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%I #13 Nov 26 2019 04:36:19
%S 1,2,4,7,1,12,4,21,10,1,37,22,5,65,47,15,1,114,98,38,6,200,199,91,21,
%T 1,351,396,210,60,7,616,777,468,158,28,1,1081,1508,1014,396,89,8,1897,
%U 2900,2151,952,255,36,1,3329,5534,4487,2212,687,126,9,5842,10492,9229
%N Triangle read by rows: T(n,k) is the number of binary sequences of length n containing k subsequences 010 (n,k >= 0).
%C Row n has ceiling(n/2) terms (n >= 1).
%C Sum of entries in row n is 2^n (A000079).
%C T(n,0) = A005251(n+3), T(n,1) = A118430(n).
%C Sum_{k=0..n-1} k*T(n,k) = (n-2)*2^(n-3) (A001787).
%H Alois P. Heinz, <a href="/A118429/b118429.txt">Rows n = 0..199, flattened</a>
%F G.f.: G(t,z) = (1+(1-t)z^2)/(1 - 2z + (1-t)z^2 - (1-t)z^3).
%F Recurrence relation: T(n,k) = 2T(n-1,k) - T(n-2,k) + T(n-3,k) + T(n-2,k-1) - T(n-3,k-1) for n >= 3.
%e T(6,2) = 5 because we have 010010, 010100, 010101, 001010 and 101010.
%e Triangle starts:
%e 1;
%e 2;
%e 4;
%e 7, 1;
%e 12, 4;
%e 21, 10, 1;
%e 37, 22, 5;
%p G:=(1+(1-t)*z^2)/(1-2*z+(1-t)*z^2-(1-t)*z^3): Gser:=simplify(series(G,z=0,18)): P[0]:=1: for n from 1 to 16 do P[n]:=sort(coeff(Gser,z^n)) od: 1; for n from 1 to 16 do seq(coeff(P[n],t,j),j=0..ceil(n/2)-1) od; # yields sequence in triangular form
%t nn=15;Map[Select[#,#>0&]&,CoefficientList[Series[1/(1-2z-(u-1)z^3/(1-(u-1)z^2)),{z,0,nn}],{z,u}]]//Grid (* _Geoffrey Critzer_, Dec 03 2013 *)
%Y Cf. A000079, A005251, A001787, A118430.
%K nonn,tabf
%O 0,2
%A _Emeric Deutsch_, Apr 27 2006
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