A352687 Triangle read by rows, a Narayana related triangle whose rows are refinements of twice the Catalan numbers (for n >= 2).

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 4, 4, 1, 0, 1, 7, 12, 7, 1, 0, 1, 11, 30, 30, 11, 1, 0, 1, 16, 65, 100, 65, 16, 1, 0, 1, 22, 126, 280, 280, 126, 22, 1, 0, 1, 29, 224, 686, 980, 686, 224, 29, 1, 0, 1, 37, 372, 1512, 2940, 2940, 1512, 372, 37, 1

Offset

0,9

Comments

This is the second triangle in a sequence of Narayana triangles. The first is A090181, whose n-th row is a refinement of Catalan(n), whereas here the n-th row of T is a refinement of 2*Catalan(n-1). We can show that T(n, k) <= A090181(n, k) for all n, k. The third triangle in this sequence is A353279, where also a recurrence for the general case is given.

Here we give a recurrence for the row polynomials, which correspond to the recurrence of the classical Narayana polynomials combinatorially proved by Sulanke (see link).

The polynomials have only real zeros and form a Sturm sequence. This follows from the recurrence along the lines given in the Chen et al. paper.

Some interesting sequences turn out to be the evaluation of the polynomial sequence at a fixed point (see the cross-references), for example the reversion of the Jacobsthal numbers A001045 essentially is -(-2)^n*P(n, -1/2).

The polynomials can also be represented as the difference between generalized Narayana polynomials, see the formula section.

Links

Table of n, a(n) for n=0..65.

Xi Chen, Arthur Li Bo Yang, James Jing Yu Zhao, Recurrences for Callan's Generalization of Narayana Polynomials, J. Syst. Sci. Complex (2021).

Peter Luschny, Illustration of the polynomials.

Robert A. Sulanke, The Narayana distribution, J. Statist. Plann. Inference, 2002, 101: 311-326, formula 2.

Formula

Explicit formula (additive form):

T(n, n) = 1, T(n > 0, 0) = 0 and otherwise T(n, k) = binomial(n, k)*binomial(n - 1, k - 1)/(n - k + 1) - 2*binomial(n - 1, k)*binomial(n - 1, k - 2)/(n - 1).

Multiplicative formula with the same boundary conditions:

T(n, k) = binomial(n, k)^2*(k*(2*k^2 + (n + 1)*(n - 2*k)))/(n^2*(n-1)*(n- k + 1)).

Bivariate generating function:

T(n, k) = [x^n] [y^k](1 - x + (1+y)*(1-x*(y-1) - sqrt((x*y+x-1)^2 - 4*x^2*y))/2).

Recursion based on polynomials:

T(n, k) = [x^k] (((2*n - 3)*(x + 1)*P(n - 1, x) - (n - 3)*(x - 1)^2*P(n - 2, x)) / n) with P(0, x) = 1, P(1, x) = x, and P(2, x) = x + x^2.

Recursion based on rows (see the second Python program):

T(n, k) = (((B(k) + B(k-1)) * (2*n - 3) - (A(k) - 2*A(k-1) + A(k-2))*(n-3))/n), where A(k) = T(n-2, k) and B(k) = T(n-1, k), for n >= 3.

Hypergeometric representation:

T(n, k) = [x^k] x*(x + 1)*hypergeom([1 - n, 2 - n], [2], x) for n >= 2.

Row sums:

Sum_{k=0..n} T(n, k) = (2/n)*binomial(2*(n - 1), n - 1) = A068875(n-1) for n >= 2.

A generalization of the Narayana polynomials is given by

N{n, k}(x) = Sum_{j=0..n-2*k}(((k + 1)/(n - k)) * binomial(n - k, j - 1) * binomial(n - k, j + k) * x^(j + k)).

N{n, 0}(x) are the classical Narayana polynomials A001263 and N{n, 1}(x) is a shifted version of A145596 based in (3, 2). Our polynomials are the difference P(n, x) = N{n, 0}(x) - N{n, 1}(x) for n >= 1.

Let RS(T, n) denote the row sum of the n-th row of T, then RS(T, n) - RS(A090181, n) = -4*binomial(2*n - 3, n - 3)/(n + 1) = A115143(n + 1) for n >= 3.

Example

Triangle starts:
[0] 1;
[1] 0, 1;
[2] 0, 1,  1;
[3] 0, 1,  2,   1;
[4] 0, 1,  4,   4,   1;
[5] 0, 1,  7,  12,   7,   1;
[6] 0, 1, 11,  30,  30,  11,   1;
[7] 0, 1, 16,  65, 100,  65,  16,   1;
[8] 0, 1, 22, 126, 280, 280, 126,  22,  1;
[9] 0, 1, 29, 224, 686, 980, 686, 224, 29, 1;

Maple

T := (n, k) -> if n = k then 1 elif k = 0 then 0 else
binomial(n, k)^2*(k*(2*k^2 + (n + 1)*(n - 2*k))) / (n^2*(n - 1)*(n - k + 1)) fi:
seq(seq(T(n, k), k = 0..n), n = 0..10);
# Alternative:
gf := 1 - x + (1 + y)*(1 - x*(y - 1) - sqrt((x*y + x - 1)^2 - 4*x^2*y))/2:
serx := expand(series(gf, x, 16)): coeffy := n -> coeff(serx, x, n):
seq(seq(coeff(coeffy(n), y, k), k = 0..n), n = 0..10);
# Using polynomial recurrence:
P := proc(n, x) option remember; if n < 3 then [1, x, x + x^2] [n + 1] else
((2*n - 3)*(x + 1)*P(n - 1, x) - (n - 3)*(x - 1)^2*P(n - 2, x)) / n fi end:
Trow := n -> seq(coeff(P(n, x), x, k), k = 0..n): seq(Trow(n), n = 0..10);
# Represented by generalized Narayana polynomials:
N := (n, k, x) -> add(((k+1)/(n-k))*binomial(n-k, j-1)*binomial(n-k, j+k)*x^(j+k), j=0..n-2*k): seq(print(ifelse(n=0, 1, expand(N(n, 0, x) - N(n, 1, x)))), n=0..7);

Mathematica

H[0, _] := 1; H[1, x_] := x;
H[n_, x_] := x*(x + 1)*Hypergeometric2F1[1 - n, 2 - n, 2, x];
Hrow[n_] := CoefficientList[H[n, x], x]; Table[Hrow[n], {n, 0, 9}] // TableForm

Prog

(Python)
from math import comb as binomial
def T(n, k):
    if k == n: return 1
    if k == 0: return 0
    return ((binomial(n, k)**2 * (k * (2 * k**2 + (n + 1) * (n - 2 * k))))
           // (n**2 * (n - 1) * (n - k + 1)))
def Trow(n): return [T(n, k) for k in range(n + 1)]
for n in range(10): print(Trow(n))
(Python) # The recursion with cache is (much) faster:
from functools import cache
@cache
def T_row(n):
    if n < 3: return ([1], [0, 1], [0, 1, 1])[n]
    A = T_row(n - 2) + [0, 0]
    B = T_row(n - 1) + [1]
    for k in range(n - 1, 1, -1):
        B[k] = (((B[k] + B[k - 1]) * (2 * n - 3)
               - (A[k] - 2 * A[k - 1] + A[k - 2]) * (n - 3)) // n)
    return B
for n in range(10): print(T_row(n))

Crossrefs

Cf. A090181 and A001263 (Narayana), A353279 (case 3), A000108 (Catalan), A145596, A172392 (central terms), A000124 (subdiagonal, column 2), A115143.

Essentially twice the Catalan numbers: A284016 (also A068875, A002420).

Values of the polynomial sequence: A068875 (row sums): P(1), A154955: P(-1), A238113: P(2)/2, A125695 (also A152681): P(-2), A054872: P(3)/2, P(3)/6 probable A234939, A336729: P(-3)/6, A082298: P(4)/5, A238113: 2^n*P(1/2), A154825 and A091593: 2^n*P(-1/2).

Sequence in context: A253628 A102728 A262495 * A336703 A323174 A295683

Adjacent sequences: A352684 A352685 A352686 * A352688 A352689 A352690

Keyword

nonn,tabl

Author

Peter Luschny, Apr 26 2022

Status

approved