Search: id:A119851
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%I A119851
%S A119851 1,3,9,26,1,75,6,216,27,622,106,1,1791,387,9,5157,1350,54,14849,4566,
%T A119851 267,1,42756,15102,1179,12,123111,49113,4833,90,354484,157622,18798,536,
%U A119851 1,1020696,500520,70317,2775,15,2938977,1575558,255231,13068,135
%N A119851 Triangle read by rows: T(n,k) is the number of ternary words of length 
               n containing k 012's (n>=0, 0<=k<=floor(n/3)).
%C A119851 Row n has 1+floor(n/3) terms. Sum of entries in row n is 3^n (A000244). 
               T(n,0)=A076264(n). T(n,1)=A119852(n). Sum(k*T(n,k),k>=0)=(n-2)*3^(n-3)=A027741(n-1).
%F A119851 G.f.=G(t,z)=1/(1-3z+z^3-tz^3). Recurrence relation: T(n,k)=3*T(n-1,k)-T(n-3,
               k)+T(n-3,k-1) for n>=3.
%e A119851 T(4,1)=6 because we have 0012, 0120, 0121, 0122, 1012 and 2012.
%e A119851 Triangle starts:
%e A119851 1;
%e A119851 3;
%e A119851 9;
%e A119851 26,1;
%e A119851 75,6;
%e A119851 216,27;
%e A119851 622,106,1;
%p A119851 G:=1/(1-3*z+z^3-t*z^3): Gser:=simplify(series(G,z=0,20)): P[0]:=1: for 
               n from 1 to 15 do P[n]:=sort(coeff(Gser,z^n)) od: for n from 0 to 
               15 do seq(coeff(P[n],t,j),j=0..floor(n/3)) od; # yields sequence 
               in triangular form
%Y A119851 Cf. A000244, A076264, A119852, A027741.
%Y A119851 Sequence in context: A074440 A006204 A013572 this_sequence A119825 A037260 
               A035313
%Y A119851 Adjacent sequences: A119848 A119849 A119850 this_sequence A119852 A119853 
               A119854
%K A119851 nonn,tabf
%O A119851 0,2
%A A119851 Emeric Deutsch (deutsch(AT)duke.poly.edu), May 26 2006

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