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%I #25 Sep 08 2022 08:46:15
%S 0,1,4,10,26,48,87,135,208,293,410,542,714,904,1141,1399,1712,2049,
%T 2448,2874,3370,3896,4499,5135,5856,6613,7462,8350,9338,10368,11505,
%U 12687,13984,15329,16796,18314,19962,21664,23503,25399,27440,29541,31794,34110,36586,39128,41837,44615,47568
%N At stage 1, start with a unit equilateral triangle. At each successive stage add 3*(n-1) new triangles around outside with vertex-to-vertex contacts. Sequence gives number of triangles at n-th stage.
%C At stage n, we count (6*n^2-6*n+5-3*(2*n-1)*(-1)^n)/8 unit up-pointing triangles and 3*(2*n^2-2*n+1+(2*n-1)*(-1)^n)/8 unit down-pointing triangles.
%C At stage n, the total number of unit triangles is (3*n^2-3*n+2)/2 = A005448(n). It is the same total as for A064412. Note also that A064412 gives number of triangles in a geometrical structure according to expansion side-side (mode S-S).
%C The edges of several unit triangles can form larger size triangles, and these are also up- or down-pointing. The number of all such larger is given by :(14*n^3-9*n^2+11*n+18-(9*n^2+3*n+14)*(-1)^n-4*((-1)^((2*n+1-(-1)^n)/4)))/64 up-pointing triangles and (14*n^3-15*n^2+35*n-6+(9*n^2+21*n+2)*(-1)^n+4*((-1)^((2*n-1+(-1)^n)/4)))/64 down-pointing triangles.
%C As for A265282 we observe that starting with n = 4 we can see and count hexagonal and dodecagonal forms for example in a reticular system (incomplete with hexagonal holes) by opposition to a compact shape A064412.
%H Colin Barker, <a href="/A269064/b269064.txt">Table of n, a(n) for n = 0..1000</a>
%H Luce ETIENNE, <a href="/A269064/a269064.pdf">Illustration of initial terms</a>
%H Kival Ngaokrajang, <a href="/A005448/a005448.pdf">Illustration of triangles expansion</a>
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,2,-2,0,2,-1).
%F a(n) = (7*n^3-3*n^2+4*n)/2 for n even.
%F a(n) = (28*n^3+30*n^2+16*n+7+(-1)^n)/8 for n odd.
%F a(n) = (14*n^3-12*n^2+23*n+6+3*(3*n-2)*(-1)^n+2*((-1)^((2*n-1+(-1)^n)/4)-(-1)^((6*n-1+(-1)^n)/4)))/32.
%F G.f.: x*(1+2*x+2*x^2+8*x^3+2*x^4+5*x^5+x^6) / ((1-x)^4*(1+x)^2*(1+x^2)). - _Colin Barker_, Feb 24 2016
%e a(0)= 0, a(1) = 1, a(2) = 4, a(3) = 7+3 = 10, a(4) = 19 + 6 + 1 = 26, a(5) = 31 + 12 + 4 + 1 = 48.
%t Table[(14 n^3 - 12 n^2 + 23 n + 6 + 3 (3 n - 2) (-1)^n + 2 ((-1)^((2*n - 1 + (-1)^n) / 4) - (-1)^((6 n - 1 + (-1)^n) / 4))) / 32, {n, 0, 45}] (* _Vincenzo Librandi_, Feb 19 2016 *)
%o (Magma) [(14*n^3-12*n^2+23*n+6+3*(3*n-2)*(-1)^n+2*((-1)^((2*n-1+(-1)^n) div 4)-(-1)^((6*n-1+(-1)^n) div 4)))/32: n in [0..50]]; // _Vincenzo Librandi_, Feb 19 2016
%o (PARI) concat(0, Vec(x*(1+2*x+2*x^2+8*x^3+2*x^4+5*x^5+x^6)/((1-x)^4*(1+x)^2*(1+x^2)) + O(x^50))) \\ _Colin Barker_, Feb 24 2016
%Y Cf. A003215, A005448, A005993, A033428, A064412, A061600, A006003, A039623, A102214, A265282.
%K nonn,easy
%O 0,3
%A _Luce ETIENNE_, Feb 18 2016
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