Independent researcher, no institution. ORCID 0009-0007-8570-2536 (opens in a new tab).
The preprint, four notes and a data sheet. Read them here or download them.
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The Pentametric Formula 𝔓(n) = (n² + 4(n | 5))/5, its ten equivalent forms, the ±4/5 and ±2π/5 laws, the consonance indicator H(n) with its 25-term palindrome of silences and the four Pythagorean consonances, and the Pentametric–Lucas Theorem 𝔓(L_k) = F_k² with its converse.
Holden's line meets the first Fibonacci arc at a 3-4-5 point; in every square the crossing lies 2F_{k+1} origins units from P. The 5 : 4 rhombus it determines can be drawn in any square from its thirds, and its corner lands on grids of 15 and 30 but never on a halving grid. Version 2 added the distances from P along the spiral; version 3 renames the Holden unit the origins unit and changes no result.
From the centre of the unit square alone: exactly one circle about it has tangents, at its crossings with the diagonal arcs, that pass through a corner of the square. It is the circle of radius √0.1 and area π/10, the circle of origins. Its eight crossings (±0.3, ±0.1), (±0.1, ±0.3) are the origins of eight Fibonacci spirals, four and their mirror images.
Corner-to-midpoint lines in an empty square give a square of area 1/5 and the origins grid of cells of area 1/10, with no grid drawn. The area 𝔓(n) is then 2(n² + 4(n|5)) cells: the blue square of n, plus or minus the blue square of 2. Fibonacci squares are entered and left at Lucas multiples of thirds of the origins unit; n times Holden's crossing lands a quarter turn on from one Lucas number to the next; and the square of area 2/5 between a square and its mirror is centred on the crossing.
The digits H(n) = 𝔓(n) mod 5 for n = 0 … 50, position by position: the palindrome as a table.
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a(n) = (n^2 + 4*Legendre(n,5))/5 (opens in a new tab)
approveda(n) = A400796(n) mod 5 (opens in a new tab)
approvedAll my papers and notes: Zenodo (opens in a new tab). My OEIS author page: OEIS (opens in a new tab).
No notes tagged pentametric yet.