A137515 Maximal number of right triangles in n turns of Pythagoras's snail.

16, 53, 109, 185, 280, 395, 531, 685, 860, 1054, 1268, 1502, 1756, 2029, 2322, 2635, 2967, 3319, 3691, 4083, 4494, 4926, 5376, 5847, 6337, 6848, 7377, 7927, 8496, 9086, 9694, 10323, 10971, 11639, 12327, 13035, 13762, 14509, 15276, 16062, 16868, 17694

Offset: 1

Author

Sébastien Dumortier, Apr 23 2008, Apr 25 2008

Keywords

• easy
• nice
• nonn

Comments

Pythagoras's snail: begin the snail with an isosceles triangle (side = 1 unit). Then new triangle's right-angle sides are the previous hypotenuse and 1 unit length side.

From one term to the next one, the number added grows by 18, 19, 20 or 21 (tested up to 5000 terms).

To restate the comment immediately above: the second differences of the terms of the sequence consist of 18, 19, 20, or 21. - Harvey P. Dale, May 20 2019

Examples

			17 triangles are needed to close the first turn. So there are 16 triangles in this turn. From the beginning, there are 53 triangles before closing the second turn... etc.
		

Links

• Wikipedia, Spiral of Theodorus.

Crossrefs

Cf. A072895, A352741.

Programs

Mathematica

w[n_] := ArcSin[ 1/Sqrt[n+1] ]//N; s[1] = w[1]; s[n_] := s[n] = s[n-1] + w[n]; a[n_] := (an = 1; While[ s[an] < 2*Pi*n, an++]; an-1); Table[ an = a[n]; Print[an]; an, {n, 1, 42}] (* Jean-François Alcover, Feb 24 2012 *)

Python

from math import asin, sqrt, pi
hyp=2
som=0
n=1
while n<500:
    if som+asin(1/sqrt(hyp))/pi*180>n*360:
        print(hyp-2, end=', ')
        n=n+1
    som=som+asin(1/sqrt(hyp))/pi*180
    hyp=hyp+1

Formula

a(n) = A072895(n) - 1. - Robert G. Wilson v, Feb 27 2013