%I #20 Apr 12 2016 03:20:53
%S 1,5,14,32,60,103,160,238,335,459,606,786,994,1241,1520,1844,2205,
%T 2617,3070,3580,4136,4755,5424,6162,6955,7823,8750,9758,10830,11989,
%U 13216,14536,15929,17421,18990,20664,22420,24287,26240,28310,30471,32755,35134,37642
%N At stage 1, start with a unit equilateral equiangular triangle. At each successive stage add 3*(n-1) new triangles around outside with edge-to-edge contacts. Sequence gives number of triangles (regardless of size) at n-th stage.
%C Number of unit triangles at n-th stage = 3n(n-1)/2 + 1, A005448.
%D Anthony Gardiner, "Mathematical Puzzling," Dover Publications, Inc., Mineola, NY., 1987, page 88.
%H N. J. A. Sloane, <a href="/A064412/a064412.jpg">Illustration of initial terms</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 G.f.: (1+x+x^2)(1+2x+x^2+3x^3)/((1-x)^2(1-x^2)(1-x^4)).
%F a(2n+1) = (7n^3+12n^2+7n+2)/2; a(2n) = (28n^3+6n^2+4n+1+(-1)^(n+1))/8. - _Len Smiley_, Oct 07 2001
%F a(n) = (14*n^3+6*n^2+5*n+7+3*(n-1)*(-1)^n-2*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((6*n-1+(-1)^n)/4)))/32. - _Luce ETIENNE_, Jun 27 2014
%e a(4) = 32: 19 triangles of side 1, 10 of side 2 and 3 of side 3.
%p A064412:=n->(14*n^3+6*n^2+5*n+7+3*(n-1)*(-1)^n-2*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((6*n-1+(-1)^n)/4)))/32; seq(A064412(n), n=1..30); # _Wesley Ivan Hurt_, Jun 27 2014
%t CoefficientList[Series[(1 + x + x^2) (1 + 2 x + x^2 + 3 x^3)/((1 - x)^2 (1 - x^2) (1 - x^4)), {x, 0, 30}], x] (* _Wesley Ivan Hurt_, Jun 27 2014 *)
%t LinearRecurrence[{2,0,-2,2,-2,0,2,-1},{1,5,14,32,60,103,160,238},50] (* _Harvey P. Dale_, Apr 12 2016 *)
%o (PARI) a(n)=polcoeff(x*(1+x+x^2)*(1+2*x+x^2+3*x^3)/((1-x)^2*(1-x^2)*(1-x^4))+x*O(x^n),n)
%Y Cf. A056640.
%K nonn,easy
%O 1,2
%A _Robert G. Wilson v_, Sep 29 2001