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%I #12 Mar 04 2018 03:09:27
%S 1,3,9,26,1,75,6,216,27,622,106,1,1791,387,9,5157,1350,54,14849,4566,
%T 267,1,42756,15102,1179,12,123111,49113,4833,90,354484,157622,18798,
%U 536,1,1020696,500520,70317,2775,15,2938977,1575558,255231,13068,135
%N 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 Row n has 1+floor(n/3) terms.
%C Sum of entries in row n is 3^n (A000244).
%C Sum_{k>=0} k*T(n,k) = (n-2)*3^(n-3) = A027741(n-1).
%F T(n,k) = Sum_{j=0..n-3k} binomial(-(k-1),j) * binomial(j,(n-3k-j)/2) * (-3)^((3j+3k-n)/2). - _Max Alekseyev_
%F G.f.: G(t,z) = 1/(1-3z+z^3-tz^3).
%F Recurrence relation: T(n,k) = 3*T(n-1,k) - T(n-3,k) + T(n-3,k-1) for n >= 3.
%e T(4,1)=6 because we have 0012, 0120, 0121, 0122, 1012 and 2012.
%e Triangle starts:
%e 1;
%e 3;
%e 9;
%e 26, 1;
%e 75, 6;
%e 216, 27;
%e 622, 106, 1;
%p 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
%o (PARI) { T(n,k) = sum(j=0,n-3*k, if((n-3*k-j)%2,0, binomial(-(k-1),j) *
%o binomial(j,(n-3*k-j)/2) * (-3)^((3*j+3*k-n)/2) )) } \\ _Max Alekseyev_
%Y Cf. A000244, A076264 (=T(n,0)), A119852 (=T(n,1)), A027741.
%K nonn,tabf
%O 0,2
%A _Emeric Deutsch_, May 26 2006
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