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

1,1

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

Links

Table of n, a(n) for n=1..42.

Formula

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

Example

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.

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 *)

Prog

(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 n=n+1 som=som+asin(1/sqrt(hyp))/pi*180 hyp=hyp+1

Crossrefs

Sequence in context: A351584 A332591 A117273 * A164324 A334643 A197246

Adjacent sequences: A137512 A137513 A137514 * A137516 A137517 A137518

Keyword

easy,nice,nonn

Author

Sébastien Dumortier, Apr 23 2008, Apr 25 2008

Status

approved