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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).
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%I #27 Nov 02 2025 03:34:25

%S 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,

%T 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,

%U 714,0,430,0,126,0,18,0,1,0,1412,0,1408,0,652,0,160,0,20,0,1,1806,0,3534,0,2490,0,938,0,198,0,22,0,1

%N 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).

%C Row sums yield A026003. T(2n,0)=A006318(n) (the large Schroeder numbers); T(2n+1,0)=0.

%H D. Baccherini, D. Merlini and R. Sprugnoli, <a href="http://pefmath.etf.rs/vol2num1/AADM-Vol2-No1-69-91.pdf">Level generating trees and proper Riordan arrays</a>, Applicable Analysis and Discrete Mathematics, 2, 2008, 69-91 (see p. 89). [From _Emeric Deutsch_, Sep 21 2008]

%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Barry1/barry411.html">The Central Coefficients of a Family of Pascal-like Triangles and Colored Lattice Paths</a>, J. Int. Seq., Vol. 22 (2019), Article 19.1.3.

%F 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.

%F 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.

%F 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

%e T(3,1)=4 because we have HU, UDU, UH and UUD.

%e Triangle begins:

%e 1;

%e 0, 1;

%e 2, 0, 1;

%e 0, 4, 0, 1;

%e 6, 0, 6, 0, 1;

%e 0, 16, 0, 8, 0, 1;

%p T:=proc(n,k) if k<n and n-k mod 2 = 0 then (2*(k+1)/(n-k))*sum(binomial((n-k)/2,j)*binomial((n+k)/2+j,(n-k-2)/2), j=0..(n-k)/2) elif k=n then 1 else 0 fi end: for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

%t T[0, 0] = 1;

%t T[n_, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + T[n-1, k+1] + T[n-2, k];

%t T[_, _] = 0;

%t Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 31 2024, after _Philippe Deléham_ *)

%Y Cf. A006318, A026003.

%K nonn,tabl

%O 0,4

%A _Emeric Deutsch_, Jul 12 2005