A143361 Triangle read by rows: T(n,k) is the number of 010-avoiding binary words of length n containing k 00 subwords (0<=k<=n-1).
2, 3, 1, 4, 2, 1, 6, 3, 2, 1, 9, 6, 3, 2, 1, 13, 11, 7, 3, 2, 1, 19, 18, 14, 8, 3, 2, 1, 28, 30, 24, 17, 9, 3, 2, 1, 41, 50, 43, 30, 20, 10, 3, 2, 1, 60, 81, 77, 57, 36, 23, 11, 3, 2, 1, 88, 130, 132, 108, 72, 42, 26, 12, 3, 2, 1, 129, 208, 224, 193, 143, 88, 48, 29, 13, 3, 2, 1
Offset: 1
Examples
T(5,2)=3 because we have 00011, 10001 and 11000. Triangle starts: 2; 3, 1; 4, 2, 1; 6, 3, 2, 1; 9, 6, 3, 2, 1; 13, 11, 7, 3, 2, 1;
Links
- Alois P. Heinz, Rows n = 1..150, flattened
Programs
-
Maple
G:=(1+z-t*z+z^2)/(1-z-t*z+t*z^2-z^3)-1: Gser:=simplify(series(G,z=0,14)): for n to 12 do P[n]:=sort(coeff(Gser,z,n)) end do: for n to 12 do seq(coeff(P[n], t,j),j=0..n-1) end do; # yields sequence in triangular form # second Maple program: b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<3, expand(b(n-1, i+1) +b(n-1, i)*`if`(i=2, x, 1)), b(n-1, 1))) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 1)): seq(T(n), n=0..15); # Alois P. Heinz, Dec 18 2013 -
Mathematica
b[n_, i_] := b[n, i] = If[n == 0, 1, If[i<3, Expand[b[n-1, i+1] + b[n-1, i]*If[i == 2, x, 1]], b[n-1, 1]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, 1]]; Table[T[n], {n, 1, 15}] // Flatten (* Jean-François Alcover, Feb 19 2015, after Alois P. Heinz *)
Formula
G.f.: G(t,z) = (1+z-tz+z^2)/(1-z-tz+tz^2-z^3)-1.
Comments