A152681 -id:A152681 - cp's OEIS frontend

A090181 Triangle of Narayana (A001263) with 0 <= k <= n, read by rows.

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 6, 1, 0, 1, 10, 20, 10, 1, 0, 1, 15, 50, 50, 15, 1, 0, 1, 21, 105, 175, 105, 21, 1, 0, 1, 28, 196, 490, 490, 196, 28, 1, 0, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1, 0, 1, 45, 540, 2520, 5292, 5292, 2520, 540, 45, 1, 0, 1, 55, 825, 4950, 13860

Offset: 0

Philippe Deléham, Jan 19 2004

Number of Dyck n-paths with exactly k peaks. - Peter Luschny, May 10 2014

			Triangle starts:
[0] 1;
[1] 0, 1;
[2] 0, 1,  1;
[3] 0, 1,  3,   1;
[4] 0, 1,  6,   6,    1;
[5] 0, 1, 10,  20,   10,    1;
[6] 0, 1, 15,  50,   50,   15,    1;
[7] 0, 1, 21, 105,  175,  105,   21,   1;
[8] 0, 1, 28, 196,  490,  490,  196,  28,  1;
[9] 0, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1;

Indranil Ghosh, Rows 0..100, flattened
Yu Hin Au, Some Properties and Combinatorial Implications of Weighted Small Schröder Numbers, arXiv:1912.00555 [math.CO], 2019.
Paul Barry, Continued fractions and transformations of integer sequences, JIS 12 (2009) 09.7.6.
Paul Barry, Invariant number triangles, eigentriangles and Somos-4 sequences, arXiv preprint arXiv:1107.5490 [math.CO], 2011.
Paul Barry, On a Generalization of the Narayana Triangle, J. Int. Seq. 14 (2011) # 11.4.5.
Paul Barry, On a transformation of Riordan moment sequences, arXiv:1802.03443 [math.CO], 2018.
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
Paul Barry, Riordan Pseudo-Involutions, Continued Fractions and Somos 4 Sequences, arXiv:1807.05794 [math.CO], 2018.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 25, 29.
Paul Barry and Aoife Hennessy, A Note on Narayana Triangles and Related Polynomials, Riordan Arrays, and MIMO Capacity Calculations, J. Int. Seq. 14 (2011), Article 11.3.8.
FindStat - Combinatorial Statistic Finder, The number of peaks of a Dyck path, The number of double rises of a Dyck path, The number of valleys of a Dyck path, The number of left oriented leafs except the first one of a binary tree, The number of left tunnels of a Dyck path
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.

Mirror image of triangle A131198. A000108 (row sums, Catalan).
Columns: A000217, A002415, A006542, A006857.
Cf. A001263, A084938, A243752.
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000108(n), A006318(n), A047891(n+1), A082298(n), A082301(n), A082302(n), A082305(n), A082366(n), A082367(n) for x=0,1,2,3,4,5,6,7,8,9. - Philippe Deléham, Aug 10 2006
Sum_{k=0..n} x^(n-k)*T(n,k) = A090192(n+1), A000012(n), A000108(n), A001003(n), A007564(n), A059231(n), A078009(n), A078018(n), A081178(n), A082147(n), A082181(n), A082148(n), A082173(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. - Philippe Deléham, Oct 21 2006
Sum_{k=0..n} T(n,k)*x^k*(x-1)^(n-k) = A000012(n), A006318(n), A103210(n), A103211(n), A133305(n), A133306(n), A133307(n), A133308(n), A133309(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, respectively. - Philippe Deléham, Oct 20 2007

[[(&+[(-1)^(j-k)*Binomial(2*n-j,j)*Binomial(j,k)*Binomial(2*n-2*j,n-j)/(n-j+1): j in [0..n]]): k in [0..n]]: n in [0..10]];

A090181 := (n,k) -> binomial(n,n-k)*binomial(n-1,n-k)/(n-k+1):
seq(print( seq(A090181(n,k),k=0..n)),n=0..5); # Peter Luschny, May 10 2014
egf := 1+int((sqrt(t)*exp((1+t)*x)*BesselI(1,2*sqrt(t)*x))/x,x);
s := n -> n!*coeff(series(egf,x,n+2),x,n);
seq(print(seq(coeff(s(n),t,j),j=0..n)),n=0..9); # Peter Luschny, Oct 30 2014
T := proc(n, k) option remember; if k = n or k = 1 then 1 elif k < 1 then 0 else (2*n/k - 1) * T(n-1, k-1) + T(n-1, k) fi end:
for n from 0 to 8 do seq(T(n, k), k = 0..n) od;  # Peter Luschny, Dec 31 2024

Flatten[Table[Sum[(-1)^(j-k) * Binomial[2n-j,j] * Binomial[j,k] * CatalanNumber[n-j], {j, 0, n}], {n,0,11},{k,0,n}]] (* Indranil Ghosh, Mar 05 2017 *)
p[0, ] := 1; p[1, x] := x; p[n_, x_] := ((2 n - 1) (1 + x) p[n - 1, x] - (n - 2) (x - 1)^2 p[n - 2, x]) / (n + 1);
Table[CoefficientList[p[n, x], x], {n, 0, 9}] // TableForm (* Peter Luschny, Apr 26 2022 *)

c(n) = binomial(2*n,n)/ (n+1);
tabl(nn) = {for(n=0, nn, for(k=0, n, print1(sum(j=0, n, (-1)^(j-k) * binomial(2*n-j,j) * binomial(j,k) * c(n-j)),", ");); print(););};
tabl(11); \\ Indranil Ghosh, Mar 05 2017

from functools import cache
@cache
def Trow(n):
    if n == 0: return [1]
    if n == 1: return [0, 1]
    if n == 2: return [0, 1, 1]
    A = Trow(n - 2) + [0, 0]
    B = Trow(n - 1) + [1]
    for k in range(n - 1, 1, -1):
        B[k] = (((B[k] + B[k - 1]) * (2 * n - 1)
               - (A[k] - 2 * A[k - 1] + A[k - 2]) * (n - 2)) // (n + 1))
    return B
for n in range(10): print(Trow(n)) # Peter Luschny, May 02 2022

def A090181_row(n):
    U = [0]*(n+1)
    for d in DyckWords(n):
        U[d.number_of_peaks()] +=1
    return U
for n in range(8): A090181_row(n) # Peter Luschny, May 10 2014

Triangle T(n, k), read by rows, given by [0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938. T(0, 0) = 1, T(n, 0) = 0 for n>0, T(n, k) = C(n-1, k-1)*C(n, k-1)/k for k>0.
Sum_{j>=0} T(n,j)*binomial(j,k) = A060693(n,k). - Philippe Deléham, May 04 2007
Sum_{k=0..n} T(n,k)*10^k = A143749(n+1). - Philippe Deléham, Oct 14 2008
From Paul Barry, Nov 10 2008: (Start)
Coefficient array of the polynomials P(n,x) = x^n*2F1(-n,-n+1;2;1/x).
T(n,k) = Sum_{j=0..n} (-1)^(j-k)*C(2n-j,j)*C(j,k)*A000108(n-j). (End)
Sum_{k=0..n} T(n,k)*5^k*3^(n-k) = A152601(n). - Philippe Deléham, Dec 10 2008
Sum_{k=0..n} T(n,k)*(-2)^k = A152681(n); Sum_{k=0..n} T(n,k)*(-1)^k = A105523(n). - Philippe Deléham, Feb 03 2009
Sum_{k=0..n} T(n,k)*2^(n+k) = A156017(n). - Philippe Deléham, Nov 27 2011
T(n, k) = C(n,n-k)*C(n-1,n-k)/(n-k+1). - Peter Luschny, May 10 2014
E.g.f.: 1+Integral((sqrt(t)*exp((1+t)*x)*BesselI(1,2*sqrt(t)*x))/x dx). - Peter Luschny, Oct 30 2014
G.f.: (1+x-x*y-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x). - Alois P. Heinz, Nov 28 2021, edited by Ron L.J. van den Burg, Dec 19 2021
T(n, k) = [x^k] (((2*n - 1)*(1 + x)*p(n-1, x) - (n - 2)*(x - 1)^2*p(n-2, x))/(n + 1)) with p(0, x) = 1 and p(1, x) = x. - Peter Luschny, Apr 26 2022
Recursion based on rows (see the Python program):
T(n, k) = (((B(k) + B(k-1))*(2*n - 1) - (A(k) - 2*A(k-1) + A(k-2))*(n-2))/(n+1)), where A(k) = T(n-2, k) and B(k) = T(n-1, k), for n >= 3. # Peter Luschny, May 02 2022

A105523 Expansion of 1-x*c(-x^2) where c(x) is the g.f. of A000108.

Original entry on oeis.org

1, -1, 0, 1, 0, -2, 0, 5, 0, -14, 0, 42, 0, -132, 0, 429, 0, -1430, 0, 4862, 0, -16796, 0, 58786, 0, -208012, 0, 742900, 0, -2674440, 0, 9694845, 0, -35357670, 0, 129644790, 0, -477638700, 0, 1767263190, 0

Offset: 0

Paul Barry, Apr 11 2005

Row sums of A105522. Row sums of inverse of A105438.

First column of number triangle A106180.

			G.f. = 1 - x + x^3 - 2*x^5 + 5*x^7 - 14*x^9 + 42*x^11 - 132*x^13 + 429*x^15 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
R. J. Martin and M. J. Kearney, An exactly solvable self-convolutive recurrence, Aequat. Math., 80 (2010), 291-318. see p. 313.
R. J. Martin and M. J. Kearney, An exactly solvable self-convolutive recurrence, arXiv:1103.4936 [math.CO], 2011.

Cf. A000108, A097331, A090192, A090181, A105522, A105438.

m:=25; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1 + 2*x - Sqrt(1+4*x^2))/(2*x))); // G. C. Greubel, Sep 16 2018

A105523_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
for w from 1 to n do a[w]:=-a[w-1]+(-1)^w*add(a[j]*a[w-j-1],j=1..w-1) od; convert(a,list)end: A105523_list(40); # Peter Luschny, May 19 2011

a[n_?EvenQ] := 0; a[n_?OddQ] := 4^n*Gamma[n/2] / (Gamma[-n/2]*(n+1)!); a[0] = 1; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 14 2011, after Vladimir Kruchinin *)
CoefficientList[Series[(1 + 2 x - Sqrt[1 + 4 x^2])/(2 x), {x, 0, 50}], x] (* Vincenzo Librandi, Nov 01 2014 *)
a[ n_] := SeriesCoefficient[ (1 + 2 x - Sqrt[ 1 + 4 x^2]) / (2 x), {x, 0, n}]; (* Michael Somos, Jun 17 2015 *)
a[ n_] := If[ n < 1, Boole[n == 0], a[n] = -2 a[n - 1] + Sum[ a[j] a[n - j - 1], {j, 0, n - 1}]]; (* Michael Somos, Jun 17 2015 *)

{a(n) = local(A); if( n<0, 0, n++; A = vector(n); A[1] = 1; for( k=2, n, A[k] = -2 * A[k-1] + sum( j=1, k-1, A[j] * A[k-j])); A[n])}; /* Michael Somos, Jul 24 2011 */

def A105523(n):
    if is_even(n): return 0 if n>0 else 1
    return -(sqrt(pi)*2^(n-1))/(gamma(1-n/2)*gamma((n+3)/2))
[A105523(n) for n in (0..29)] # Peter Luschny, Oct 31 2014

G.f.: (1 + 2*x - sqrt(1+4*x^2))/(2*x).
a(n) = 0^n + sin(Pi*(n-2)/2)(C((n-1)/2)(1-(-1)^n)/2).
G.f.: 1/(1+x/(1-x/(1+x/(1-x/(1+x/(1-x.... (continued fraction). - Paul Barry, Jan 15 2009
a(n) = Sum{k = 0..n} A090181(n,k)*(-1)^k. - Philippe Deléham, Feb 02 2009
a(n) = (1/n)*sum_{i = 0..n-1} (-2)^i*binomial(n, i)*binomial(2*n-i-2, n-1). - Vladimir Kruchinin, Dec 26 2010
With offset 1, a(n) = -2 * a(n-1) + Sum_{k=1..n-1} a(k) * a(n-k), for n>1. - Michael Somos, Jul 25 2011
D-finite with recurrence: (n+3)*a(n+2) = -4*n*a(n), a(0)=1, a(1)=-1. - Fung Lam, Mar 18 2014
For nonzero terms, a(n) ~ (-1)^((n+1)/2)/sqrt(2*Pi)*2^(n+1)/(n+1)^(3/2). - Fung Lam, Mar 17 2014
a(n) = -(sqrt(Pi)*2^(n-1))/(Gamma(1-n/2)*Gamma((n+3)/2)) for n odd. - Peter Luschny, Oct 31 2014
From Peter Bala, Apr 20 2024: (Start)
a(n) = Sum_{k = 0..n} (-2)^(n-k)*binomial(n + k, 2*k)*Catalan(k), where Catalan(k) = A000108(k).
a(n) = (-2)^n * hypergeom([-n, n+1], [2], 1/2).
O.g.f.: A(x) = 1/x * series reversion of x*(1 - x)/(1 - 2*x).  Cf. A152681. (End)

Typo in definition corrected by Robert Israel, Oct 31 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.

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A090181 Triangle of Narayana (A001263) with 0 <= k <= n, read by rows.

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A105523 Expansion of 1-x*c(-x^2) where c(x) is the g.f. of A000108.

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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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