A224269 Consider the spiral of Theodorus (A072895). This sequence gives the number of k successive triangles which is closer to 360 degrees than any previous k triangles.
17, 53, 185, 396, 4926, 9086, 20291, 28083, 440835, 579644, 1819320, 3032895, 8305458, 15159436, 29824343, 46104922, 88019569, 89143145, 94929121, 107958869, 227428224, 402409536, 527154160
Offset: 1
Keywords
Examples
a(1) = 17 because the first 16 right triangles result in 351.15042° (8.84958° before the original axis) and the first 17 right triangles result in 364.78344°. 17 right triangles are within 4.78344° of the original axis. a(2) = 53 because the first 54 right triangles result in 727.48834° and the first 53 right triangles result in 719.73897°. This is closer to the original axis than 16 and is within 0.2610252°. a(3) # 109 nor 110 because the first 109 right triangles result in 1079.12463° and the first 110 right triangles result in 1084.57110°. Neither angle is closer to the original axis (1080°) than 53. Therefore the third turn around the center is not close to the original axis than twice around. a(3) = 185 because the first 186 right triangles result in 1444.08227° (4.08227° after the original axis) and the first 185 right triangles result in 1439.88864°. This is closer to the original axis than 53 and is within 0.11136°.
Links
- Herbert Kociemba, Table of n, a(n) for n = 1..40
- The MacTutor History of Mathematics archive, Theodorus of Cyrene.
- The National Museum of American History, Kenneth E. Behring Center, Painting - Square Roots to Sixteen.
- Wikipedia, Theodorus of Cyrene
Crossrefs
Cf. A072895.
Programs
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Mathematica
lmt = Infinity; lst = {}; k = n = 1; s = 0; While[n < 1001, While[s < 2Pi*n, s = N[s + ArcTan[ 1 / Sqrt@ k], 32]; k++]; a = s - 2Pi*n; b = 2Pi*n - (s - ArcTan[1/Sqrt[k - 1]]); If[Min[a, b] < lmt, lmt = Min[a, b]; If[a < b, AppendTo[lst, {n, k - 1}]; Print[{n, k - 1}], AppendTo[lst, {n, k - 2}]; Print[{n, k - 2}]]]; n++]; Last@ Transpose@ lst k=minDist=1;lst={};K=-2.1577829966594462209291427868295777235; num[n_]:=Module[{a=-(K/2)+n Pi,b},b=a^2-1/6;If[Floor[b]==Floor[b+1/(144 a^2)],Floor[b],Undefined]]; While[k<40000000,n=num[k];If[!NumberQ[n],Print[k," Stop"];Break[]]; a=2Pi-Mod[K+2 Sqrt[n]+1/(6 Sqrt[n]),2Pi]; b=Mod[K+2 Sqrt[n+1]+1/(6 Sqrt[n+1]),2Pi]; If[a
Herbert Kociemba, Jul 18 2013 *)
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