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A129179
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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).
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4
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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, 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, 1, 1, 1, 6, 15, 25, 40, 60, 77, 92, 106, 117, 122, 121, 120, 116, 107, 98, 91, 82, 71, 62, 54, 45
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OFFSET
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0,5
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COMMENTS
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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.
Row n has 1+n^2 terms.
Row sums are the large Schroeder numbers (A006318).
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LINKS
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FORMULA
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G.f.: G(t,z) satisfies G(t,z) = 1 + z*G(t,z) + t*z*G(t,t^2*z)*G(t,z).
O.g.f. as a continued fraction: (t marks the area and z marks the semilength of the path)
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) + ....
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/( (...) ))))))).
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)
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
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EXAMPLE
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T(3,5) = 3 because we have UDUUDD, UUDDUD and UHHD.
Triangle starts:
1;
1,1;
1,2,1,1,1;
1,3,3,3,4,3,2,1,1,1;
1,4,6,7,10,11,10,9,8,7,5,4,3,2,1,1,1;
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MAPLE
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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
# second Maple program:
b:= proc(x, y) option remember; `if`(y>x or 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))))
end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(2*n, 0)):
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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