A269064 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.

0, 1, 4, 10, 26, 48, 87, 135, 208, 293, 410, 542, 714, 904, 1141, 1399, 1712, 2049, 2448, 2874, 3370, 3896, 4499, 5135, 5856, 6613, 7462, 8350, 9338, 10368, 11505, 12687, 13984, 15329, 16796, 18314, 19962, 21664, 23503, 25399, 27440, 29541, 31794, 34110, 36586, 39128, 41837, 44615, 47568

Offset: 0

Author

Luce ETIENNE, Feb 18 2016

Keywords

• nonn
• easy

Comments

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.

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).

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.

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.

Examples

			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.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Luce ETIENNE, Illustration of initial terms
• Kival Ngaokrajang, Illustration of triangles expansion
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,2,-2,0,2,-1).

Crossrefs

Cf. A003215, A005448, A005993, A033428, A064412, A061600, A006003, A039623, A102214, A265282.

Programs

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

Mathematica

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 *)

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

Formula

a(n) = (7*n^3-3*n^2+4*n)/2 for n even.

a(n) = (28*n^3+30*n^2+16*n+7+(-1)^n)/8 for n odd.

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.

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