A125695 -id:A125695 - cp's OEIS frontend

A125694 Riordan array ((1+3x-sqrt(1+2x+9x^2))/(2x),(1+3x-sqrt(1+2x+9x^2))/2).

1, -2, 1, 2, -4, 1, 2, 8, -6, 1, -10, -4, 18, -8, 1, 6, -24, -26, 32, -10, 1, 42, 60, -18, -72, 50, -12, 1, -102, 24, 162, 48, -150, 72, -14, 1, -82, -388, -214, 248, 230, -268, 98, -16, 1, 782, 536, -546, -800, 158, 600, -434, 128, -18, 1

Offset: 0

Paul Barry, Nov 30 2006

First column is A125695. Row sums are A091593. Inverse of A125693.

			Triangle begins
1,
-2, 1,
2, -4, 1,
2, 8, -6, 1,
-10, -4, 18, -8, 1,
6, -24, -26, 32, -10, 1,
42, 60, -18, -72, 50, -12, 1

m = 9;
T[0, 0] = 1; T[1, 0] = 2; T[1, 1] = 1; T[n_, k_] /; 0 <= k <= n := T[n, k] = 3 T[n - 1, k] + T[n - 1, k - 1] - T[n - 2, k - 1]; T[, ] = 0;
M = Table[T[n, k], {n, 0, m}, {k, 0, m}] // Inverse;
A[n_, k_] := M[[n + 1, k + 1]];
Table[A[n, k], {n, 0, m}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 13 2019 *)

A125694 = lambda n,k : (-3)^(n-k)*binomial(n,k)*hypergeometric([k-n, n+1], [k+2], 1/3)
for n in (0..6): [round(A125694(n,k).n(100)) for k in (0..n)] # Peter Luschny, Sep 17 2014

T(n-1,k-1) = (k/n)*sum_{i=0..n-k} binomial(n,n-k-i) *(-3)^(n-k-i) *binomial(i+n-1,n-1). - Vladimir Kruchinin, Feb 12 2011

T(n, k) = (-3)^(n-k)*C(n,k)*hypergeometric([k-n, n+1], [k+2], 1/3). - Peter Luschny, Sep 17 2014

A152681 [x^(n+1)]Reversion[x(1-x)/(1-3x)].

Original entry on oeis.org

1, -2, 2, 2, -10, 6, 42, -102, -82, 782, -814, -3854, 12454, 5014, -98694, 142218, 472158, -1932258, -19038, 14816994, -27370410, -64159962, 334154442, -121279878, -2418497010, 5523511086, 8914677362, -61259567662, 44249714438

Offset: 0

Paul Barry, Dec 10 2008

Hankel transform is (-2)^C(n+1,2).

			G.f. = 1 - 2*x + 2*x^2 + 2*x^3 - 10*x^4 + 6*x^5 + 42*x^6 + ... - _Michael Somos_, Sep 18 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A125695.

m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1 +3*x -Sqrt(1+2*x+9*x^2))/(2*x))); // G. C. Greubel, Sep 14 2018

A152681_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
for w from 1 to n do a[w] := -2*a[w-1]+add(a[j]*a[w-j-1],j=1..w-1) od;convert(a,list)end: A152681_list(28); # Peter Luschny, May 19 2011

CoefficientList[Series[(1+3*x-Sqrt[1+2*x+9*x^2])/(2*x), {x,0,50}], x] (* G. C. Greubel, Sep 14 2018 *)

x='x+O('x^30); Vec((1 +3*x -sqrt(1+2*x+9*x^2))/(2*x)) \\ G. C. Greubel, Sep 14 2018

A152681 = lambda n : (-3)^n*hypergeometric([-n, n+1], [2], 1/3)
[round(A152681(n).n(100)) for n in (0..20)] # Peter Luschny, Sep 17 2014

G.f.: (1 + 3*x - sqrt(1 + 2*x + 9*x^2))/(2*x).
a(n) = Sum_{k=0..n} C(n+k,2k)*A000108(k)*(-3)^(n-k).
a(n) = 0^n - 2*Sum_{k=0..floor((n-1)/2)} C(n-1,2k)*A000108(k)(-1)^(n-k-1)*2^k.
a(n) = Sum_{k=0..n} A090181(n,k)*(-2)^k. - Philippe Deléham, Feb 02 2009
D-finite with recurrence (n+1)*a(n) + (2*n-1)*a(n-1) + 9*(n-2)*a(n-2) = 0. - R. J. Mathar, Oct 25 2012
a(n) = (-3)^n*hypergeometric([-n, n+1], [2], 1/3). - Peter Luschny, Sep 17 2014

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

A125694 Riordan array ((1+3x-sqrt(1+2x+9x^2))/(2x),(1+3x-sqrt(1+2x+9x^2))/2).

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A152681 [x^(n+1)]Reversion[x(1-x)/(1-3x)].

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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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cp's OEIS Frontend

A125694 Riordan array ((1+3x-sqrt(1+2x+9x^2))/(2x),(1+3x-sqrt(1+2x+9x^2))/2).

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Programs

Mathematica

Sage

Formula

A152681 [x^(n+1)]Reversion[x*(1-x)/(1-3*x)].

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Author

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Examples

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Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Python

Formula

A152681 [x^(n+1)]Reversion[x(1-x)/(1-3x)].