A336729 -id:A336729 - cp's OEIS frontend

A336727 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(0,k) = 1 and T(n,k) = (1/n) * Sum_{j=1..n} (-k)^(n-j) * binomial(n,j) * binomial(n,j-1) for n > 0.

1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, -1, -1, 1, 1, 1, -2, -1, 0, 1, 1, 1, -3, 1, 5, 2, 1, 1, 1, -4, 5, 10, -3, 0, 1, 1, 1, -5, 11, 9, -38, -21, -5, 1, 1, 1, -6, 19, -4, -103, 28, 51, 0, 1, 1, 1, -7, 29, -35, -174, 357, 289, 41, 14, 1, 1, 1, -8, 41, -90, -203, 1176, -131, -1262, -391, 0, 1

Offset: 0

Seiichi Manyama, Aug 02 2020

			  1,  1,   1,   1,    1,    1,    1, ...
  1,  1,   1,   1,    1,    1,    1, ...
  1,  0,  -1,  -2,   -3,   -4,   -5, ...
  1, -1,  -1,   1,    5,   11,   19, ...
  1,  0,   5,  10,    9,   -4,  -35, ...
  1,  2,  -3, -38, -103, -174, -203, ...
  1,  0, -21,  28,  357, 1176, 2575, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0-3 give: A000012, A090192, (-1)^n * A154825(n), A336729.
Main diagonal gives A336728.
Cf. A243631, A307884, A336708, A336709.

T[0, k_] := 1; T[n_, k_] := Sum[If[k == 0, Boole[n == j],(-k)^(n - j)] * Binomial[n, j] * Binomial[n , j - 1], {j, 1, n}] / n; Table[T[k, n- k], {n, 0, 11}, {k, 0, n}] //Flatten (* Amiram Eldar, Aug 02 2020 *)

{T(n, k) = if(n==0, 1, sum(j=1, n, (-k)^(n-j)*binomial(n, j)*binomial(n, j-1))/n)}

{T(n, k) = local(A=1+x*O(x^n)); for(i=0, n, A=1+x*A/(1+k*x*A)); polcoef(A, n)}

{T(n, k) = sum(j=0, n, (-k)^j*(k+1)^(n-j)*binomial(n, j)*binomial(n+j, n)/(j+1))}

G.f. A_k(x) of column k satisfies A_k(x) = 1 + x * A_k(x) / (1 + k * x * A_k(x)).
A_k(x) = 2/(1 - (k+1)*x + sqrt(1 + 2*(k-1)*x + ((k+1)*x)^2)).
T(n, k) = Sum_{j=0..n} (-k)^j * (k+1)^(n-j) * binomial(n,j) * binomial(n+j,n)/(j+1).
(n+1) * T(n,k) = -(k-1) * (2*n-1) * T(n-1,k) - (k+1)^2 * (n-2) * T(n-2,k) for n>1. - Seiichi Manyama, Aug 08 2020

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

Original entry on oeis.org

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

Peter Luschny, Apr 26 2022

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.

			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;

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.

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).

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);

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

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))

# 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))

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.

cp's OEIS Frontend

A336727 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(0,k) = 1 and T(n,k) = (1/n) * Sum_{j=1..n} (-k)^(n-j) * binomial(n,j) * binomial(n,j-1) for n > 0.

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

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