2n} F_{2n+2}}? \quad \text{No: } F_{2n+1} F_{2n+2} - 1 = F_{2n+1} F_{2n+2} - (F_{2n+1}^2 - F_{2n} F_{2n+2}) = F_{2n}(F_{2n+1} + F_{2n+2}),\]

canceling \((F_{2n+1} + F_{2n+2})\) identically to leave \(1 / F_{2n}\)!

• Ko Hayashi’s Exact Integer-Lattice Proof: Over horizontal axis OX to X \((16, 0)\), plotting integer lattice points E \((10, 10)\) (slope \(1 = 1/F_2\)), M \((15, 5)\) (slope \(1/3 = 1/F_4\)), and N \((16, 2)\) (slope \(1/8 = 1/F_6\), vertical leg XN): right triangle OME has exact perpendicularity \(s(OE, EM) = 1\) between OE and EM with \(Q(E, M)/Q(O, E) = 50/200 = (1/2)^2\), while right triangle ONM has \(s(OM, MN) = 1\) between OM and MN with \(Q(M, N)/Q(O, M) = 10/250 = (1/5)^2\), proving geometrically along ray ON that \(\frac{\pi}{4} = \arctan\frac{1}{2} + \arctan\frac{1}{5} + \arctan\frac{1}{8}\).

Machine-Verified Exact Symbolic Invariants

Invariant	Exact Symbolic Value
twist 1 over 2	\(\frac{1}{4}\)
twist 1 over 3	\(\frac{1}{9}\)
twist 1 over 5	\(\frac{1}{25}\)
twist 1 over 8	\(\frac{1}{64}\)
remainder after 8 steps	\(\operatorname{atan}{\left(\frac{1}{2584} \right)}\)