A321396 Square array read by ascending antidiagonals, A(n, k) for n >= 0 and k >= 0, related to a class of Motzkin trees.
0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 0, 1, 0, 1, 1, 3, 2, 5, 0, 1, 0, 1, 1, 3, 2, 7, 0, 0, 1, 0, 1, 1, 3, 3, 9, 5, 14, 0, 1, 0, 1, 1, 3, 3, 9, 7, 20, 0, 0, 1, 0, 1, 1, 3, 3, 10, 9, 27, 19, 42
Offset: 0
Examples
Array begins: [0] 0, 1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, 0, 132, ... A126120 [1] 0, 1, 0, 1, 1, 2, 2, 7, 5, 20, 19, 60, 62, 202, ... A300126 [2] 0, 1, 0, 1, 1, 3, 2, 9, 7, 27, 25, 85, 86, 287, ... A321572 [3] 0, 1, 0, 1, 1, 3, 3, 9, 9, 29, 32, 93, 111, 317, ... [4] 0, 1, 0, 1, 1, 3, 3, 10, 9, 31, 34, 100, 119, 344, ... [5] 0, 1, 0, 1, 1, 3, 3, 10, 10, 31, 36, 102, 126, 352, ... [6] 0, 1, 0, 1, 1, 3, 3, 10, 10, 32, 36, 104, 128, 359, ... [7] 0, 1, 0, 1, 1, 3, 3, 10, 10, 32, 37, 104, 130, 361, ... [8] 0, 1, 0, 1, 1, 3, 3, 10, 10, 32, 37, 105, 130, 363, ... [9] 0, 1, 0, 1, 1, 3, 3, 10, 10, 32, 37, 105, 131, 363, ... Array read by ascending diagonals: [0] 0 [1] 0, 1 [2] 0, 1, 0 [3] 0, 1, 0, 1 [4] 0, 1, 0, 1, 0 [5] 0, 1, 0, 1, 1, 2 [6] 0, 1, 0, 1, 1, 2, 0 [7] 0, 1, 0, 1, 1, 3, 2, 5 [8] 0, 1, 0, 1, 1, 3, 2, 7, 0 [9] 0, 1, 0, 1, 1, 3, 3, 9, 5, 14
Links
- Olivier Bodini, Danièle Gardy, Bernhard Gittenberger, Zbigniew Gołębiewski, On the number of unary-binary tree-like structures with restrictions on the unary height, arXiv:1510.01167v1 [math.CO], 2015.
Crossrefs
Programs
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Maple
Arow := proc(n, len) local rowgf, ser; rowgf := proc(n) option remember; `if`(n = 0, (1-sqrt(1-4*z^2))/(2*z), expand((1 - sqrt(1 - 4*z^2*rowgf(n-1)))/(2*z))) end: ser := series(rowgf(n)/z^n, z, 2*(2+max(len, n))); seq(coeff(ser, z, k), k=0..len) end: seq(Arow(n, 13), n=0..9);
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Mathematica
nmax = 11; gf[-1] = 1; gf[n_] := gf[n] = (1-Sqrt[1 - 4z^2 gf[n-1]])/(2z); row[n_] := row[n] = gf[n]/z^n + O[z]^(nmax+1) // CoefficientList[#, z]&; A[n_, k_] := row[n][[k + 1]]; Table[A[n - k, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 08 2018 *)
Formula
Define a sequence of generating functions recursively gf(-1) = 1 and for n >= 0
gf(n) = (1 - sqrt(1 - 4*z^2*gf(n-1)))/(2*z).
Row n of the array has the generating function gf(n)/z^n. For fixed k column k differs only for finitely many indices from the limit value A321397(k).
Comments