A291889 -id:A291889 - cp's OEIS frontend

A291883 Number T(n,k) of symmetrically unique Dyck paths of semilength n and height k; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 5, 3, 1, 0, 1, 9, 11, 4, 1, 0, 1, 19, 31, 19, 5, 1, 0, 1, 35, 91, 69, 29, 6, 1, 0, 1, 71, 250, 252, 127, 41, 7, 1, 0, 1, 135, 690, 855, 540, 209, 55, 8, 1, 0, 1, 271, 1863, 2867, 2117, 1005, 319, 71, 9, 1, 0, 1, 527, 5017, 9339, 8063, 4411, 1705, 461, 89, 10, 1

Offset: 0

Alois P. Heinz, Sep 05 2017

			: T(4,2) = 5:       /\      /\        /\/\    /\  /\    /\/\/\
:              /\/\/  \  /\/  \/\  /\/    \  /  \/  \  /      \
:
Triangle T(n,k) begins:
  1;
  0, 1;
  0, 1,   1;
  0, 1,   2,   1;
  0, 1,   5,   3,   1;
  0, 1,   9,  11,   4,   1;
  0, 1,  19,  31,  19,   5,   1;
  0, 1,  35,  91,  69,  29,   6,  1;
  0, 1,  71, 250, 252, 127,  41,  7, 1;
  0, 1, 135, 690, 855, 540, 209, 55, 8, 1;
  ...

Alois P. Heinz, Rows n = 0..140, flattened

Columns k=0-10 give: A000007, A057427, A056326, A291887, A291888, A291889, A291890, A291891, A291892, A291893, A291894.
Main and first two lower diagonals give A000012, A001477, A028387(n-1) for n>0.
Row sums give A007123(n+1).
T(2n,n) give A291885.
Cf. A080936, A132890, A291886.

b:= proc(x, y, k) option remember; `if`(x=0, z^k, `if`(y0, b(x-1, y-1, k), 0))
    end:
g:= proc(x, y, k) option remember; `if`(x=0, z^k, `if`(y>0,
      g(x-2, y-1, k), 0)+ g(x-2, y+1, max(y+1, k)))
    end:
T:= n-> (p-> seq(coeff(p, z, i)/2, i=0..n))(b(2*n, 0$2)+g(2*n, 0$2)):
seq(T(n), n=0..14);

b[x_, y_, k_] := b[x, y, k] = If[x == 0, z^k, If[y < x - 1, b[x - 1, y + 1, Max[y + 1, k]], 0] + If[y > 0, b[x - 1, y - 1, k], 0]];
g[x_, y_, k_] := g[x, y, k] = If[x == 0, z^k, If[y > 0, g[x - 2, y - 1, k], 0] + g[x - 2, y + 1, Max[y + 1, k]]];
T[n_] := Function[p, Table[Coefficient[p, z, i]/2, {i, 0, n}]][b[2*n, 0, 0] + g[2*n, 0, 0]];
Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Jun 03 2018, from Maple *)

from sympy.core.cache import cacheit
from sympy import Poly, Symbol, flatten
z=Symbol('z')
@cacheit
def b(x, y, k): return z**k if x==0 else (b(x - 1, y + 1, max(y + 1, k)) if y0 else 0)
@cacheit
def g(x, y, k): return z**k if x==0 else (g(x - 2, y - 1, k) if y>0 else 0) + g(x - 2, y + 1, max(y + 1, k))
def T(n): return 1 if n==0 else [i//2 for i in Poly(b(2*n, 0, 0) + g(2*n, 0, 0)).all_coeffs()[::-1]]
print(flatten(map(T, range(15)))) # Indranil Ghosh, Sep 06 2017

T(n,k) = (A080936(n,k) + A132890(n,k))/2.

Sum_{k=1..n} k * T(n,k) = A291886(n).

cp's OEIS Frontend

A291883 Number T(n,k) of symmetrically unique Dyck paths of semilength n and height k; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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