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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).
2
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).
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
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, May 29 2006
STATUS
approved