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 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). 4
 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, 0, 0, 0, 36, 504, 1107, 504, 36, 0, 0, 0, 9, 414, 2304, 2907, 882, 45, 0, 0, 0, 1, 210, 2850, 8350, 6765, 1452, 55, 0, 0, 0, 0, 66, 2277, 14355, 25653, 14355, 2277, 66, 0, 0, 0, 0, 12 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS 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=0..n} k*T(n,k) = (2n+1)*3^(n-1) = 3*A081038(n-1) for n >= 1. T(n,k) = A120987(n,n-k). REFERENCES R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 24, p. 154. LINKS G. C. Greubel, Table of n, a(n) for n = 0..11475 Giuliano Cabrele, Words Partitioned according to Number of Strictly Increasing Runs MathPages, Balls In Bins With Limited Capacity. FORMULA T(n,k) = trinomial(n+1,3n-3k+2) = trinomial(n+1,3k-n) (conjecture). G.f.: 1/(1-3tz-3t(1-t)z^2-t(1-t)^2*z^3). Can anyone prove the conjecture (either from the g.f. or combinatorially from the definition)? From Giuliano Cabrele, 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). 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). 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) Sum_{k=0..n} T(n,k) *(-1)^(n-k) = A157241(n+1). - Philippe Deléham, Oct 25 2011 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 EXAMPLE 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 |). Triangle starts:   1;   0,   3;   0,   3,   6;   0,   1,  16,  10;   0,   0,  15,  51,  15;   0,   0,   6,  90, 126,  21; MAPLE 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:=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 MATHEMATICA 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 *) PROG (MuPAD) // binomial c. defined as in linked document Cb:=(x, m)->_if(0<=m and is(m in Z_), binomial(x, m), 0): // closed formula derived and proved in the linked document // Qsc(r, q, m) with r=2 T(n, k):=(n, k)->_plus((-1)^(k-j)*Cb(n+1, k-j)*Cb(3*j, n)\$j=0..k): // Giuliano Cabrele, Dec 11 2015 CROSSREFS Cf. A000244, A008287, A081038, A099464, A119900, A120987, A265644. Nb(s,2,q) = A027907(q,s). - Giuliano Cabrele, Dec 11 2015 Sequence in context: A194136 A194480 A194485 * A281293 A258108 A282610 Adjacent sequences:  A120984 A120985 A120986 * A120988 A120989 A120990 KEYWORD nonn,tabl AUTHOR Emeric Deutsch, Jul 23 2006 STATUS approved

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Last modified December 5 10:45 EST 2019. Contains 329751 sequences. (Running on oeis4.)