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%I #18 Nov 26 2019 04:35:56
%S 1,2,4,8,15,1,28,4,52,12,97,30,1,181,70,5,338,156,18,631,339,53,1,
%T 1178,722,142,6,2199,1515,357,25,4105,3140,862,84,1,7663,6444,2018,
%U 252,7,14305,13116,4614,700,33,26704,26513,10348,1846,124,1,49850,53280,22844
%N Triangle read by rows: T(n,k) is the number of binary sequences of length n containing k subsequences 0110 (n,k >= 0).
%C Row n has ceiling(n/3) terms (n>=1).
%C Sum of entries in row n is 2^n (A000079).
%C T(n,0) = A049864(n).
%C T(n,1) = A118892(n).
%C Sum_{n>=0} k*T(n,k) = (n-3)*2^(n-4) (A001787).
%H Alois P. Heinz, <a href="/A118890/b118890.txt">Rows n = 0..250, flattened</a>
%F G.f.: G(t,z) = (1+(1-t)z^3)/(1 - 2z + (1-t)(1-z)z^3).
%e T(8,2) = 5 because we have 01100110, 01101100, 01101101, 00110110 and 10110110.
%e Triangle starts:
%e 1;
%e 2;
%e 4;
%e 8;
%e 15, 1;
%e 28, 4;
%e 52, 12;
%e 97, 30, 1;
%e 181, 70, 5;
%e 338, 156, 18;
%e 631, 339, 53, 1;
%p G:=(1+(1-t)*z^3)/(1-2*z+(1-t)*(1-z)*z^3): Gser:=simplify(series(G,z=0,24)): P[0]:=1: for n from 1 to 18 do P[n]:=sort(coeff(Gser,z^n)) od: 1; for n from 1 to 18 do seq(coeff(P[n],t,j),j=0..ceil(n/3)-1) od; # yields sequence in triangular form
%t nn=18;c=x^3;Map[Select[#,#>0&]&,CoefficientList[Series[1/(1-2x - (y-1)x^4/ (1-(y-1)c)),{x,0,nn}],{x,y}]]//Flatten (* _Geoffrey Critzer_, Dec 25 2013 *)
%Y Cf. A000079, A049864, A118892, A011787.
%K nonn,tabf
%O 0,2
%A _Emeric Deutsch_, May 04 2006
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