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A129179 Triangle read by rows: T(n, k) is the number of Schroeder paths of semilength n such that the area between the x-axis and the path is k (n >= 0; 0 <= k <= n^2).

%I A129179 #47 Oct 16 2024 21:52:13
%S A129179 1,1,1,1,2,1,1,1,1,3,3,3,4,3,2,1,1,1,1,4,6,7,10,11,10,9,8,7,5,4,3,2,1,
%T A129179 1,1,1,5,10,14,21,28,31,33,34,34,31,27,25,22,17,14,13,10,7,5,4,3,2,1,
%U A129179 1,1,1,6,15,25,40,60,77,92,106,117,122,121,120,116,107,98,91,82,71,62,54,45
%N A129179 Triangle read by rows: T(n, k) is the number of Schroeder paths of semilength n such that the area between the x-axis and the path is k (n >= 0; 0 <= k <= n^2).
%C A129179 A Schroeder path of semilength n 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 A129179 Row n has 1+n^2 terms.
%C A129179 Row sums are the large Schroeder numbers (A006318).
%H A129179 Alois P. Heinz, <a href="/A129179/b129179.txt">Rows n = 0..32, flattened</a>
%F A129179 G.f.: G(t,z) satisfies G(t,z) = 1 + z*G(t,z) + t*z*G(t,t^2*z)*G(t,z).
%F A129179 Sum_{k>=0} k*T(n,k) = A129180(n).
%F A129179 From _Peter Bala_, Aug 02 2019: (Start)
%F A129179 O.g.f. as a continued fraction: (t marks the area and z marks the semilength of the path)
%F A129179 G(t,z) = 1/(1 - z - t*z/(1 - t^2*z - t^3*z/(1 - t^4*z - t^5*z/(1 - t^6*z - (...) )))) = 1 + z*(1 + t) + z^2*(1 + 2*t + t^2 + t^3 + t^4) + ....
%F A129179 G(t,z) = 1/(1 - (1 + t)*z/(1 - t^3*z/(1 - (t^2 + t^5)*z/(1 - t^7*z/(1 - (t^4 + t^9)*z/(1 - t^11*z/( (...) ))))))).
%F A129179 O.g.f. as a ratio of q-series: N(t,z)/D(t,z), where N(t,z) = Sum_{n >= 0} (-1)^n*t^(2*n^2+n)*z^n/( (Product_{k = 1..n} 1 - t^(2*k)) * (Product_{k = 1..n+1} 1 - t^(2*k-2)*z) ) and D(t,z) = Sum_{n >= 0} (-1)^n*t^(2*n^2-n)*z^n/( (Product_{k = 1..n} 1 - t^(2*k)) * (Product_{k = 1..n} 1 - t^(2*k-2)*z) ). (End)
%F A129179 Conjecture: T(n, k) = [z^k] R(n, 0) for n >= 0, k >= 0 where R(n, q) = Sum_{j=0..q + (q mod 2) + 1} z^j*R(n-1, j) for n > 0, q >= 0 with R(0, q) = 1 for q >= 0. - _Mikhail Kurkov_, Aug 03 2023
%F A129179 Conjecture above can be simplified as follows: T(n, k) = [z^k] P(n+1, n+1) for n >= 0, k >= 0 where P(n, k) = P(n-1, k) + z^(n-k)*P(n + ((n-k) mod 2), k-1) for 0 < k <= n with P(n, k) = 0 for k > n, P(n, 0) = 1 for n >= 0. - _Mikhail Kurkov_, Oct 02 2024
%e A129179 T(3,5) = 3 because we have UDUUDD, UUDDUD and UHHD.
%e A129179 Triangle starts:
%e A129179   1;
%e A129179   1,1;
%e A129179   1,2,1,1,1;
%e A129179   1,3,3,3,4,3,2,1,1,1;
%e A129179   1,4,6,7,10,11,10,9,8,7,5,4,3,2,1,1,1;
%p A129179 G:=1/(1-z-t*z*g[1]): for i from 1 to 11 do g[i]:=1/(1-t^(2*i)*z-t^(2*i+1)*z*g[i+1]) od: g[12]:=0: Gser:=simplify(series(G,z=0,13)): for n from 0 to 11 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 6 do seq(coeff(P[n],t,j),j=0..n^2) od; # yields sequence in triangular form
%p A129179 # second Maple program:
%p A129179 b:= proc(x, y) option remember; `if`(y>x or y<0, 0,
%p A129179       `if`(x=0, 1, expand(b(x-1, y-1)*z^(y-1/2)
%p A129179       +b(x-2, y)*z^(2*y) +b(x-1, y+1)*z^(y+1/2))))
%p A129179     end:
%p A129179 T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(2*n, 0)):
%p A129179 seq(T(n), n=0..7);  # _Alois P. Heinz_, May 27 2015
%t A129179 b[x_, y_] := b[x, y] = If[y>x || y<0, 0, If[x==0, 1, Expand[b[x-1, y-1]*z^(y-1/2)  + b[x-2, y]*z^(2*y) + b[x-1, y+1]*z^(y+1/2)]]]; T[n_] := Function[{p}, Table[ Coefficient[p, z, i], {i, 0, Exponent[p, z]}]][b[2*n, 0]]; Table[T[n], {n, 0, 7}] // Flatten (* _Jean-François Alcover_, Jun 29 2015, after _Alois P. Heinz_ *)
%Y A129179 Cf. A006318 (row sums), A129180, A326676.
%K A129179 nonn,tabf
%O A129179 0,5
%A A129179 _Emeric Deutsch_, Apr 08 2007