A353279 Triangle read by rows, a Narayana related triangle whose rows are refinements of four times the Catalan numbers (for n >= 3).

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 5, 8, 5, 1, 0, 1, 8, 19, 19, 8, 1, 0, 1, 12, 41, 60, 41, 12, 1, 0, 1, 17, 81, 165, 165, 81, 17, 1, 0, 1, 23, 148, 406, 560, 406, 148, 23, 1, 0, 1, 30, 253, 910, 1666, 1666, 910, 253, 30, 1

Offset: 0

Author

Peter Luschny, Apr 29 2022

Keywords

• nonn
• tabl

Comments

This is the third term of a sequence of generalized Narayana triangles (respectively Narayana polynomials). See A090181 for the classical case and A352687 for a discussion of the case k = 2. Many of the relations given there can be directly transferred to the present case. Here we emphasize the recurrence for the general case (see the formula section).

Examples

			Triangle starts:
[0] 1;
[1] 0, 1;
[2] 0, 1,  1;
[3] 0, 1,  2,   1
[4] 0, 1,  3,   3,   1
[5] 0, 1,  5,   8,   5,   1
[6] 0, 1,  8,  19,  19,   8,   1
[7] 0, 1, 12,  41,  60,  41,  12,   1
[8] 0, 1, 17,  81, 165, 165,  81,  17,  1
[9] 0, 1, 23, 148, 406, 560, 406, 148, 23, 1
		

Links

• Peter Luschny, Illustration of the polynomials.

Crossrefs

Cf. A090181 (and A001263), A352687, A097805, A000108.

Programs

Maple

Q := proc(n, k) option remember; local A, B, j;
if n <= k then return [seq(binomial(n-1, j-1), j = 0..n)] fi; # A097805
A := [op(Q(n - 2, k)), 0, 0]; B := [op(Q(n - 1, k)), 1];
for j from n by -1 to 3 do
    B[j] := ((B[j] + B[j-1])*(2*(n - k) + 1)
           - (A[j] - 2*A[j-1] + A[j-2])*(n - k - 1)) / (n - k + 2);
od: B end:
Trow := n -> Q(n, 3): for n from 0 to 9 do print(Trow(n)) od:

Mathematica

Q[n_, k_] := Q[n, k] = Module[{A, B, j},
If[n <= k, Return[Table[Binomial[n-1, j-1], {j, 0, n}]]];
A = Join[Q[n-2, k], {0, 0}]; B = Join[Q[n-1, k], {1}];
For[j = n, j >= 3, j--,
   B[[j]] = ((B[[j]] + B[[j-1]])*(2*(n-k)+1)-
   (A[[j]]-2*A[[j-1]]+A[[j-2]])*(n-k-1))/(n-k+2)];
B];
Trow[n_] := Q[n, 3];
Table[Trow[n], {n, 0, 9}] // Flatten (* Jean-François Alcover, Jul 07 2022, translated from Maple code *)

Python

from functools import cache
from math import comb
def comp(n, k):  # compositions A097805
    return comb(n-1, k-1) if k != 0 else k**n
@cache
def Trow(n, k):
    if n <= k:
        return [comp(n, j) for j in range(n + 1)]
    A = Trow(n - 2, k) + [0, 0]
    B = Trow(n - 1, k) + [1]
    for j in range(n - 1, 1, -1):
        B[j] = ((B[j] + B[j - 1]) * (2 * (n - k) + 1)
              - (A[j] - 2 * A[j - 1] + A[j - 2]) * (n - k - 1)) // (n - k + 2)
    return B
for n in range(10): print(Trow(n, 3)) # k=1 -> A090181, k=2 -> A352687

Formula

Define Q(n, k) recursively as [A097805(n, k) for k = 0..n] if n <= k, and otherwise Q(n, k) = [(B(j) + B(j-1))*(2*(n - k) + 1) - (A(j) - 2*A(j-1) + A(j-2))*(n - k - 1)) / (n - k + 2), for j from n by -1 down to 3], where A(n) = Q(n - 2, k) '+' [0, 0] and B(n) = Q(n - 1, k) '+' [1]. a '+' b denotes the concatenation of the lists a and b. Then T(n) = Q(n, 3) is the n-th row of this triangle and the row sum equals 4*CatalanNumber(n - 2) if n >= 3.

Q(n, 1) are the rows of the Narayana triangle A090181 and Q(n, 2) the rows of A352687. It can be shown that Q(n, k)(m) >= Q(n, k + 1)(m) for k >= 1; thus A090181(n, k) >= A352687(n, k) >= T(n, k) >= Q(n, 4)(k) >= ... is an infinite weakly descending sequence for all terms of the sequence of triangles Q(n, k).