A118890 Triangle read by rows: T(n,k) is the number of binary sequences of length n containing k subsequences 0110 (n,k >= 0).

1, 2, 4, 8, 15, 1, 28, 4, 52, 12, 97, 30, 1, 181, 70, 5, 338, 156, 18, 631, 339, 53, 1, 1178, 722, 142, 6, 2199, 1515, 357, 25, 4105, 3140, 862, 84, 1, 7663, 6444, 2018, 252, 7, 14305, 13116, 4614, 700, 33, 26704, 26513, 10348, 1846, 124, 1, 49850, 53280, 22844

Offset: 0

Author

Emeric Deutsch, May 04 2006

Keywords

• nonn
• tabf

Comments

Row n has ceiling(n/3) terms (n>=1).

Sum of entries in row n is 2^n (A000079).

T(n,0) = A049864(n).

T(n,1) = A118892(n).

Sum_{n>=0} k*T(n,k) = (n-3)*2^(n-4) (A001787).

Examples

			T(8,2) = 5 because we have 01100110, 01101100, 01101101, 00110110 and 10110110.
Triangle starts:
    1;
    2;
    4;
    8;
   15,   1;
   28,   4;
   52,  12;
   97,  30,  1;
  181,  70,  5;
  338, 156, 18;
  631, 339, 53, 1;
		

Links

• Alois P. Heinz, Rows n = 0..250, flattened

Crossrefs

Cf. A000079, A049864, A118892, A001787.

Programs

Maple

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

Mathematica

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 *)

Formula

G.f.: G(t,z) = (1+(1-t)z^3)/(1 - 2z + (1-t)(1-z)z^3).