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 A137515 Maximal number of right-angled triangles in n turns of the Pythagoras's snail. 2
 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 (list; graph; refs; listen; history; text; internal format)
 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 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 = w; 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: A232509 A087973 A117273 * A164324 A197246 A187104 Adjacent sequences:  A137512 A137513 A137514 * A137516 A137517 A137518 KEYWORD easy,nice,nonn AUTHOR Sébastien Dumortier, Apr 23 2008, Apr 25 2008 STATUS approved

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Last modified November 13 08:18 EST 2019. Contains 329093 sequences. (Running on oeis4.)