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 A110109 Triangle read by rows: T(n,k) (0 <= k <= n) is the number of left factors of Schroeder paths, going from (0,0) to (n,k) (a Schroeder path of length 2n is a lattice path from (0,0) to (2n,0) consisting of U=(1,1), D=(1,-1) and H=(2,0) steps and never going below the x-axis). 0
 1, 0, 1, 2, 0, 1, 0, 4, 0, 1, 6, 0, 6, 0, 1, 0, 16, 0, 8, 0, 1, 22, 0, 30, 0, 10, 0, 1, 0, 68, 0, 48, 0, 12, 0, 1, 90, 0, 146, 0, 70, 0, 14, 0, 1, 0, 304, 0, 264, 0, 96, 0, 16, 0, 1, 394, 0, 714, 0, 430, 0, 126, 0, 18, 0, 1, 0, 1412, 0, 1408, 0, 652, 0, 160, 0, 20, 0, 1, 1806, 0, 3534, 0 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS Row sums yield A026003. T(2n,0)=A006318(n) (the large Schroeder numbers); T(2n+1,0)=0. LINKS D. Baccherini, D. Merlini and R. Sprugnoli, Level generating trees and proper Riordan arrays, Applicable Analysis and Discrete Mathematics, 2, 2008, 69-91 (see p. 89). [From Emeric Deutsch, Sep 21 2008] Paul Barry, The Central Coefficients of a Family of Pascal-like Triangles and Colored Lattice Paths, J. Int. Seq., Vol. 22 (2019), Article 19.1.3. FORMULA T(n,k) = (2(k+1)/(n-k))*Sum_{j=0..(n-k)/2} binomial((n-k)/2, j)*binomial((n+k)/2+j, (n-k-2)/2) if k < n and n-k is even; T(n,n)=1; T(n,k)=0 if n-k is odd. G.f. = R(z^2)/(1-tzR(z^2)), where R = 1 + zR + zR^2 = (1-z-sqrt(1-6z+z^2))/(2z) is the g.f. of the large Schroeder numbers. T(n,k) = T(n-1,k-1) + T(n-1,k+1) + T(n-2,k), T(0,0)=1. - Philippe Deléham, Nov 18 2009 EXAMPLE T(3,1)=4 because we have HU, UDU, UH and UUD. Triangle begins:   1;   0,  1;   2,  0,  1;   0,  4,  0,  1;   6,  0,  6,  0,  1;   0, 16,  0,  8,  0,  1; MAPLE T:=proc(n, k) if k

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Last modified May 31 02:51 EDT 2020. Contains 334747 sequences. (Running on oeis4.)