A119914 Triangle read by rows: T(n,k) is number of ternary words of length n and having k runs of 0's of odd length (0 <= k <= ceiling(n/2); a run of 0's is a subsequence of consecutive 0's of maximal length).

1, 2, 1, 5, 4, 12, 13, 2, 29, 40, 12, 70, 117, 52, 4, 169, 332, 196, 32, 408, 921, 678, 172, 8, 985, 2512, 2216, 768, 80, 2378, 6761, 6952, 3064, 512, 16, 5741, 18004, 21144, 11328, 2640, 192, 13860, 47525, 62762, 39624, 11920, 1424, 32, 33461, 124536

Offset

0,2

Comments

Row n has 1+ceiling(n/2) terms.

Sum of entries in row n is 3^n (A000244).

T(n,0) = A000129(n+1) (Pell numbers).

T(n,1) = A119915(n).

Sum_{k>=0} k*T(n,k) = A119916(n).

Links

Table of n, a(n) for n=0..49.

Formula

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

T(n,k) = 2T(n-1,k) + T(n-2,k) + 2T(n-2,k-1) (n >= 2).

Example

T(4,2)=12 because we have 0101, 0102, 0110, 0120, 0201, 0202, 0210, 0220, 1010, 1020, 2010 and 2020.
Triangle starts:
   1;
   2,   1;
   5,   4;
  12,  13,  2;
  29,  40, 12;
  70, 117, 52, 4;

Maple

G:=(1+t*z)/(1-2*z-z^2-2*t*z^2): Gser:=simplify(series(G, z=0, 14)): 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

Crossrefs

Cf. A000244, A000129, A119915, A119916.

Sequence in context: A233525 A209143 A243274 * A152192 A290889 A120924

Adjacent sequences: A119911 A119912 A119913 * A119915 A119916 A119917

Keyword

nonn,tabf

Author

Emeric Deutsch, May 29 2006

Status

approved