The angle of 0.8
The correction in 𝔓(n) = n²/5 + 0.8·(n|5) is always +0.8, −0.8 or 0. Drawn as a slope, 0.8 is an angle: arctan(4/5) = 38.6598°, and twice that is 77.3196°. This page follows that angle: the triangle it makes, the three ways 4/5 can be read in a right triangle, and the family of angles it belongs to, which leads back to the floor cycles of other primes.
1 · The ±0.8 triangle
From P, one step to the right, then up 0.8 when (n|5) = +1 (n ends in 1, 4, 6, 9), down 0.8 when it is −1 (2, 3, 7, 8), and straight on when n is a multiple of 5. The correction itself is explained on The ±0.8 correction.
The two outer points are 1.6 apart, the full span of the correction. Each half is a right triangle with legs 1 and 0.8, that is 5 and 4: the triangle 4-5-√41, with the angle θ = arctan(4/5) at O. The whole triangle has apex angle 2θ = 77.3196°, with tan 2θ = 40/9, sin 2θ = 40/41, cos 2θ = 9/41, so doubling the angle gives the whole-number triangle 9-40-41. Each long side is √1.64 = √41/5.
exact All values above are exact (fractions and the double-angle formulas); the degrees are rounded to four places.
2 · One triangle, many names
A right triangle can be read from either acute corner. Drag x: the blue leg is 1, the red leg is x, the green side is the hypotenuse. On the right the same triangle is flipped, so the red leg is now next to the angle. Colours follow the physical side, not the name sine, cosine or tangent.
θ + φ = 90° for every x, so the sine of one is the cosine of the other, and their tangents are reciprocals.
3 · 0.8, three ways
The same 4/5 is the sine of the larger angle of the 3-4-5 triangle, the cosine of its smaller angle, and the tangent of θ in the 4-5-√41 triangle of section 1. So sin(arctan 4/3) = cos(arctan 3/4) = tan(arctan 4/5) = 0.8. The 3-4-5 triangle is itself a doubled angle: 2·arctan(1/2) = arctan(4/3), and arctan(1/2) is the angle θ₂ of the paper's construction, so sin 2θ₂ = 0.8. Two different members of the same family (next section) give the same 0.8. With numbers this small, that is best read as a coincidence.
4 · The family arctan(k/(k+1))
arctan(1/2) is the first of a family: θk = arctan(k/(k+1)). Our angle arctan(4/5) is k = 4. Two exact facts hold for every k:
- θk + arctan(1/(2k+1)) = 45°, because tan of the sum is (2k² + 2k + 1)/(2k² + 2k + 1) = 1.
- Doubling θk gives the whole-number triangle (2k+1, 2k(k+1), p) with p = k² + (k+1)²: 3-4-5, 5-12-13, 7-24-25, 9-40-41, 11-60-61, …
θk never reaches 45°: the gap is always arctan(1/(2k+1)), which shrinks as k grows but is never zero. In the same way 2θk creeps towards 90° and the doubled triangle gets thinner and thinner, without ever becoming a line. The slider goes to k = 100, where the gap is 0.2851°.
The link to the floor cycles of other primes
On Other primes and The 13-cycle as music, the sequence ⌊n²/p⌋ mod p has period p². For a prime p, every non-zero value occurs p − 1 times per period and 0 occurs 2p − 1 times, so
(2p − 1) + (p − 1)² = p².
This is a Pythagorean triple exactly when 2p − 1 is a perfect square, and 2p − 1 = (2k+1)² means p = k² + (k+1)². So the primes whose silences form a square are exactly the hypotenuses of the doubled triangles of this family, and the triple of the counts is the doubled triangle itself: for k = 1, 9 silences and 16 = 4² for the rest give 3-4-5; for our k = 4, p = 41, 81 silences and 40² give 9-40-41, the triangle of the ±0.8 apex. When p = k² + (k+1)² is not prime (25, 85, 145, …) the even counts break. The table counts every period in your browser: choose how far it goes, and click a row to draw that triangle above.
proved the 45° identity, the doubled triple and the step from the counts to p = k² + (k+1)² are short algebra. computed the counts in the table, for the rows shown (up to k = 30, p = 1861). The uniform counts for every prime are shown on Other primes.
5 · And in the geometry?
We first looked for the angle between the marked points of the two pictures (centres, corners and O), and there the honest answer is: nowhere special. (It does appear exactly where a line meets an arc: see section 6 below.) See for yourself: choose a picture and a slope, and every segment between two marked points (centres, corners and O) that makes that angle is drawn in gold. With custom you can choose any number of squares, not only multiples of 5 (up to 300 around the pentagon, 40 in the tiling, where the squares grow so fast that the small ones vanish).
- Golden-rectangle tiling (all corners up to square 40): two segments have slope 4/5, from the corner of squares 1|2 to the corner of 5|8 (length √41) and from 3|4 to 6|9 (length 2√41), and none further out. Slope 3/4 also has exactly two, while the Fibonacci slopes 1/2, 2/3, 3/5, 5/8, … have 36 to 41 each.
- The Pentametric spiral (squares 1 to 60 around the pentagon): fourteen segments make the angle with a spoke or with the line across it, for example from the centre of square 5 to the side corners of square 25 (ratio 20/25). All of them lie on the silent ray. From the centre of square 5, the side corners of square 5(k+1) are seen at exactly θk = arctan(k/(k+1)) from the line across the spoke (they are 5k steps of √0.1 further along it and 5(k+1) steps to the side), so the whole family of section 4 appears there, but so do other slopes. A multiple of 5 has 𝔓(n) = n²/5 exactly, so its square has its inner corner at P, and along that ray every rational slope appears; 1/2 and 1/3 appear even more often.
6 · Found: where Holden's line meets the arc
The search of section 5 joins marked points, and there the angle is nowhere special. Lluis found it where that search never looked: where one of Holden's lines crosses an arc of the spiral. In his GeoGebra frame (E at the origin of the axes, the arc of radius 1 centred at (0, −1)), Holden's green line x + 3y = −2 runs through C₂ = (1, −1) on the arc and leaves it again at one more point, Q₅. The vertical through Q₅ is at x = −0.8, so the triangle from E down to it has legs 1 and 0.8: the ±0.8 triangle of section 1 standing upright, with the angle of 0.8 at E.
proved Put the line into the circle: x² + (y + 1)² = 1 with y + 1 = (1 − x)/3 gives 5x² − x − 4 = 0, that is (5x + 4)(x − 1) = 0. So the line meets the arc at C₂ (x = 1) and at Q₅ = (−4/5, −2/5) exactly. Seen from the arc's centre, Q₅ is the 3-4-5 point (−4/5, 3/5): a chord of slope 1/3 through the edge of a circle doubles its angle at the centre, tan 2θ₁ = (2/3)/(1 − 1/9) = 3/4. Holden's arctan(1/3) makes the 3-4-5 triangle, and the 0.8 comes with it. Four such triangles make the Epitetartic Rhombus; the file is in the GeoGebra gallery.
The 0.8 lives in the arithmetic, in the correction, in the doubled angle of the paper's θ₂ and in the family of triangles whose primes give square silences, and, as Lluis found, in the geometry too: where Holden's line meets an arc of the spiral, and in the Epitetartic Rhombus that every square holds.