%I #23 Apr 13 2017 17:38:04
%S 1,3,3,6,12,15,21,18,30,27,39,30,48,39,57,42,66,51,75,54,84,63,93,66,
%T 102,75,111,78,120,87,129,90,138,99,147,102,156,111,165,114,174,123,
%U 183,126,192,135,201,138,210,147,219
%N Start with a single equilateral triangle for n=0; for the odd n-th generation add a triangle at each expandable side of the triangles of the (n-1)-th generation (this is the "side to side" version); for the even n-th generation use the "side to vertex" version; a(n) is the number of triangles added in the n-th generation.
%C See a comment on V-V and V-S at A249246.
%C There are a total of 16 combinations as shown in the table below:
%C +-------------------------------------------------------+
%C | Even n-th version V-V S-V V-S S-S |
%C +-------------------------------------------------------+
%C | Odd n-th version |
%C | V-V A008486 A248969 A261951 A261952 |
%C | S-V A261950 A008486 A008486 A261956 |
%C | V-S A249246 A008486 A008486 A261957 |
%C | S-S A261953 a(n) A261955 A008486 |
%C +-------------------------------------------------------+
%C Note: V-V = vertex to vertex, S-V = side to vertex,
%C V-S = vertex to side, S-S = side to side.
%H Kival Ngaokrajang, <a href="/A261954/a261954.pdf">Illustration of initial terms</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,1,0,-1).
%F a(0) = 1, a(1) = 3; for even n >= 2, a(n) = 9*(n/2-1) + 3 or a(n) = A017197(n/2-1); for odd n >= 3, a(n) = a(n-2) + 9, if mod(n,4) = 1 otherwise a(n) = a(n-2) + 3.
%F Conjectures from _Colin Barker_, Sep 10 2015: (Start)
%F a(n) = a(n-2)+a(n-4)-a(n-6) for n>6.
%F G.f.: (7*x^6+6*x^5+8*x^4+3*x^3+2*x^2+3*x+1) / ((x-1)^2*(x+1)^2*(x^2+1)).
%F (End)
%o (PARI) a=3; print1("1, ", a, ", "); for (n=2, 100, if (Mod(n,4)==0||Mod(n,4)==2, print1(9*(n/2-1)+3, ", "), if (Mod(n,4)==1, a=a+9, a=a+3); print1(a, ", ")))
%Y Cf. A008486, A017197, A248969, A249246.
%K nonn,easy
%O 0,2
%A _Kival Ngaokrajang_, Sep 06 2015