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 A107131 A Motzkin related triangle. 11
 1, 0, 1, 0, 1, 1, 0, 0, 3, 1, 0, 0, 2, 6, 1, 0, 0, 0, 10, 10, 1, 0, 0, 0, 5, 30, 15, 1, 0, 0, 0, 0, 35, 70, 21, 1, 0, 0, 0, 0, 14, 140, 140, 28, 1, 0, 0, 0, 0, 0, 126, 420, 252, 36, 1, 0, 0, 0, 0, 0, 42, 630, 1050, 420, 45, 1, 0, 0, 0, 0, 0, 0, 462, 2310, 2310, 660, 55, 1 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,9 COMMENTS Row sums are Motzkin numbers A001006. Diagonal sums are A025250(n+1). Inverse binomial transform of Narayana number triangle A001263. - Paul Barry, May 15 2005 T(n,k)=number of Motzkin paths of length n with k steps U=(1,1) or H=(1,0). Example: T(3,2)=3 because we have HUD, UDH and UHD (here D=(1,-1)). T(n,k) = number of bushes with n+1 edges and k+1 leaves (a bush is an ordered tree in which the outdegree of each nonroot node is at least two). - Emeric Deutsch, May 29 2005 Row reverse of A055151. - Peter Bala, May 07 2012 Rows of A088617 are shifted columns of A107131, whose reversed rows are the Motzkin polynomials of A055151, which give the row polynomials (mod signs) of the o.g.f. that is the compositional inverse for an o.g.f. of the Fibonacci polynomials of A011973. The diagonals of A055151 give the rows of A088671, and the antidiagonals (top to bottom) of A088617 give the rows of A107131. The diagonals of A107131 give the columns of A055151. From the relation between A088617 and A107131, the o.g.f. of this entry is (1 - t*x - sqrt((1-t*x)^2 - 4*t*x^2))/(2*t*x^2). - Tom Copeland, Jan 21 2016 LINKS Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened). Marilena Barnabei, Flavio Bonetti, Niccolò Castronuovo, and Matteo Silimbani, Consecutive patterns in restricted permutations and involutions, arXiv:1902.02213 [math.CO], 2019. Paul Barry, On the duals of the Fibonacci and Catalan-Fibonacci polynomials and Motzkin paths, arXiv:2101.10218 [math.CO], 2021. Paul Barry and A. Hennessy, A Note on Narayana Triangles and Related Polynomials, Riordan Arrays, and MIMO Capacity Calculations , J. Int. Seq. 14 (2011) # 11.3.8 FORMULA Number triangle T(n, k) = binomial(k+1, n-k+1)*binomial(n, k)/(k+1). T(n, k) = Sum_{j=0..n} (-1)^(n-j)C(n, j)*C(j+1, k)*C(j+1, k+1)/(j+1). - Paul Barry, May 15 2005 G.f.: G = G(t, z) satisfies G = 1 + t*z*G + t*z^2*G^2. - Emeric Deutsch, May 29 2005 Coefficient array for the polynomials x^n*Hypergeometric2F1((1-n)/2, -n/2; 2; 4/x). - Paul Barry, Oct 04 2008 From Paul Barry, Jan 12 2009: (Start) G.f.: 1/(1-xy(1+x)/(1-x^2*y/(1-xy(1+x)/(1-x^2y/(1-xy(1+x).... (continued fraction). T(n,k) = C(n, 2n-2k)*A000108(n-k). (End) EXAMPLE Triangle begins   1;   0,  1;   0,  1,  1;   0,  0,  3,  1;   0,  0,  2,  6,  1;   0,  0,  0, 10, 10,   1;   0,  0,  0,  5, 30,  15,   1;   0,  0,  0,  0, 35,  70,  21,   1;   0,  0,  0,  0, 14, 140, 140,  28,  1;   0,  0,  0,  0,  0, 126, 420, 252, 36, 1; MAPLE egf := exp(t*x)*hypergeom([], , t*x^2); s := n -> n!*coeff(series(egf, x, n+2), x, n); seq(print(seq(coeff(s(n), t, j), j=0..n)), n=0..9); # Peter Luschny, Oct 29 2014 MATHEMATICA T[n_, k_] := Binomial[k+1, n-k+1] Binomial[n, k]/(k+1); Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* Jean-François Alcover, Jun 19 2018 *) PROG (Magma) [Binomial(n, 2*(n-k))*Catalan(n-k): k in [0..n], n in [0..13]]; // G. C. Greubel, May 22 2022 (SageMath) flatten([[binomial(n, 2*(n-k))*catalan_number(n-k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 22 2022 CROSSREFS Cf. A001006 (row sums), A025250 (diag. sums), A055151 (row reverse). Cf. A001263, A011973, A025250, A055151, A088617, A088671, A107131. Sequence in context: A229143 A330018 A065413 * A027200 A035654 A170846 Adjacent sequences:  A107128 A107129 A107130 * A107132 A107133 A107134 KEYWORD easy,nonn,tabl AUTHOR Paul Barry, May 12 2005 STATUS approved

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Last modified July 6 12:26 EDT 2022. Contains 355110 sequences. (Running on oeis4.)