%I
%S 1,1,1,1,2,2,4,3,1,7,6,3,12,13,6,1,21,26,13,4,37,50,30,10,1,65,96,66,
%T 24,5,114,184,139,59,15,1,200,350,288,140,40,6,351,661,591,318,105,21,
%U 1,616,1242,1199,704,266,62,7,1081,2324,2406,1533,645,174,28,1,1897,4332
%N Triangle read by rows: T(n,k) is the number of compositions of n having exactly k parts equal to 2.
%C Row n has 1+floor(n/2) terms. Row sums are the powers of 2 (A000079). Column 0 yields A005251.
%C Number of binary words of length n1 having k isolated 0's. Example: T(5,1)=6 because we have 0111, 0100, 1011, 1101, 0010 and 1110.  _Emeric Deutsch_, May 21 2006
%H Alois P. Heinz, <a href="/A105114/b105114.txt">Rows n = 0..200, flattened</a>
%F G.f.: (1z)/(12z+z^2z^3tz^2+tz^3).
%e T(7,3) = 4 because we have (1,2,2,2), (2,1,2,2), (2,2,1,2) and (2,2,2,1).
%e Triangle begins:
%e 1;
%e 1;
%e 1, 1;
%e 2, 2;
%e 4, 3, 1;
%e 7, 6, 3;
%e 12, 13, 6, 1;
%e 21, 26, 13, 4;
%e 37, 50, 30, 10, 1;
%e 65, 96, 66, 24, 5;
%e 114, 184, 139, 59, 15, 1;
%e 200, 350, 288, 140, 40, 6;
%e 351, 661, 591, 318, 105, 21, 1;
%e 616, 1242, 1199, 704, 266, 62, 7;
%p G:=(1z)/(12*zz^2*t+z^3*t+z^2z^3):Gser:=simplify(series(G,z=0,18)): P[0]:=1: for n from 1 to 16 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 16 do seq(coeff(t*P[n],t^k),k=1..1+floor(n/2)) od;# yields sequence in triangular form
%t nn=15;f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[1/(1(x/(1x)x^2+y x^2)),{x,0,nn}],{x,y}]]//Grid (* _Geoffrey Critzer_, Nov 05 2012 *)
%Y Cf. A000079, A005251.
%K nonn,tabf
%O 0,5
%A _Emeric Deutsch_, Apr 07 2005
