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%I A120987 #58 Jul 16 2019 03:24:50
%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,
%T A120987 28,0,0,0,36,504,1107,504,36,0,0,0,9,414,2304,2907,882,45,0,0,0,1,210,
%U A120987 2850,8350,6765,1452,55,0,0,0,0,66,2277,14355,25653,14355,2277,66,0,0,0,0,12
%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).
%C A120987 Sum of entries in column k is A099464(k+1) (a trisection of the tribonacci numbers).
%C A120987 Row n contains 1 + floor(2n/3) nonzero terms.
%C A120987 T(n,n) = (n+1)*(n+2)/2 (the triangular numbers (A000217)).
%C A120987 Sum_{k=0..n} k*T(n,k) = (2n+1)*3^(n-1) = 3*A081038(n-1) for n >= 1.
%C A120987 T(n,k) = A120987(n,n-k).
%D A120987 R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 24, p. 154.
%H A120987 G. C. Greubel, Table of n, a(n) for n = 0..11475
%H A120987 Giuliano Cabrele, Words Partitioned according to Number of Strictly Increasing Runs
%H A120987 MathPages, Balls In Bins With Limited Capacity.
%F A120987 T(n,k) = trinomial(n+1,3n-3k+2) = trinomial(n+1,3k-n) (conjecture).
%F A120987 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 From _Giuliano Cabrele_, Mar 02 2008: (Start)
%F A120987 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. is (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)
%F A120987 Sum_{k=0..n} T(n,k) *(-1)^(n-k) = A157241(n+1). - _Philippe Deléham_, Oct 25 2011
%F A120987 The generalized conjecture above can in fact be proved, as described in the file "Words Partitioned according to Number of Strictly Increasing Runs" linked above. - _Giuliano Cabrele_, Dec 11 2015
%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
%t A120987 Flatten[Table[Sum[(-1)^j*Binomial[n + 1, j]*Binomial[3 k - 3 j, n], {j, 0, k}], {n, 0, 10}, {k, 0, n}]] (* _G. C. Greubel_, Dec 20 2015 *)
%o A120987 (MuPAD)
%o A120987 // binomial c. defined as in linked document
%o A120987 Cb:=(x,m)->_if(0<=m and is(m in Z_), binomial(x,m), 0):
%o A120987 // closed formula derived and proved in the linked document
%o A120987 // Qsc(r,q,m) with r=2
%o A120987 T(n,k):=(n,k)->_plus((-1)^(k-j)*Cb(n+1,k-j)*Cb(3*j, n)$j=0..k):
%o A120987 // _Giuliano Cabrele_, Dec 11 2015
%Y A120987 Cf. A000244, A008287, A081038, A099464, A119900, A120987, A265644.
%Y A120987 Nb(s,2,q) = A027907(q,s). - _Giuliano Cabrele_, Dec 11 2015
%K A120987 nonn,tabl
%O A120987 0,3
%A A120987 _Emeric Deutsch_, Jul 23 2006
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