%I #17 Aug 02 2019 08:48:44
%S 1,0,1,0,1,1,0,0,2,1,0,0,1,3,1,0,0,1,3,4,1,0,0,1,3,6,5,1,0,0,0,4,7,10,
%T 6,1,0,0,0,3,10,14,15,7,1,0,0,0,2,11,21,25,21,8,1,0,0,0,1,10,28,40,41,
%U 28,9,1,0,0,0,1,9,31,60,71,63,36,10,1
%N Triangular array: T(n,k) equals the number of n triangle stacks of large Schröder type with k down-triangles in the bottom row of the stack.
%C We define two types of plane triangles of unit area - up-triangles with vertices at the lattice points (x, y), (x+1, y+1) and (x+2, y) and down-triangles with vertices at the lattice points (x, y), (x-1, y+1) and (x+1, y+1).
%C To construct a triangle stack of large Schröder type we start with a horizontal row of k contiguous down-triangles forming the base row of the stack. Subsequent rows of the stack are formed by placing up-triangles on some, all or none of the down-triangles of the previous row. In the spaces between pairs of adjacent up-triangles further down-triangles may be placed. For an example, see the illustration in the Links section. There is an obvious bijective correspondence between triangle stacks of large Schröder type with a base of k down-triangles and large Schröder paths of semilength k. For another version of this array see A129179.
%C For triangle stacks of small Schröder type, where the base row consists of contiguous up-triangles, see A224704.
%H P. Bala, <a href="/A326676/a326676.png">Example of triangle stack of large Schröder type</a>
%H P. Bala, <a href="/A326676/a326676.pdf">Triangle Stacks of large Schröder type</a>
%F O.g.f. as a continued fraction: (q marks the area of the stack and b marks down-triangles in the base of the stack)
%F A(q,b) = 1/(1 - q*b - q^2*b/(1 - q^3*b - q^4*b/(1 - q^5*b - q^6*b/( (...) )))) = 1 + b*q + (b + b^2)*q^2 + (2*b^2 + b^3)*q^3 + (b^2 + 3*b^3 + b^4)*q^4 + ....
%F A(q,b) = 1/(1 - (q + q^2)*b/(1 - q^4*b/(1 - (q^3 + q^6)*b/(1 - q^8*b/(1 - (q^5 + q^10)*b/(1 - q^12*b/( (...) ))))))).
%F O.g.f. as a ratio of q-series: N(q,b)/D(q,b), where N(q,b) = Sum_{n >= 0} (-1)^n*q^(2*n^2+2*n)*b^n/( (Product_{k = 1..n} 1 - q^(2*k)) * (Product_{k = 1..n+1} 1 - q^(2*k-1)*b) ) and D(q,b) = Sum_{n >= 0} (-1)^n*q^(2*n^2)*b^n/( (Product_{k = 1..n} 1 - q^(2*k)) * (Product_{k = 1..n} 1 - q^(2*k-1)*b) ).
%e Triangle begins
%e n\k 0 1 2 3 4 5 6 7 8 9 10
%e - - - - - - - - - - - - - - - - - - - - - - -
%e 0 | 1
%e 1 | 0 1
%e 2 | 0 1 1
%e 3 | 0 0 2 1
%e 4 | 0 0 1 3 1
%e 5 | 0 0 1 3 4 1
%e 6 | 0 0 1 3 6 5 1
%e 7 | 0 0 0 4 7 10 6 1
%e 8 | 0 0 0 3 10 14 15 7 1
%e 9 | 0 0 0 2 11 21 25 21 8 1
%e 10 | 0 0 0 1 10 28 40 41 28 9 1
%e ...
%Y Row sums A088352. Column sums A006318. Cf. A047998, A129179, A224704.
%K nonn,easy,tabl
%O 0,9
%A _Peter Bala_, Jul 17 2019