

A137515


Maximal number of rightangled 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
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OFFSET

1,1


COMMENTS

Pythagoras's snail: begin the snail with an isosceles triangle (side = 1 unit). Then new triangle's rightangle 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[n1] + w[n]; a[n_] := (an = 1; While[ s[an] < 2*Pi*n, an++]; an1); Table[ an = a[n]; Print[an]; an, {n, 1, 42}] (* JeanFrançois 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 hyp2 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



