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A335411 a(n) is the number of vertices formed by n-secting the angles of an equilateral triangle.

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%I A335411 #17 Jul 16 2020 03:54:09
%S A335411 3,7,21,25,63,67,129,133,219,199,333,337,471,475,633,637,819,823,1029,
%T A335411 1009,1263,1267,1521,1525,1803,1807,2109,2113,2439,2419,2793,2797,
%U A335411 3171,3175,3573,3577,3999,4003,4449,4429,4923,4927,5421,5425,5943,5947,6489
%N A335411 a(n) is the number of vertices formed by n-secting the angles of an equilateral triangle.
%C A335411 See A277402 for illustrations.
%H A335411 Lars Blomberg, <a href="/A335411/b335411.txt">Table of n, a(n) for n = 1..500</a>
%F A335411 Empirically for 12 < n < 500: a(n) = a(n-2) + a(n-10) - a(n-12) + 120.
%F A335411 Conjectures from _Colin Barker_, Jun 08 2020: (Start)
%F A335411 G.f.: x*(3 + 4*x + 11*x^2 + 24*x^4 + 24*x^6 + 24*x^8 - 24*x^9 + 45*x^10 + 20*x^11 - 11*x^12) / ((1 - x)^3*(1 + x)^2*(1 - x + x^2 - x^3 + x^4)*(1 + x + x^2 + x^3 + x^4)).
%F A335411 a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-10) - a(n-11) - a(n-12) + a(n-13) for n>13.
%F A335411 (End)
%F A335411 Colin Barker's recurrence conjecture holds for 13 < n <= 500. _Lars Blomberg_, Jun 12 2020
%F A335411 Empirical: a(2*k - 1) = 3*(4*k^2 - 6*k + 3), for k >= 1. - _Ivan N. Ianakiev_, Jul 15 2020
%Y A335411 Cf. A331782, A277402 (regions), A335412 (edges), A335413 (ngons).
%K A335411 nonn
%O A335411 1,1
%A A335411 _Lars Blomberg_, Jun 08 2020