cp's OEIS Frontend

A277402 "3-Portolan numbers": number of regions formed by n-secting the angles of an equilateral triangle.

%I A277402 #40 Jul 14 2020 23:42:15
%S A277402 1,6,19,30,61,78,127,150,217,234,331,366,469,510,631,678,817,870,1027,
%T A277402 1074,1261,1326,1519,1590,1801,1878,2107,2190,2437,2514,2791,2886,
%U A277402 3169,3270,3571,3678,3997,4110,4447,4554,4921,5046,5419,5550,5941,6078,6487,6630,7057,7194
%N A277402 "3-Portolan numbers": number of regions formed by n-secting the angles of an equilateral triangle.
%C A277402 I like the name "portolan numbers": cf. the rhumbline designs on medieval maps, constructed in a similar way.
%C A277402 The regions can be counted using an adaptation of Smiley and Wick's method in A092098: count regions assuming there are no redundant intersections, then subtract the number of regions that Ceva's Theorem says must vanish.
%C A277402 Off-diagonal redundant intersections occur for triples of integers 1 <= i, j, k < floor(n/2)-1 such that M(i)*M(j) = M(k), where M(x) is the ratio (sin(Pi(n-x)/(3n)))/(sin(Pi*x/(3n))). In the case 10|n, this corresponds to the charming identity (sin(7*Pi/30)*sin(8*Pi/30))/(sin(3*Pi/30)*sin(2*Pi/30)) = sin(9*Pi/30)/sin(1*Pi/30).
%C A277402 Differs from A092098 (which counts regions when *sides*, not angles, are n-sected) for the first time at the tenth term.
%C A277402 The above equation has solutions if and only if 10|n. This can be shown by rewriting the equation in exponential form, and using facts about vanishing sums of roots of unity to narrow the possibilities for n. (See Conway and Jones, 1976.) This is computationally feasible because A273096(6) = 1. - _Ethan Beihl_, Nov 26 2016
%H A277402 Lars Blomberg, <a href="/A277402/b277402.txt">Table of n, a(n) for n = 1..500</a>
%H A277402 Lars Blomberg, <a href="/A277402/a277402.png">Coloured illustration for n=3</a>
%H A277402 Lars Blomberg, <a href="/A277402/a277402_1.png">Coloured illustration for n=4</a>
%H A277402 Lars Blomberg, <a href="/A277402/a277402_2.png">Coloured illustration for n=19</a>
%H A277402 Lars Blomberg, <a href="/A277402/a277402_3.png">Coloured illustration for n=20</a>
%H A277402 J. H. Conway and A. J. Jones, <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa30/aa3033.pdf">Trigonometric diophantine equations (On vanishing sums of roots of unity)</a>, Acta Arithmetica 30(3), 229-240 (1976).
%H A277402 Wikipedia, <a href="https://en.wikipedia.org/wiki/Rhumbline_network">Rhumbline network</a>
%F A277402 Empirical g.f.: x*(1 + 5*x + 12*x^2 + 6*x^3 + 18*x^4 + 6*x^5 + 18*x^6 + 6*x^7 + 18*x^8 - 6*x^9 + 29*x^10 + 13*x^11 - 6*x^12) / ((1-x)^3*(1+x)^2*(1 - x + x^2 - x^3 + x^4)*(1 + x + x^2 + x^3 + x^4)). - _Colin Barker_, Oct 14 2016
%F A277402 Empirically for 12 < n <= 500: a(n) = a(n-2) + a(n-10) - a(n-12) + 120. - _Lars Blomberg_, Jun 08 2020
%F A277402 Empirical: a(2*k + 1) = 6*k*(2*k + 1) + 1, for k >= 0. - _Ivan N. Ianakiev_, Jun 27 2020
%F A277402 Empirical: 10*a(n) = 30*n^2 -45*n +23 +13*(-1)^n -15*(-1)^n*n - 24*b(n) where b(n) is the 10-periodic sequence 4, 0, -1, 0, -1, 0, -1, 0, -1, 0, 4, 0 .... of offset 0. - _R. J. Mathar_, Jul 05 2020
%e A277402 For n=3, a(n) gives the 19 regions formed by the intersection of 3*2 lines: 3 pentagons, 3 quadrilaterals, 12 triangles, and 1 big central hexagon.
%t A277402 regions[n_]:=
%t A277402 If[Mod[n,2]==0, 3n^2-6n+6, 3n^2-3n+1]-
%t A277402   6*Length@
%t A277402     Select[
%t A277402      Flatten@
%t A277402       With[
%t A277402        {b=N@
%t A277402           Table[
%t A277402              1/2 - (Sqrt[3]/2)Tan[(60Degree/n)(n/2-i)],
%t A277402              {i, 1, Floor[n/2]- 1}
%t A277402              ]},
%t A277402        Table[
%t A277402         Abs[(1-b[[k]])b[[l]]b[[j]] - b[[k]](1-b[[l]])(1-b[[j]])],
%t A277402         {j, 1, Floor[n/2]-1},
%t A277402         {k, 1, Floor[n/2]-1},
%t A277402         {l, 1, Floor[n/2]-1}]
%t A277402        ],
%t A277402      Chop@#==0&]
%Y A277402 Cf. A092098, A335411 (vertices), A335412 (edges), A335413 (ngons).
%K A277402 nonn
%O A277402 1,2
%A A277402 _Ethan Beihl_, Oct 13 2016