A120907 - OEIS

%I #3 Mar 30 2012 17:36:10

%S 1,3,6,2,1,10,10,7,15,30,31,4,1,21,70,105,36,11,28,140,294,184,76,6,1,

%T 36,252,714,696,396,78,15,45,420,1554,2160,1666,566,141,8,1,55,660,

%U 3102,5808,5918,2990,995,136,19,66,990,5775,13992,18348,12746,5615,1280,226

%N Triangle read by rows: T(n,k) is the number of ternary words of length n on {0,1,2} having sum of the lengths of the drops equal to k (n>=0, k>=0). The drops of a ternary word on {0,1,2} are the subwords 10,20 and 21, their lengths being the differences 1, 2 and 1, respectively.

%C Row n has 2*floor(n/2)+1 terms (i.e. each of the rows 2n and 2n+1 has 2n+1 terms). Row sums are the powers of 3 (A000244). T(n,0)=A000217(n+1) (the triangular numbers). Sum(k*T(n,k),k>=0)=4(n-1)3^(n-2)=A120908(n)=4*A027471(n).

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

%e T(4,3)=4 because we have 1020,2010,2021 and 2120.

%e Triangle starts:

%e 1;

%e 3;

%e 6,2,1;

%e 10,10,7;

%e 15,30,31,4,1;

%e 21,70,105,36,11;

%p G:=1/(1-z+t*z)/(1-2*z+z^2-t*z-t*z^2): Gser:=simplify(series(G,z=0,15)): 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..2*floor(n/2)) od; # yields sequence in triangular form

%Y Cf. A000244, A000217, A120908, A027471, A120906.

%K nonn,tabf

%O 0,2

%A _Emeric Deutsch_, Jul 15 2006