%I #16 Sep 17 2019 10:25:09

%S 1,2,4,7,1,13,2,1,24,5,2,1,44,12,5,2,1,81,26,13,5,2,1,149,56,29,14,5,

%T 2,1,274,118,65,32,15,5,2,1,504,244,143,74,35,16,5,2,1,927,499,307,

%U 169,83,38,17,5,2,1,1705,1010,652,374,196,92,41,18,5,2,1,3136,2027,1369,819

%N Triangle read by rows: T(n,k) is the number of binary sequences of length n containing k subsequences 000 (n, k >= 0).

%C Row n has n-1 terms (n >= 2). Sum of entries in row n is 2^n (A000079). T(n,0) = A000073(n+3) (the tribonacci numbers). T(n,1) = A073778(n-1). Sum_{k=0..n-1} k*T(n,k) = (n-2)*2^(n-3) (A001787).

%H Alois P. Heinz, <a href="/A118390/b118390.txt">Rows n = 0..142, flattened</a>

%F G.f.: G(t,z) = (1 + (1-t)z + (1-t)z^2)/(1 - (1+t)z - (1-t)z^2 - (1-t)z^3). Recurrence relation: T(n,k) = T(n-1,k) + T(n-2,k) + T(n-3,k) + T(n-1,k-1) - T(n-2,k-1) - T(n-3,k-1) for n >= 3.

%e T(6,2) = 5 because we have 000010, 000011, 010000, 100001 and 110000.

%e Triangle starts:

%e 1;

%e 2;

%e 4;

%e 7, 1;

%e 13, 2, 1;

%e 24, 5, 2, 1;

%e 44, 12, 5, 2, 1;

%e 81, 26, 13, 5, 2, 1;

%p G:=(1+(1-t)*z+(1-t)*z^2)/(1-(1+t)*z-(1-t)*z^2-(1-t)*z^3): Gser:=simplify(series(G,z=0,32)): P[0]:=1: for n from 1 to 13 do P[n]:=coeff(Gser,z^n) od: P[0]; P[1]; for n from 2 to 13 do seq(coeff(P[n],t,k),k=0..n-2) od; # yields sequence in triangular form

%p # second Maple program:

%p b:= proc(n, t) option remember; `if`(n=0, 1,

%p expand(b(n-1, min(2, t+1))*`if`(t>1, x, 1))+b(n-1, 0))

%p end:

%p T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0)):

%p seq(T(n), n=0..14); # _Alois P. Heinz_, Sep 17 2019

%t nn=15;a=x^2/(1-y x)+x;b=1/(1-x);f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[b (1+a)/(1-a x/(1-x)) ,{x,0,nn}],{x,y}]]//Grid (* _Geoffrey Critzer_, Nov 18 2012 *)

%Y Cf. A000073, A000079, A001787, A073778, A076791.

%K nonn,tabf

%O 0,2

%A _Emeric Deutsch_, Apr 27 2006