A119808 - OEIS

A119808

Triangle read by rows: T(n,k) is the number of ternary words of length n having k runs of consecutive 0's (0<=k<=ceiling(n/2)).

1, 2, 1, 4, 5, 8, 17, 2, 16, 49, 16, 32, 129, 78, 4, 64, 321, 300, 44, 128, 769, 1002, 280, 8, 256, 1793, 3048, 1352, 112, 512, 4097, 8678, 5500, 880, 16, 1024, 9217, 23524, 19892, 5120, 272, 2048, 20481, 61410, 66032, 24600, 2544, 32, 4096, 45057, 155616, 205360

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OFFSET

0,2

COMMENTS

Row n contains 1+ceiling(n/2) terms. T(n,0) = 2^n (A000079). T(n,1) = 1+(n-1)*2^n = A000337(n) for n>=1. T(n,2) = 2*A055581(n-3) (n>=3). Sum(k*T(n,k),k>=0) = A081038(n-1).

LINKS

Alois P. Heinz, Rows n = 0..200, flattened

FORMULA

G.f.: (1-z+tz)/(1-3z+2z^2-2tz^2). G.f. of column k: 2^(k-1)*z^(2k-1)*/ [(1-z)^k*(1-2z)^(k+1)] (k>=1). Recurrence relation: T(n,k) = 3T(n-1,k) -2T(n-2,k) +2T(n-2,k-1) for n>=2.

EXAMPLE

T(2,1) = 5 because we have 00, 01, 02, 10 and 20.

Triangle starts:

2,1;

4,5;

8,17,2;

16,49,16;

32,129,78,4;

MAPLE

G:=(1-z+t*z)/(1-3*z+2*z^2-2*t*z^2): Gser:=simplify(series(G, z=0, 15)): P[0]:=1: for n from 1 to 12 do P[n]:=sort(coeff(Gser, z^n)) od: for n from 0 to 12 do seq(coeff(P[n], t, j), j=0..ceil(n/2)) od; # yields sequence in triangular form

MATHEMATICA

nn=15; f[list_]:=Select[list, #>0&]; a = y x/(1-x) +1; Map[f, CoefficientList[ Series[a/(1-2x a), {x, 0, nn}], {x, y}]]//Grid (* Geoffrey Critzer, Nov 19 2012 *)

CROSSREFS

Cf. A000079, A000337, A055581, A081038, A034839.

Sequence in context: A080030 A337589 A125156 * A021470 A138205 A137224

Adjacent sequences: A119805 A119806 A119807 * A119809 A119810 A119811

KEYWORD

nonn,tabf

AUTHOR

Emeric Deutsch, May 25 2006

STATUS

approved