A001263 -id:A001263 - cp's OEIS frontend

A134290 Ninth column (and diagonal) of Narayana triangle A001263.

1, 45, 825, 9075, 70785, 429429, 2147145, 9202050, 34763300, 118195220, 367479684, 1057896060, 2848181700, 7229999700, 17420856420, 40067969766, 88385227425, 187746398125, 385374185625, 766691800875, 1482270815025, 2791289197125, 5130235085625, 9219552907500

Offset: 0

Wolfdieter Lang, Nov 13 2007

See a comment under A134288 on the coincidence of column and diagonal sequences.

Kekulé numbers K(O(1,8,n)) for certain benzenoids (see the Cyvin-Gutman reference, p. 105, eq. (i)).

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988.

T. D. Noe, Table of n, a(n) for n = 0..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 25.
W. F. Wheatley and James Ethridge (Proposers), Comment from Alan H. Rapoport, Problem 84, Missouri Journal of Mathematical Sciences, volume 8, #2, spring 1996, pages 97-102.

Cf. A002378.
Cf. A134289 (eighth column of Narayana triangle).
Cf. A134291 (tenth column of Narayana triangle).

List([0..25], n-> Binomial(n+9,9)*Binomial(n+8,7)/8); # G. C. Greubel, Aug 28 2019

[Binomial(n+9,9)*Binomial(n+8,7)/8: n in [0..25]]; // G. C. Greubel, Aug 28 2019

a := n -> ((n+1)*((n+2)*(n+3)*(n+4)*(n+5)*(n+6)*(n+7)*(n+8))^2*(n+9))/14631321600:
seq(a(n), n=0..23); # Peter Luschny, Sep 01 2016

Table[Binomial[n+9,9]*Binomial[n+8,7]/8, {n,0,25}] (* G. C. Greubel, Aug 28 2019 *)

Vec((1+28*x+196*x^2+490*x^3+490*x^4+196*x^5+28*x^6+x^7)/(1-x)^17 + O(x^25)) \\ Altug Alkan, Sep 01 2016

vector(25, n, binomial(n+8,9)*binomial(n+7,7)/8) \\ G. C. Greubel, Aug 28 2019

[binomial(n+9,9)*binomial(n+8,7)/8 for n in (0..25)] # G. C. Greubel, Aug 28 2019

a(n) = A001263(n+9,9) = binomial(n+9,9)*binomial(n+9,8)/(n+9).
O.g.f.: P(8,x)/(1-x)^17 with the numerator polynomial P(8,x) = Sum_{k=1..8} A001263(8,k)*x^(k-1), the eighth row polynomial of the Narayana triangle: P(8,x) = 1 + 28*x + 196*x^2 + 490*x^3 + 490*x^4 + 196*x^5 + 28*x^6 + x^7.
a(n) = Product_{i=1..8} A002378(n+i)/A002378(i). - Bruno Berselli, Sep 01 2016
From Amiram Eldar, Oct 19 2020: (Start)
Sum_{n>=0} 1/a(n) = 497925669/175 - 288288*Pi^2.
Sum_{n>=0} (-1)^n/a(n) = 580367/35 - 1680*Pi^2. (End)

A082148 a(0)=1; for n >= 1, a(n) = Sum_{k=0..n} 10^kN(n,k), where N(n,k) = (1/n)C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

Original entry on oeis.org

1, 1, 11, 131, 1661, 22101, 305151, 4335711, 63009881, 932449961, 14004694451, 212944033051, 3271618296661, 50711564152381, 792088104593511, 12454801769554551, 196991734871121201, 3131967533789345361, 50026642742943415131, 802406215117502069811

Offset: 0

Benoit Cloitre, May 10 2003

nonn

More generally, coefficients of (1+m*x-sqrt(m^2*x^2-(2*m+4)*x+1))/((2*m+2)*x) are given by: a(n) = Sum_{k=0..n} (m+1)^k*N(n,k)).
The Hankel transform of this sequence is 10^C(n+1,2). - Philippe Deléham, Oct 29 2007
a(n) = upper left term in M^n, M = the production matrix:
   1,  1;
  10, 10, 10;
   1,  1,  1,  1;
  10, 10, 10, 10, 10;
   1,  1,  1,  1,  1,  1;
  ...
- Gary W. Adamson, Jul 08 2011
Shifts left when INVERT transform applied ten times. - Benedict W. J. Irwin, Feb 07 2016
For fixed m > 0, if g.f. = (1+m*x-sqrt(m^2*x^2-(2*m+4)*x+1))/((2*m+2)*x) then a(n,m) ~ (m + 2 + 2*sqrt(m+1))^(n + 1/2) / (2*sqrt(Pi) * (m+1)^(3/4) * n^(3/2)). - Vaclav Kotesovec, Mar 19 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

Cf. A001003, A007564, A059231.

I:=[1,11]; [1] cat [n le 2 select I[n] else (11*(2*n-1)*Self(n-1) - 81*(n-2)*Self(n-2))/(n+1): n in [1..30]]; // G. C. Greubel, Feb 10 2018

A082148_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]+10*add(a[j]*a[w-j-1],j=1..w-1) od;
convert(a, list) end: A082148_list(17); # Peter Luschny, May 19 2011

Table[SeriesCoefficient[(1+9*x-Sqrt[81*x^2-22*x+1])/(20*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
a[n_] := Sum[10^k*1/n*Binomial[n, k]*Binomial[n, k + 1], {k, 0, n}];
a[0] = 1; Array[a, 20, 0] (* Robert G. Wilson v, Feb 10 2018 *)
a[n_] := Hypergeometric2F1[1 - n, -n, 2, 10];
Table[a[n], {n, 0, 18}] (* Peter Luschny, Mar 19 2018 *)

a(n)=if(n<1,1,sum(k=0,n,10^k/n*binomial(n,k)*binomial(n,k+1)))

G.f.: (1+9*x-sqrt(81*x^2-22*x+1))/(20*x).
a(n) = Sum_{k=0..n} A088617(n, k)*10^k*(-9)^(n-k). - Philippe Deléham, Jan 21 2004
a(n) = (11*(2n-1)*a(n-1) - 81*(n-2)*a(n-2)) / (n+1) for n>=2, a(0)=a(1)=1. - Philippe Deléham, Aug 19 2005
a(n) ~ sqrt(20+11*sqrt(10))*(11+2*sqrt(10))^n/(20*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2012
G.f.: 1/(1 - x/(1 - 10*x/(1 - x/(1 - 10*x/(1 - x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, Aug 10 2017
a(n) = hypergeom([1 - n, -n], [2], 10). - Peter Luschny, Mar 19 2018

A082173 a(0)=1; for n >= 1, a(n) = Sum_{k=0..n} 11^kN(n,k) where N(n,k) = (1/n)C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

Original entry on oeis.org

1, 1, 12, 155, 2124, 30482, 453432, 6936799, 108507180, 1727970542, 27924685416, 456820603086, 7550600079672, 125905525750500, 2115511349837040, 35782547891727495, 608787760350045420, 10411451736723707990

Offset: 0

Benoit Cloitre, May 10 2003

nonn

More generally coefficients of (1 + m*x - sqrt(m^2*x^2 -(2*m+4)*x + 1) )/((2*m+2)*x) are given by a(n) = Sum_{k=0..n} (m+1)^k*N(n,k).
The Hankel transform of this sequence is 11^C(n+1,2). - Philippe Deléham, Oct 29 2007
For fixed m > 0, if g.f. = (1 + m*x - sqrt(m^2*x^2 -(2*m+4)*x + 1) )/((2*m+2)*x) then a(n,m) ~ (m + 2 + 2*sqrt(m+1))^(n + 1/2) / (2*sqrt(Pi) * (m+1)^(3/4) * n^(3/2)). - Vaclav Kotesovec, Mar 19 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A001003, A001263, A007564, A059231, A088617.

[1] cat [&+[11^k*Binomial(n, k)*Binomial(n, k+1)/n:k in [0..n]]:n in [1..18]]; // Marius A. Burtea, Jan 22 2020

A082173_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]+11*add(a[j]*a[w-j-1],j=1..w-1)od;
convert(a, list) end: A082173_list(17); # Peter Luschny, May 19 2011

Table[SeriesCoefficient[(1+10*x-Sqrt[100*x^2-24*x+1])/(22*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
a[n_] := Hypergeometric2F1[1 - n, -n, 2, 11];
Table[a[n], {n, 0, 18}] (* Peter Luschny, Mar 19 2018 *)

a(n)=if(n<1,1,sum(k=0,n,11^k/n*binomial(n,k)*binomial(n,k+1)))

def A082173_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1+10*x-sqrt(100*x^2-24*x+1))/(22*x) ).list()
A082173_list(30) # G. C. Greubel, Jan 21 2024

G.f.: (1+10*x-sqrt(100*x^2-24*x+1))/(22*x).
a(n) = Sum_{k=0..n} A088617(n, k)*11^k*(-10)^(n-k). - Philippe Deléham, Jan 21 2004
a(n) = (12*(2n-1)*a(n-1) - 100*(n-2)*a(n-2)) / (n+1) for n >= 2, a(0) = a(1) = 1. - Philippe Deléham, Aug 19 2005
From Gary W. Adamson, Jul 08 2011: (Start)
a(n) = upper left term in M^n, M = the production matrix:
   1,  1
  11, 11, 11
   1,  1,  1,  1
  11, 11, 11, 11, 11
   1,  1,  1,  1,  1,  1
  ... (End)
a(n) ~ sqrt(22+12*sqrt(11))*(12+2*sqrt(11))^n/(22*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2012
G.f.: 1/(1 - x/(1 - 11*x/(1 - x/(1 - 11*x/(1 - x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, Aug 10 2017
a(n) = hypergeom([1 - n, -n], [2], 11). - Peter Luschny, Mar 19 2018

A082181 a(0) = 1, for n>=1, a(n) = Sum_{k=0..n} 9^kN(n,k), where N(n,k) = (1/n)C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

Original entry on oeis.org

1, 1, 10, 109, 1270, 15562, 198100, 2596645, 34825150, 475697854, 6595646860, 92590323058, 1313427716380, 18798095833012, 271118225915560, 3936516861402901, 57494017447915150, 844109420603623030

Offset: 0

Benoit Cloitre, May 10 2003

nonn

More generally, coefficients of (1+m*x-sqrt(m^2*x^2-(2*m+4)*x+1))/((2*m+2)*x) are given by: a(n) = Sum_{k=0..n} (m+1)^k*N(n,k).
The Hankel transform of this sequence is 9^C(n+1,2). - Philippe Deléham, Oct 29 2007
From Gary W. Adamson, Jul 08 2011: (Start)
a(n) = upper left term in M^n, M = the production matrix:
  1, 1
  9, 9, 9
  1, 1, 1, 1
  9, 9, 9, 9, 9
  1, 1, 1, 1, 1, 1
  ... (End)
Shifts left when INVERT transform applied nine times. - Benedict W. J. Irwin, Feb 07 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

Cf. A001003, A001263, A007564, A059231, A088617.

[(&+[Binomial(n,k)*Binomial(n-1,k)*9^k/(k+1): k in [0..n]]): n in [0..30]]; // G. C. Greubel, May 23 2022

A082181_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]+9*add(a[j]*a[w-j-1],j=1..w-1) od;
convert(a, list) end: A082181_list(17); # Peter Luschny, May 19 2011

Table[SeriesCoefficient[(1+8*x-Sqrt[64*x^2-20*x+1])/(18*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
a[n_] := Hypergeometric2F1[1 - n, -n, 2, 9];
Table[a[n], {n, 0, 18}] (* Peter Luschny, Mar 19 2018 *)

a(n)=if(n<1,1,sum(k=0,n,9^k/n*binomial(n,k)*binomial(n,k+1)))

[sum(binomial(n,k)*binomial(n-1,k)*9^k/(k+1) for k in (0..n)) for n in (0..30)] # G. C. Greubel, May 23 2022

G.f.: (1+8*x-sqrt(64*x^2-20*x+1))/(18*x).
a(n) = Sum_{k=0..n} A088617(n, k)*9^k*(-8)^(n-k). - Philippe Deléham, Jan 21 2004
a(n) = (10*(2*n-1)*a(n-1) - 64*(n-2)*a(n-2)) / (n+1) for n>=2, a(0)=a(1)=1. - Philippe Deléham, Aug 19 2005
a(n) ~ 2^(4*n+1)/(3*sqrt(3*Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2012
G.f.: 1/(1 - x/(1 - 9*x/(1 - x/(1 - 9*x/(1 - x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, Apr 21 2017
a(n) = hypergeom([1 - n, -n], [2], 9). - Peter Luschny, Mar 19 2018

Corrected by T. D. Noe, Oct 25 2006

A103366 Second column of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).

Original entry on oeis.org

1, -3, 12, -70, 585, -6741, 103068, -2023092, 49615695, -1487006785, 53477384268, -2272859942574, 112699083156751, -6447858833644515, 421601806346674320, -31242589674606301224, 2604618355967848204587, -242686626407684239621113, 25124942798118355122058980

Offset: 2

Paul D. Hanna, Feb 02 2005

sign

Vaclav Kotesovec, Table of n, a(n) for n = 2..200

Cf. A001263, A103364, A103365, A103367, A115369.

a(n)=if(n<2,0,(matrix(n,n,m,j,binomial(m-1,j-1)*binomial(m,j-1)/j)^-1)[n,2])

a(n) ~ (-1)^n * 2^(2*n-5) * n! * (n-1)! / (BesselJ(2, BesselJZero(1, 1)) * BesselJZero(1, 1)^(2*n-4)), where BesselJZero(1, 1) = A115369. - Vaclav Kotesovec, Jul 11 2025

A145904 Square array read by antidiagonals: Hilbert transform of the Narayana numbers A001263.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 16, 7, 1, 1, 15, 40, 31, 9, 1, 1, 21, 85, 105, 51, 11, 1, 1, 28, 161, 295, 219, 76, 13, 1, 1, 36, 280, 721, 771, 396, 106, 15, 1, 1, 45, 456, 1582, 2331, 1681, 650, 141, 17, 1

Offset: 0

Peter Bala, Oct 31 2008

Refer to A145905 for the definition of the Hilbert transform of a lower triangular array. For the Hilbert transform of A008459, the array of type B Narayana numbers, see A108625.

This seems to be a duplicate of A273350. - Alois P. Heinz, Jun 04 2016. This could probably be proved by showing that the g.f.s are the same. - N. J. A. Sloane, Jul 02 2016

			The array begins
n\k|..0.....1.....2.....3.....4.....5
=====================================
0..|..1.....1.....1.....1.....1.....1
1..|..1.....3.....5.....7.....9....11
2..|..1.....6....16....31....51....76
3..|..1....10....40...105...219...396
4..|..1....15....85...295...771..1681
5..|..1....21...161...721..2331..6083
...
Row 2: (1 + 3x + x^2)/(1 - x)^3 = 1 + 6x + 16x^2 + 31x^3 + ... .
Row 3: (1 + 6x + 6x^2 + x^3)/(1 - x)^4 = 1 + 10x + 40x^2 + 105x^3 + ... .

Cf. A001263, A005891 (row 2), A063490 (row 3), A108625 (Hilbert transform of h-vectors of type B associahedra).

Cf. also A273350.

Table[1/(# + 1)*Sum[Binomial[# + 1, i - 1] Binomial[# + 1, i] Binomial[# + k - i + 1, k + 1 - i], {i, 0, k + 1}] &[m - k], {m, 0, 9}, {k, 0, m}] // Flatten (* Michael De Vlieger, Jan 15 2018 *)

taylor(((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y),x,0,10,y,0,10);
T(n,m,k):=1/(n+1)*sum(binomial(n+1,i-1)*binomial(n+1,i)*binomial(n+m-i+1,m+1-i),i,0,m+1); /* Vladimir Kruchinin, Jan 15 2018 */

Row n generating function: 1/(n+1) * 1/(1-x) * Jacobi_P(n,1,1,(1+x)/(1-x)) = N_n(x)/(1-x)^n where N_n(x) denotes the shifted Narayana polynomial N_n(x) = sum{k = 1..n} A001263(k)*x^(k-1) of degree n-1.
Conjectural column n generating function: N_n(x^2)/(1-x)^(2n+1).
The entries in row n are given by the values of a polynomial function p_n(x) at x = 0,1,2,... . The first few are p_1(x) = 2x + 1, p_2(x) = (5x^2 + 5x + 2)/2, p_3(x) = (2x + 1)*(7x^2 + 7x + 6)/6 and p_4(x) = (7x^4 + 14x^3 + 21x^2 + 14x + 4)/4. These polynomials appear to have their zeros on the line Re x = -1/2; that is, the polynomials p_n(-x) appear to satisfy a Riemann hypothesis. The corresponding result for A108625 is true (see A142995 for details).
Contribution from Paul Barry, Jan 06 2009: (Start)
The g.f. for the corresponding number triangle is:
1/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y.... (a continued fraction). (End)
This g.f. satisfies x^2*y*g^2 - (1-x-x*y)*g + 1 = 0. - R. J. Mathar, Jun 16 2016
G.f.: ((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y). - Vladimir Kruchinin, Jan 15 2018
T(n,m) = 1/(n+1)*Sum_{i=0..m+1} C(n+1,i-1)*C(n+1,i)*C(n+m-i+1,m+1-i). - Vladimir Kruchinin, Jan 15 2018

A126182 Let P be Pascal's triangle A007318 and let N be Narayana's triangle A001263, both regarded as lower triangular matrices. Sequence gives triangle obtained from P*N, read by rows.

Original entry on oeis.org

1, 2, 1, 4, 5, 1, 8, 18, 9, 1, 16, 56, 50, 14, 1, 32, 160, 220, 110, 20, 1, 64, 432, 840, 645, 210, 27, 1, 128, 1120, 2912, 3150, 1575, 364, 35, 1, 256, 2816, 9408, 13552, 9534, 3388, 588, 44, 1, 512, 6912, 28800, 53088, 49644, 24822, 6636, 900, 54, 1

Offset: 0

Emeric Deutsch, Dec 19 2006, Mar 30 2007

Also T(n,k) is number of hex trees with n edges and k left edges (0<=k<=n). A hex tree is a rooted tree where each vertex has 0, 1, or 2 children and, when only one child is present, it is either a left child, or a median child, or a right child (name due to an obvious bijection with certain tree-like polyhexes; see the Harary-Read reference). Accordingly, one can have left, vertical, or right edges.
Also (with a different offset) T(n,k) is the number of skew Dyck paths of semilength n and having k peaks (1<=k<=n). A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of the path is defined to be the number of its steps. E.g., T(3,2)=5 because we have (UD)U(UD)D, (UD)U(UD)L, U(UD)D(UD), U(UD)(UD)D and U(UD)(UD)L (the peaks are shown between parentheses).
Sum of terms in row n = A002212(n+1). T(n,1) = A001793(n); T(n,2) = A006974(n-2); Sum_{k=0..n}kT(n,k) = A026379(n+1).
A126216 = N * P. - Gary W. Adamson, Nov 30 2007

			The triangle P begins
  1,
  1, 1
  1, 2, 1
  1, 3, 3, 1, ...
and T begins
  1,
  1,  1,
  1,  3,  1,
  1,  6,  6,  1,
  1, 10, 20, 10, 1, ...
The product P*T gives
   1,
   2,  1,
   4,  5,  1,
   8, 18,  9,  1,
  16, 56, 50, 14, 1, ...

E. Deutsch, E. Munarini, S. Rinaldi, Skew Dyck paths, J. Stat. Plann. Infer. 140 (8) (2010) 2191-2203.
M. Dziemianczuk, Enumerations of plane trees with multiple edges and Raney lattice paths, Discrete Mathematics 337 (2014): 9-24.
F. Harary and R. C. Read, The enumeration of tree-like polyhexes, Proc. Edinburgh Math. Soc. (2) 17 (1970), 1-13.

Cf. A001263, A007318, A002212, A001793, A006974, A026379.

Cf. A128718, A026378, A126216.

T:=proc(n,k) if k=0 then 2^n elif k<=n then binomial(n+1,k)*sum(binomial(k,n-k-j)*binomial(n+1-k,j)*2^j,j=n-2*k..n-k)/(n+1) else 0 fi end: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

t[n_, 0] := 2^n; t[n_, k_] := Binomial[n+1, k]*Sum[Binomial[k, n-k-j]*Binomial[n+1-k, j]*2^j, {j, n-2*k, n-k}]/(n+1); Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 12 2013 *)
nmax = 10; n[x_, y_] := (1-x*(1+y) - Sqrt[(1-x*(1+y))^2 - 4*y*x^2])/(2*x); s = Series[(n[x/(1-x), y]-1)/x, {x, 0, nmax+1}, {y, 0, nmax+1}];t[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {y, 0, k}]; Table[t[n, k], {n, 0, nmax}, {k, 1, n+1}] // Flatten (* Jean-François Alcover, Apr 16 2015, after Vladimir Kruchinin *)

tabl(nn) = {mP = matrix(nn, nn, n, k, binomial(n-1, k-1)); mN = matrix(nn, nn, n, k, binomial(n-1, k-1) * binomial(n, k-1) / k); mprod = mP*mN; for (n=1, nn, for (k=1, n, print1(mprod[n, k], ", ");); print(););} \\ Michel Marcus, Apr 16 2015

T(n,k) = (1/(n+1))*binomial(n+1,k)*Sum_{j=n-2k..n-k}2^j*binomial(k,n-k-j)*binomial(n+1-k,j) if 0 < k <= n; T(n,0) = 2^n.
G.f. G=G(t,z) satisfies G  = 1 + (t+2)*z*G + t*z^2*G^2.
E.g.f.: exp((t+2)*x)*BesselI_{1}(2*sqrt(t)*x)/(sqrt(t)*x). - Peter Luschny, Oct 29 2014
G.f.: N(x/(1-x),y)-1)/x, where N(x,y) is the g.f. of Narayana's triangle A001263. - Vladimir Kruchinin, Apr 06 2015.

New definition in terms of P and N from Philippe Deléham, Jun 30 2007

Edited by N. J. A. Sloane, Jul 22 2007

A128088 a(n) = Sum_{k=0..n} A000108(k)*A001263(n+1,k+1), where A000108 is the Catalan numbers and A001263 is the Narayana triangle.

Original entry on oeis.org

1, 2, 6, 24, 115, 618, 3591, 22088, 141903, 943590, 6452490, 45159480, 322305165, 2339100078, 17223121350, 128428689888, 968383277791, 7374380672718, 56655414930642, 438741242896680, 3422125459579869, 26866961380274598, 212191772351034249, 1685036376746788392

Offset: 0

Paul D. Hanna, Feb 23 2007

nonn

a(n) is the number of permutations of length n+1 avoiding the partially ordered pattern (POP) {1>2>3>4} of length 5. That is, the number of length n+1 permutations having no subsequences of length 5 in which the element in position 1 is larger than the element in position 2, which in turn is larger than the element in position 3, and that element is larger than the element in position 4. - Sergey Kitaev, Dec 13 2020

			Illustrate a(n) = Sum_{k=0..n} A000108(k)*A001263(n+1,k+1) by:
a(2) = 1*(1) + 1*(3) + 2*(1) = 6;
a(3) = 1*(1) + 1*(6) + 2*(6) + 5*(1) = 24;
a(4) = 1*(1) + 1*(10)+ 2*(20)+ 5*(10)+ 14*(1) = 115.
The Narayana triangle A001263(n+1,k+1) = C(n,k)*C(n+1,k)/(k+1) begins:
1;
1, 1;
1, 3, 1;
1, 6, 6, 1;
1, 10, 20, 10, 1;
1, 15, 50, 50, 15, 1; ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, arXiv:1903.08946 [math.CO], 2019.
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, The Electronic Journal of Combinatorics 26(3) (2019), P3.26.

Cf. A005802, A000108, A001263, A259833.

a := n -> hypergeom([1/2, -n - 1, -n], [2, 2], 4):
seq(simplify(a(n)), n = 0..23);  # Peter Luschny, Nov 06 2023

Table[Sum[Binomial[2*k,k]*Binomial[n,k]*Binomial[n+1,k]/(k+1)^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 20 2012 *)
Table[HypergeometricPFQ[{1/2, -1 - n, -n}, {2, 2}, 4], {n, 0, 20}] (* Vaclav Kotesovec, May 14 2016 *)

{a(n)=sum(k=0,n,binomial(2*k,k)*binomial(n,k)*binomial(n+1,k)/(k+1)^2)}

a(n) = (n+1)*A005802(n), where A005802(n) = number of permutations in S_n with longest increasing subsequence of length <= 3.
a(n) = Sum_{k=0..n} C(2k,k)*C(n,k)*C(n+1,k)/(k+1)^2.
Recurrence: (n+2)^2*a(n) = (n+1)*(7*n+2)*a(n-1) + 3*(n-2)*(7*n-4)*a(n-2) - 27*(n-2)*(n-1)*a(n-3) . - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 3^(2*n+9/2)/(16*Pi*n^3). - Vaclav Kotesovec, Oct 20 2012
a(n) = hypergeom([1/2, -n - 1, -n], [2, 2], 4). - Vaclav Kotesovec, May 14 2016

A166360 Triangle of Narayana numbers mod 2, T(n,k) = A001263(n,k) mod 2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1

Offset: 1

Gerald McGarvey, Oct 12 2009

			Triangle begins:
  1
  1 1
  1 1 1
  1 0 0 1
  1 0 0 0 1
  1 1 0 0 1 1
  1 1 1 1 1 1 1
  1 0 0 0 0 0 0 1
  1 0 0 0 0 0 0 0 1
  1 1 0 0 0 0 0 0 1 1
  1 1 1 0 0 0 0 0 1 1 1
  1 0 0 1 0 0 0 0 1 0 0 1
  1 0 0 0 1 0 0 0 1 0 0 0 1
  1 1 0 0 1 1 0 0 1 1 0 0 1 1
  1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
  ...

Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened

Cf. A001263, A047999, A007318.

Cf. A230116 (rows seen as binary numbers).

a166360 n k = a166360_tabl !! (n-1) !! (k-1)
a166360_row n = a166360_tabl !! (n-1)
a166360_tabl = map (map (flip mod 2)) a001263_tabl
-- Reinhard Zumkeller, Oct 10 2013

T[n_, k_] := Mod[Binomial[n-1, k-1] * Binomial[n, k-1] / k, 2]; Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Amiram Eldar, May 13 2025 *)

p = 2; s=14; NT = matrix(s,s,n,k, binomial(n-1, k-1)*binomial(n, k-1)/k);
NTMP = matrix(s,s,n,k, NT[n,k]%p);
for(n=1,s,for(k=1,n,print1(NTMP[n,k]," "));print())

A174128 Irregular triangle read by rows: row n (n > 0) is the expansion of Sum_{m=1..n} A001263(n,m)x^(m - 1)(1 - x)^(n - m).

Original entry on oeis.org

1, 1, 1, 1, -1, 1, 3, -3, 1, 6, -4, -4, 2, 1, 10, 0, -20, 10, 1, 15, 15, -55, 15, 15, -5, 1, 21, 49, -105, -35, 105, -35, 1, 28, 112, -140, -266, 364, -56, -56, 14, 1, 36, 216, -84, -882, 756, 336, -504, 126, 1, 45, 375, 210, -2100, 672, 2520, -2100, 210, 210, -42

Offset: 1

Roger L. Bagula, Mar 09 2010

Row n gives the coefficients in the expansion of (1/x)*(1 - x)^n*N(n,x/(1 - x)), where N(n,x) is the n-th row polynomial for the triangle of Narayana numbers A001263.

			Triangle begins
    1;
    1;
    1,  1,  -1;
    1,  3,  -3;
    1,  6,  -4,   -4,    2;
    1, 10,   0,  -20,   10;
    1, 15,  15,  -55,   15,  15,  -5;
    1, 21,  49, -105,  -35, 105, -35;
    1, 28, 112, -140, -266, 364, -56,  -56,  14;
    1, 36, 216,  -84, -882, 756, 336, -504, 126;
    ...

G. C. Greubel, Rows n = 1..50 of the irregular triangle, flattened
Michael Albert, Cheyne Homberger and Jay Pantone, Equipopularity Classes in the Separable Permutations, arXiv:1410.7312 [math.CO], 2014; see p. 13.
Wikipedia, Hypergeometric function

Cf. A122753, A123018, A123019, A123021, A123027, A123199, A123202, A123217, A123221, A141720, A144387, A144400.

p[x_, n_]:= p[x, n]= Sum[(Binomial[n, j]*Binomial[n, j-1]/n)*x^j*(1-x)^(n-j), {j, 1, n}]/x;
Table[CoefficientList[p[x, n], x], {n, 1, 12}]//Flatten

def p(n,x): return (1/(n*x))*sum( binomial(n,j)*binomial(n,j-1)*x^j*(1-x)^(n-j) for j in (1..n) )
def T(n): return ( p(n,x) ).full_simplify().coefficients(sparse=False)
[T(n) for n in (1..12)] # G. C. Greubel, Jul 14 2021

The n-th row of the triangle is generated by the coefficients of (1 - x)^(n - 1)*F(-n, 1 - n; 2; x/(1 - x)), where F(a, b ; c; z) is the ordinary hypergeometric function.

G.f.: (1 - y - sqrt(1 - 2*y + ((1 - 2*x)*y)^2))/(2*(1 - x)*x*y). - Franck Maminirina Ramaharo, Oct 23 2018

Edited and new name by Joerg Arndt, Oct 28 2014

Comments and formula clarified by Franck Maminirina Ramaharo, Oct 23 2018

cp's OEIS Frontend

A134290 Ninth column (and diagonal) of Narayana triangle A001263.

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A082148 a(0)=1; for n >= 1, a(n) = Sum_{k=0..n} 10^k*N(n,k), where N(n,k) = (1/n)*C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

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A082173 a(0)=1; for n >= 1, a(n) = Sum_{k=0..n} 11^k*N(n,k) where N(n,k) = (1/n)*C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

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A082181 a(0) = 1, for n>=1, a(n) = Sum_{k=0..n} 9^k*N(n,k), where N(n,k) = (1/n)*C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

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A103366 Second column of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).

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A145904 Square array read by antidiagonals: Hilbert transform of the Narayana numbers A001263.

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A082148 a(0)=1; for n >= 1, a(n) = Sum_{k=0..n} 10^kN(n,k), where N(n,k) = (1/n)C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

A082173 a(0)=1; for n >= 1, a(n) = Sum_{k=0..n} 11^kN(n,k) where N(n,k) = (1/n)C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

A082181 a(0) = 1, for n>=1, a(n) = Sum_{k=0..n} 9^kN(n,k), where N(n,k) = (1/n)C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

A174128 Irregular triangle read by rows: row n (n > 0) is the expansion of Sum_{m=1..n} A001263(n,m)x^(m - 1)(1 - x)^(n - m).