A118897 Triangle read by rows: T(n,k) is the number of binary sequences of length n containing k subsequences 0000 (n,k>=0).
1, 2, 4, 8, 15, 1, 29, 2, 1, 56, 5, 2, 1, 108, 12, 5, 2, 1, 208, 28, 12, 5, 2, 1, 401, 62, 29, 12, 5, 2, 1, 773, 136, 65, 30, 12, 5, 2, 1, 1490, 294, 145, 68, 31, 12, 5, 2, 1, 2872, 628, 319, 154, 71, 32, 12, 5, 2, 1, 5536, 1328, 694, 344, 163, 74, 33, 12, 5, 2, 1, 10671, 2787
Offset: 0
Examples
T(7,2) = 5 because we have 0000010, 0000011, 0100000, 1100000 and 1000001. Triangle starts: 1; 2; 4; 8; 15, 1; 29, 2, 1; 56, 5, 2, 1; 108, 12, 5, 2, 1; ...
Links
- Alois P. Heinz, Rows n = 0..143, flattened
Programs
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Maple
G:=(1+(1-t)*(z+z^2+z^3))/(1-(1+t)*z-(1-t)*(z^2+z^3+z^4)): Gser:=simplify(series(G,z=0,17)): P[0]:=1: for n from 1 to 14 do P[n]:=sort(coeff(Gser,z^n)) od: 1;2;4;8; for n from 4 to 14 do seq(coeff(P[n],t,j),j=0..n-3) od; # yields sequence in triangular form # second Maple program: b:= proc(n, t) option remember; `if`(n=0, 1, expand(b(n-1, min(3, t+1))*`if`(t>2, x, 1))+b(n-1, 0)) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0)): seq(T(n), n=0..14); # Alois P. Heinz, Sep 17 2019
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Mathematica
nn=15;a=x^3/(1-y x)+x+x^2;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 *)
Formula
G.f.: G(t,z) = [1+(1-t)(z+z^2+z^3)]/[1-(1+t)z-(1-t)(z^2+z^3+z^4)].
Comments