Search: id:A120987
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%I A120987
%S A120987 1,0,3,0,3,6,0,1,16,10,0,0,15,51,15,0,0,6,90,126,21,0,0,1,77,357,266,28,
%T A120987 0,0,0,36,504,1107,504,36,0,0,0,9,414,2304,2907,882,45,0,0,0,1,210,2850,
%U A120987 8350,6765,1452,55,0,0,0,0,66,2277,14355,25653,14355,2277,66,0,0,0,0,12
%N A120987 Triangle read by rows: T(n,k) is the number of ternary words of length 
               n with k strictly increasing runs (0<=k<=n; for example, the ternary 
               word 2|01|12|02|1|1|012|2 has 8 strictly increasing runs).
%C A120987 Sum of entries in row n is 3^n (A000244). Sum of entries in column k 
               is A099464(k+1) (a trisection of the tribonacci numbers). Row n contains 
               1+floor(2n/3) nonzero terms. T(n,n)=(n+1)(n+2)/2 (the triangular 
               numbers (A000217). Sum(k*T(n,k),k=0..n)=(2n+1)*3^(n-1)=3*A081038(n-1) 
               for n>=1. T(n,k)=A120987(n,n-k).
%F A120987 T(n,k)=trinom(n+1,3n-3k+2)=trinom(n+1,3k-n) (conjecture). G.f.=1/[1-3tz-3t(1-t)z^2-t(1-t)^2*z^3].
%F A120987 Can anyone prove the conjecture (either from the g.f. or combinatorially 
               from the definition)?
%F A120987 Comment from Giuliano Cabrele (giulianocabrele(AT)tin.it), Mar 02 2008: 
               (Start) The conjecture is compatible with the g.f., which can be 
               rewritten as (1-t)/(1-t[1+(1-t)z]^3) and expanded to give T(n,k) 
               = sum{j=0..k, (-1)^(k-j)*C(3j, n)*C(n+1, k-j)} = sum{j=0..k, (-1)^j*C(n+1,
               j)*C(3k-3j,n )} = trinomial(n+1,3k-n) = A027907(n+1,3k-n).
%F A120987 Also (1-t)/(1-t[1+(1-t)z]^2) equals the G.f. for the case of binary words, 
               A119900, where sum{j=0..k, (-1)^(k-j)*C(2j,n)*C(n+1,k-j)} = C(n+1,
               2k-n). Changing the exponent to 1 gives 1/(1-zt), the G.f. for the 
               case of unary words, the expansion coefficients of which can be written 
               as kronecker delta(k-n)^(n+1) = sum{j=0..k, (-1)^(k-j)*C(j, n)*C(n+1,
               k-j)}.
%F A120987 So the conjecture shifts to that the g.f.=(1-t)/(1-t[1+(1-t)z]^m) and 
               coefficients T(m,n,k)=sum{j=0..k, (-1)^(k-j)*C(mj,n)*C(n+1, k-j)} 
               may apply to the general case of m-ary words. (End)
%e A120987 T(5,2)=6 because we have 012|01, 012|02, 012|12, 01|012, 02|012 and 12|012 
               (the runs are separated by |).
%e A120987 Triangle starts:
%e A120987 1;
%e A120987 0,3;
%e A120987 0,3,6;
%e A120987 0,1,16,10;
%e A120987 0,0,15,51,15;
%e A120987 0,0,6,90,126,21;
%p A120987 G:=1/(1-3*t*z-3*t*(1-t)*z^2-t*(1-t)^2*z^3): Gser:=simplify(series(G,z=0,
               33)): P[0]:=1: for n from 1 to 13 do P[n]:=sort(coeff(Gser,z^n)) 
               od: for n from 0 to 12 do seq(coeff(P[n],t,j),j=0..n) od; # yields 
               sequence in triangular form
%Y A120987 Cf. A000244, A099464, A081038, A120987, A119900.
%Y A120987 Sequence in context: A138188 A014715 A131656 this_sequence A011076 A010599 
               A038517
%Y A120987 Adjacent sequences: A120984 A120985 A120986 this_sequence A120988 A120989 
               A120990
%K A120987 nonn,tabl
%O A120987 0,3
%A A120987 Emeric Deutsch (deutsch(AT)duke.poly.edu), Jul 23 2006

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