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A078391 Triangle read by rows: T(n,k) = Catalan(k)*Catalan(n-k).

%I A078391 #68 Aug 26 2024 12:12:40
%S A078391 1,1,1,2,1,2,5,2,2,5,14,5,4,5,14,42,14,10,10,14,42,132,42,28,25,28,42,
%T A078391 132,429,132,84,70,70,84,132,429,1430,429,264,210,196,210,264,429,
%U A078391 1430,4862,1430,858,660,588,588,660,858,1430,4862,16796,4862,2860,2145,1848
%N A078391 Triangle read by rows: T(n,k) = Catalan(k)*Catalan(n-k).
%C A078391 T(n,k) is the number of Dyck paths of semilength n+1 whose first return point to the axis have abscissa 2k+2. - _Emeric Deutsch_, Mar 01 2004
%C A078391 With offset = 1, T(n,k) is the number of binary trees with n internal nodes that have exactly k internal nodes in the left subtree, n>=1, 0<=k<=n-1. - _Geoffrey Critzer_, Feb 24 2013
%C A078391 T(n-1,k) is also the number of tilings of a triangular shape T_n (row k has length k for k=1, 2, ..., n) with n rectangular tiles (including squares) with contain a rectangular tile (n-k,k+1) for k = 0, 1, ... ,n-1, n >= 1. Let the number of tilings of T_n with n rectangular tiles (including squares) be A(n) and take A(0) = 1. Decompose these n-tilings of T_n into n disjoint and exhaustive classes C(n, k), for k = 0, 1, ..., n-1, n >= 1. In class C(n, k) one takes a fixed rectangular tile (n-k,k+1) leaving triangles T_(n-1-k) and T_k to be tiled (but for the k=0 class T_0 is not shown). Then A(n) = A(n-1)*A(0) + A(n-2)*A(1) + ... + A(0)*A(n-1) = sum(A(n-1-k)*A(k), k=0..n-1), n >= 1, with A(0)=1. But this is the recurrence for the Catalan numbers, hence A(n) = C(n). See the link with examples n = 1..7. - _Wolfdieter Lang_ and _Kival Ngaokrajang_, Dec 27 2014
%C A078391 T(n,k) is the number of triangulations of an (n+3)-polygon using a (0,1,k+2)-triangle. - _Yuchun Ji_, Jan 21 2021
%C A078391 Alternating sum of even index 2n is A079489(n). - _F. Chapoton_, Aug 26 2024
%D A078391 R. Sedgewick and P. Flajolet, Analysis of Algorithms, Addison Wesley, first edition, page 225.
%H A078391 FindStat - Combinatorial Statistic Finder, <a href="http://www.findstat.org/StatisticsDatabase/St000026">The position of the first return of a Dyck path.</a>, <a href="http://www.findstat.org/StatisticsDatabase/St000051">The size of the left subtree.</a>
%H A078391 Kival Ngaokrajang, <a href="/A078391/a078391_4.pdf">Illustration of C(n,k), for n = 1..7, k = 0..n-1</a>
%H A078391 Richard P. Stanley, <a href="http://www-math.mit.edu/~rstan/ec/catadd.pdf"> Catalan Addendum (k8)</a>
%F A078391 G.f.: C(z)*C(tz), where C(z) = (1-sqrt(1-4z))/(2z) is the g.f. of the Catalan numbers (A000108). - _Emeric Deutsch_, Mar 01 2004
%F A078391 T(n,k) = A118921(n+1,k+1)/(2*(n-k+1)). - _Philippe Deléham_, Dec 13 2006
%F A078391 When viewed as a square array, for n>0 and k>0, A(n,k) = Sum_{i=0..n-1,j=0..k-1} A[i,j]*A[n-i,k-j]. - _Gerald McGarvey_, Dec 30 2007
%e A078391 The triangle T(n,k) begins:
%e A078391 n\k     0    1    2    3    4    5    6    7    8    9    10 ...
%e A078391 0:      1
%e A078391 1:      1    1
%e A078391 2:      2    1    2
%e A078391 3:      5    2    2    5
%e A078391 4:     14    5    4    5   14
%e A078391 5:     42   14   10   10   14   42
%e A078391 6:    132   42   28   25   28   42  132
%e A078391 7:    429  132   84   70   70   84  132  429
%e A078391 8:   1430  429  264  210  196  210  264  429 1430
%e A078391 9:   4862 1430  858  660  588  588  660  858 1430 4862
%e A078391 10: 16796 4862 2860 2145 1848 1764 1848 2145 2860 4862 16796
%e A078391 ...  Reformatted - _Wolfdieter Lang_, Dec 27 2014
%e A078391 ----------------------------------------------------------------
%t A078391 nn=10;r=(1-(1-4x)^(1/2))/(2x);l=(1-(1-4x y)^(1/2))/(2x y);f[list_]:=Select[list,#>0&];Map[f,Drop[CoefficientList[Series[1+x l r,{x,0,nn}],{x,y}],1]]//Grid (* _Geoffrey Critzer_, Feb 24 2013 *)
%t A078391 Table[CatalanNumber[k]CatalanNumber[n-k],{n,0,10},{k,0,n}]//Flatten (* _Harvey P. Dale_, Nov 14 2019 *)
%Y A078391 Row sums are Catalan numbers A000108(n+1). T(2*k,k) = A001246(k), k >= 0. T(n,0) = T(n,n) = A000108(n).
%Y A078391 Cf. A067804, A079489.
%K A078391 nonn,tabl
%O A078391 0,4
%A A078391 _Henry Bottomley_, Dec 24 2002