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Triangle read by rows: T(n,k) is the number of Grand Dyck paths of semilength n and having k returns to the x-axis.
2

%I #41 Nov 01 2021 12:23:27

%S 2,2,4,4,8,8,10,20,24,16,28,56,72,64,32,84,168,224,224,160,64,264,528,

%T 720,768,640,384,128,858,1716,2376,2640,2400,1728,896,256,2860,5720,

%U 8008,9152,8800,7040,4480,2048,512,9724,19448,27456,32032,32032,27456,19712,11264,4608,1024

%N Triangle read by rows: T(n,k) is the number of Grand Dyck paths of semilength n and having k returns to the x-axis.

%C A Grand Dyck path of semilength n is a path in the half-plane x >= 0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1).

%C Triangle T(n,k), 1 <= k <= n, read by rows, given by [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] DELTA [2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - _Philippe Deléham_, Jun 29 2005

%C T(n,k) is also equal to the number of grand Dyck paths of semilength n having k symmetric vertices. A symmetric vertex is a vertex in the first half of the path (not including the midpoint) that is a mirror image of a vertex in the second half, when with respect to the reflection along the vertical line through the midpoint of the path. - _Sergi Elizalde_, Feb 12 2021

%H Sergi Elizalde, <a href="https://arxiv.org/abs/2002.12874">The degree of symmetry of lattice paths</a>, arXiv:2002.12874 [math.CO], 2020.

%H Sergi Elizalde, <a href="https://www.mat.univie.ac.at/~slc/wpapers/FPSAC2020/26.html">Measuring symmetry in lattice paths and partitions</a>, Sem. Lothar. Combin. 84B.26, 12 pp (2020).

%F T(n,1) = 2*A000108(n-1).

%F T(n,n) = 2^n.

%F T(n,k) = k * 2^k * binomial(2*n-k,n)/(2*n-k) (1 <= k <= n).

%F G.f.: 1/(1-2*t*z*C), where C = (1-sqrt(1-4*z))/(2*z) is the Catalan function.

%F T(n,k) = 2^k * A106566(n,k). - _Philippe Deléham_, Jun 29 2005

%e T(2,2)=4 because we have u(d)u(d), u(d)d(u), d(u)d(u) and d(u)u(d) (return steps to x-axis shown between parentheses).

%e Triangle begins:

%e 2;

%e 2, 4;

%e 4, 8, 8;

%e 10, 20, 24, 16;

%e 28, 56, 72, 64, 32;

%p T:= (n,k)-> 2^k*k*binomial(2*n-k,n)/(2*n-k): for n from 1 to 10 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form

%t nn=10;c=(1-(1-4x)^(1/2))/(2x);f[list_]:=Select[list,#>0&];Map[f,Drop[CoefficientList[Series[1/(1-2y x c),{x,0,nn}],{x,y}],1]]//Flatten (* _Geoffrey Critzer_, Jan 30 2012 *)

%Y Cf. A000984 (row sums), A000108.

%K nonn,tabl

%O 1,1

%A _Emeric Deutsch_, Jun 23 2005