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%I #6 Nov 16 2019 20:08:51
%S 1,2,1,5,4,12,13,2,29,40,12,70,117,52,4,169,332,196,32,408,921,678,
%T 172,8,985,2512,2216,768,80,2378,6761,6952,3064,512,16,5741,18004,
%U 21144,11328,2640,192,13860,47525,62762,39624,11920,1424,32,33461,124536
%N 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).
%C Row n has 1+ceiling(n/2) terms.
%C Sum of entries in row n is 3^n (A000244).
%C T(n,0) = A000129(n+1) (Pell numbers).
%C T(n,1) = A119915(n).
%C Sum_{k>=0} k*T(n,k) = A119916(n).
%F G.f. = G(t,z) = (1+tz)/(1-2z-z^2-2tz^2).
%F T(n,k) = 2T(n-1,k) + T(n-2,k) + 2T(n-2,k-1) (n >= 2).
%e T(4,2)=12 because we have 0101, 0102, 0110, 0120, 0201, 0202, 0210, 0220, 1010, 1020, 2010 and 2020.
%e Triangle starts:
%e 1;
%e 2, 1;
%e 5, 4;
%e 12, 13, 2;
%e 29, 40, 12;
%e 70, 117, 52, 4;
%p 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
%Y Cf. A000244, A000129, A119915, A119916.
%K nonn,tabf
%O 0,2
%A _Emeric Deutsch_, May 29 2006
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