Distance to the spiral

Gold: the crossings with Holden's lines (whole numbers of units). Blue: the joins, where one quarter-circle meets the next. With the Pentametric spiral chosen, its curve (purple) is drawn over the Fibonacci one (pale): it meets Holden's lines at the Lucas numbers. On the logarithmic scale the curve becomes almost a straight staircase: each quarter turn multiplies the distance by about φ = 1.618…, the golden ratio, as for a true golden spiral.

In the pentagon picture of the Pentametric spiral, square n sits n origins units from the Origin with its diagonal along its ray. A square of area n²/5 would touch the Origin exactly with its inner corner. The real square has area 𝔓(n) = (n² + 4χ)/5, where χ = (n | 5) is 0, +1 or −1, so its half-diagonal is √(n² + 4χ) units, and its inner corner misses the Origin by n − √(n² + 4χ) origins units: past it when χ = +1, short of it when χ = −1, exactly on it when 5 divides n. This is the ±0.8 of the formula, seen as a distance.

proved At the Lucas numbers the miss is exactly 2ψ^k = ±2/φ^k origins units (ψ = (1 − √5)/2): −1.236, +0.764, −0.472, +0.292, −0.180, +0.111, … It changes side every time and shrinks by the golden ratio. Proof: χ(L_k) = −(−1)^k and L_k² − 4(−1)^k = 5F_k², so the miss is L_k − √5·F_k = (φ^k + ψ^k) − (φ^k − ψ^k) = 2ψ^k. For the other n it is close to ∓2/n.

• Crossings: twice the next Fibonacci number. In the square of side F_k the spiral crosses Holden's line once, at exactly 2F_k+1 origins units from the Origin: 2, 4, 6, 10, 16, 26, 42, … Each crossing sits at the same place in its quarter turn, 53.13° in, the angle arctan(4/3) of the 3-4-5 triangle: seen from the centre of its arc, every crossing is the same 3-4-5 point, scaled by F_k (the Epitetartic Rhombus; note doi:10.5281/zenodo.23325442). proved
• Joins: Pythagoras with two crossings. Where the arc of square j meets the arc of square j + 1, the distance is 2√F_2j+1 units: √8, √20, √52, √136, √356, … Since F_2j+1 = F_j² + F_j+1², this is √((2F_j)² + (2F_j+1)²): the hypotenuse of a right triangle whose legs are the distances of the crossings in squares j − 1 and j. proved (the join is the arc's end, M_j − F_ji^j; its squared distance is (2F_j)² + (F_j + L_j)² units², and F_j + L_j = 2F_j+1.) Checked for j ≤ 27.
• The closest point. The spiral starts 2 units from the Origin, comes in to its closest point, √10 − √2 = 1.748 units, at 29.5% of the first quarter turn (where the line from the arc's centre through the Origin meets the arc), and from there on it only moves away. proved for the minimum; computed that the distance only grows after it.
• The Pentametric spiral. Through the centres of all the squares n, it is exactly n origins units from the Origin everywhere, by construction, and it meets Holden's lines at the Lucas numbers (the Fibonacci squares' centres, L_k units: Holden). In the same square the Fibonacci spiral crosses the line F_k units further out: 2F_k+1 = L_k + F_k. proved
• The corners. The inner corner of square n misses the Origin by n − √(n² + 4χ) units, exactly 2ψ^k = ±2/φ^k at the Lucas numbers. proved
• Growth by φ. Distances at the same place of successive quarter turns grow in the ratio F_k+2/F_k+1 → φ: the Fibonacci spiral, made of circle arcs, approaches the golden logarithmic spiral around the Origin.

Coordinates as in the note: Gardiner's frame, first square centred at (0, 0), Origin P = (1/10, 3/10), centre of square k at P + i^k−1L_kw with w = −(1 + 3i)/10, and its arc centred at the corner M_k = C_k + F_ki^k−1(1 + i)/2 with radius F_k. Idea and question: Lluis Garcia Torcal. Written up with proofs in the note The Epitetartic Rhombus (version 2, Section 5).