A088617 -id:A088617 - cp's OEIS frontend

A107131 A Motzkin related triangle.

1, 0, 1, 0, 1, 1, 0, 0, 3, 1, 0, 0, 2, 6, 1, 0, 0, 0, 10, 10, 1, 0, 0, 0, 5, 30, 15, 1, 0, 0, 0, 0, 35, 70, 21, 1, 0, 0, 0, 0, 14, 140, 140, 28, 1, 0, 0, 0, 0, 0, 126, 420, 252, 36, 1, 0, 0, 0, 0, 0, 42, 630, 1050, 420, 45, 1, 0, 0, 0, 0, 0, 0, 462, 2310, 2310, 660, 55, 1

Offset: 0

Paul Barry, May 12 2005

Row sums are Motzkin numbers A001006. Diagonal sums are A025250(n+1).
Inverse binomial transform of Narayana number triangle A001263. - Paul Barry, May 15 2005
T(n,k)=number of Motzkin paths of length n with k steps U=(1,1) or H=(1,0). Example: T(3,2)=3 because we have HUD, UDH and UHD (here D=(1,-1)). T(n,k) = number of bushes with n+1 edges and k+1 leaves (a bush is an ordered tree in which the outdegree of each nonroot node is at least two). - Emeric Deutsch, May 29 2005
Row reverse of A055151. - Peter Bala, May 07 2012
Rows of A088617 are shifted columns of A107131, whose reversed rows are the Motzkin polynomials of A055151, which give the row polynomials (mod signs) of the o.g.f. that is the compositional inverse for an o.g.f. of the Fibonacci polynomials of A011973. The diagonals of A055151 give the rows of A088671, and the antidiagonals (top to bottom) of A088617 give the rows of A107131. The diagonals of A107131 give the columns of A055151. From the relation between A088617 and A107131, the o.g.f. of this entry is (1 - t*x - sqrt((1-t*x)^2 - 4*t*x^2))/(2*t*x^2). - Tom Copeland, Jan 21 2016

			Triangle begins
  1;
  0,  1;
  0,  1,  1;
  0,  0,  3,  1;
  0,  0,  2,  6,  1;
  0,  0,  0, 10, 10,   1;
  0,  0,  0,  5, 30,  15,   1;
  0,  0,  0,  0, 35,  70,  21,   1;
  0,  0,  0,  0, 14, 140, 140,  28,  1;
  0,  0,  0,  0,  0, 126, 420, 252, 36, 1;

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened).
Marilena Barnabei, Flavio Bonetti, Niccolò Castronuovo, and Matteo Silimbani, Consecutive patterns in restricted permutations and involutions, arXiv:1902.02213 [math.CO], 2019.
Paul Barry, On the duals of the Fibonacci and Catalan-Fibonacci polynomials and Motzkin paths, arXiv:2101.10218 [math.CO], 2021.
Paul Barry and A. Hennessy, A Note on Narayana Triangles and Related Polynomials, Riordan Arrays, and MIMO Capacity Calculations, J. Int. Seq. 14 (2011), Article 11.3.8.

Cf. A001006 (row sums), A025250 (diag. sums), A055151 (row reverse).

Cf. A001263, A011973, A025250, A055151, A088617, A088671, A107131.

[Binomial(n, 2*(n-k))*Catalan(n-k): k in [0..n], n in [0..13]]; // G. C. Greubel, May 22 2022

egf := exp(t*x)*hypergeom([],[2],t*x^2);
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 29 2014

T[n_, k_] := Binomial[k+1, n-k+1] Binomial[n, k]/(k+1);
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Jun 19 2018 *)

flatten([[binomial(n, 2*(n-k))*catalan_number(n-k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 22 2022

Number triangle T(n, k) = binomial(k+1, n-k+1)*binomial(n, k)/(k+1).
T(n, k) = Sum_{j=0..n} (-1)^(n-j)C(n, j)*C(j+1, k)*C(j+1, k+1)/(j+1). - Paul Barry, May 15 2005
G.f.: G = G(t, z) satisfies G = 1 + t*z*G + t*z^2*G^2. - Emeric Deutsch, May 29 2005
Coefficient array for the polynomials x^n*Hypergeometric2F1((1-n)/2, -n/2; 2; 4/x). - Paul Barry, Oct 04 2008
From Paul Barry, Jan 12 2009: (Start)
G.f.: 1/(1-xy(1+x)/(1-x^2*y/(1-xy(1+x)/(1-x^2y/(1-xy(1+x).... (continued fraction).
T(n,k) = C(n, 2n-2k)*A000108(n-k). (End)

A007004 a(n) = (3n)! / ((n+1)(n!)^3).

Original entry on oeis.org

1, 3, 30, 420, 6930, 126126, 2450448, 49884120, 1051723530, 22787343150, 504636071940, 11377249621920, 260363981732400, 6034149862347600, 141371511060715200, 3343436236585914480, 79726203788589122490, 1914992149823954412750, 46295775130831740013500

Offset: 0

N. J. A. Sloane, James Propp

nonn

Number of walks with steps (0,1)/North, (1,0)/East and (-1,-1)/Southwest from (0,0) to (0,0) of length 3n, and staying above the line y=x (i.e., any point (x,y) along the walk satisfies y>=x ). - Shanzhen Gao, Nov 09 2010
Number of walks in 3-dimensions using steps (1,0,0), (0,1,0), and (0,0,1) from (0,0,0) to (n,n,n) such that after each step we have y<=x. - Eric Werley, Jun 24 2011
Number of possible necklaces consisting of n white beads, n-1 red beads and n-1 black beads, where two necklaces are considered equivalent if they differ by a cyclic permutation. - Thotsaporn Thanatipanonda, Feb 20 2011

			n=1, three walks: NE(SW), (SW)NE, N(SW)E. - _Shanzhen Gao_, Nov 09 2010

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..150
Alin Bostan, Calcul Formel pour la Combinatoire des Marches [The text is in English], Habilitation à Diriger des Recherches, Laboratoire d'Informatique de Paris Nord, Université Paris 13, December 2017.

Cf. A000108, A060693.

Row n=3 of A215561.

[Factorial(3*n) / ((n+1)*Factorial(n)^3): n in [0..30]]; // Vincenzo Librandi, May 26 2011

seq(binomial(2*n,n)*binomial(3*n,n)/(n+1), n=0..20); # Zerinvary Lajos, May 27 2006

a[n_]:=(3*n)!/((n + 1)*(n!)^3); (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)
CoefficientList[Series[Hypergeometric2F1[1/3,2/3,2,27 x],{x,0,20}],x] (* Harvey P. Dale, Apr 07 2013 *)
Table[Multinomial[n, n, n]/(n + 1), {n, 0, 12}] (* Emanuele Munarini, Oct 25 2016 *)

makelist(multinomial_coeff(n,n,n)/(n+1),n,0,24); /* Emanuele Munarini, Oct 25 2016 */

a(n) = C(2*n,n)*C(3*n,n)/(n+1) = A000108(n)*C(3*n,n). - Zerinvary Lajos, May 27 2006
a(n) = A060693(2n,n) = A088617(2n,n). - Philippe Deléham, Nov 23 2011
a(n) = (3*(3*n-1)*(3*n-2)*a(n-1))/(n*(n+1)) for n>0, a(0)=1. - Alois P. Heinz, Aug 13 2013
a(n) ~ 3^(3*n+1/2)/(2*Pi*n^2). - Vaclav Kotesovec, Sep 06 2016

More terms from Zerinvary Lajos, May 27 2006

A078018 a(n) = Sum_{k=0..n} 6^kN(n,k), with a(0)=1, where N(n,k) = C(n,k) C(n,k+1)/n are the Narayana numbers (A001263).

Original entry on oeis.org

1, 1, 7, 55, 469, 4237, 39907, 387739, 3858505, 39130777, 402972031, 4202705311, 44299426717, 471189693925, 5051001609115, 54513542257795, 591858123926545, 6459813793353265, 70837427884259575, 780073647992404615

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 6^C(n+1,2). - Philippe Deléham, Oct 29 2007
Shifts left when INVERT transform applied six 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, A007564, A059231.

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (1 + 5*x - Sqrt(25*x^2-14*x+1))/(12*x) )); // G. C. Greubel, Jun 29 2019

A078018_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]+6*add(a[j]*a[w-j-1],j=1..w-1) od;
convert(a, list) end: A078018_list(19);
# Peter Luschny, May 19 2011

Table[SeriesCoefficient[(1+5*x-Sqrt[25*x^2-14*x+1])/(12*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 13 2012 *)
a[n_]:= Hypergeometric2F1[1 - n, -n, 2, 6]; Table[a[n], {n, 0, 20}] (* Peter Luschny, Mar 19 2018 *)

a(n)=if(n<1,1,sum(k=0,n,6^k/n*binomial(n,k)*binomial(n,k+1)))

a=((1 + 5*x - sqrt(25*x^2-14*x+1))/(12*x)).series(x, 30).coefficients(x, sparse=False); [1]+a[1:] # G. C. Greubel, Jun 29 2019

G.f.: (1 + 5*x - sqrt(25*x^2-14*x+1))/(12*x).
a(n) = Sum_{k=0..n} A088617(n, k)*6^k*(-5)^(n-k). - Philippe Deléham, Jan 21 2004
a(n) = ( 7*(2*n-1)*a(n-1) - 25*(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;
  6, 6, 6;
  1, 1, 1, 1;
  6, 6, 6, 6, 6;
  1, 1, 1, 1, 1, 1;
  ... (End)
a(n) ~ sqrt(12+7*sqrt(6))*(7+2*sqrt(6))^n/(12*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 13 2012
G.f.: 1/(1 - x/(1 - 6*x/(1 - x/(1 - 6*x/(1 - x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, Apr 21 2017
a(n) = hypergeom([1 - n, -n], [2], 6). - Peter Luschny, Mar 19 2018

A008550 Table T(n,k), n>=0 and k>=0, read by antidiagonals: the k-th column given by the k-th Narayana polynomial.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 3, 1, 1, 1, 14, 11, 4, 1, 1, 1, 42, 45, 19, 5, 1, 1, 1, 132, 197, 100, 29, 6, 1, 1, 1, 429, 903, 562, 185, 41, 7, 1, 1, 1, 1430, 4279, 3304, 1257, 306, 55, 8, 1, 1, 1, 4862, 20793, 20071, 8925, 2426, 469, 71, 9, 1, 1

Offset: 0

Philippe Deléham, Jan 23 2004

Mirror image of A243631. - Philippe Deléham, Sep 26 2014

			Row n=0:  1, 1,  1,   1,    1,     1,      1, ... see A000012.
Row n=1:  1, 1,  2,   5,   14,    42,    132, ... see A000108.
Row n=2:  1, 1,  3,  11,   45,   197,    903, ... see A001003.
Row n=3:  1, 1,  4,  19,  100,   562,   3304, ... see A007564.
Row n=4:  1, 1,  5,  29,  185,  1257,   8925, ... see A059231.
Row n=5:  1, 1,  6,  41,  306,  2426,  20076, ... see A078009.
Row n=6:  1, 1,  7,  55,  469,  4237,  39907, ... see A078018.
Row n=7:  1, 1,  8,  71,  680,  6882,  72528, ... see A081178.
Row n=8:  1, 1,  9,  89,  945, 10577, 123129, ... see A082147.
Row n=9:  1, 1, 10, 109, 1270, 15562, 198100, ... see A082181.
Row n=10: 1, 1, 11, 131,  161,  1661,  22101, ... see A082148.
Row n=11: 1, 1, 12, 155, 2124, 30482, 453432, ... see A082173.
... - _Philippe Deléham_, Apr 03 2013
The first few rows of the antidiagonal triangle are:
  1;
  1,  1;
  1,  1,  1;
  1,  2,  1,  1;
  1,  5,  3,  1, 1;
  1, 14, 11,  4, 1, 1;
  1, 42, 45, 19, 5, 1, 1; - _G. C. Greubel_, Feb 15 2021

Michael De Vlieger, Table of n, a(n) for n = 0..11475
H. Prodinger, On a functional difference equation of Runyon, Morrison, Carlitz, and Riordan, arXiv:math/0103149 [math.CO], 2001.
H. Prodinger, On a functional difference equation of Runyon, Morrison, Carlitz, and Riordan, Séminaire Lotharingien de Combinatoire 46 (2001), Article B46a.
L. Yang, S.-L. Yang, A relation between Schroder paths and Motzkin paths, Graphs Combinat. 36 (2020) 1489-1502, eq. (6).

Columns: A000012, A000012, A000027, A028387, A090197, A090198, A090199, A090200.
Main diagonal is A242369.
A diagonal is in A099169.
Cf. A204057 (another version), A088617, A243631.
Cf. A132745.

[Truncate(HypergeometricSeries(k-n, k-n+1, 2, k)): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 15 2021

gf := n -> 2/(sqrt((n-1)^2*x^2-2*(n+1)*x+1)+(n-1)*x+1):
for n from 0 to 11 do PolynomialTools:-CoefficientList(convert( series(gf(n),x,12),polynom),x) od; # Peter Luschny, Nov 17 2014

(* First program *)
Unprotect[Power]; Power[0 | 0, 0 | 0] = 1; Protect[Power]; Table[Function[n, Sum[Apply[Binomial[#1 + #2, #1] Binomial[#1, #2]/(#2 + 1) &, {k, j}]*n^j*(1 - n)^(k - j), {j, 0, k}]][m - k + 1] /. k_ /; k <= 0 -> 1, {m, -1, 9}, {k, m + 1, 0, -1}] // Flatten (* Michael De Vlieger, Aug 10 2017 Note: this code renders 0^0 = 1. To restore normal Power functionality: Unprotect[Power]; ClearAll[Power]; Protect[Power] *)
(* Second program *)
Table[Hypergeometric2F1[1-n+k, k-n, 2, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Feb 15 2021 *)

flatten([[hypergeometric([k-n, k-n+1], [2], k).simplify_hypergeometric() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 15 2021

T(n, k) = Sum_{j>0} A001263(k, j)*n^(j-1); T(n, 0)=1.
T(n, k) = Sum_{j, 0<=j<=k} A088617(k, j)*n^j*(1-n)^(k-j).
The o.g.f. of row n is gf(n) = 2/(sqrt((n-1)^2*x^2-2*(n+1)*x+1)+(n-1)*x+1). - Peter Luschny, Nov 17 2014
G.f. of row n: 1/(1 - x/(1 - n*x/(1 - x/(1 - n*x/(1 - x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, Aug 10 2017
T(n, k) = Hypergeometric2F1([k-n, k-n+1], [2], k), as a number triangle. - G. C. Greubel, Feb 15 2021

A082147 a(0)=1; for n >= 1, a(n) = Sum_{k=0..n} 8^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, 9, 89, 945, 10577, 123129, 1476841, 18130401, 226739489, 2878666857, 37006326777, 480750990993, 6301611631473, 83240669582937, 1106980509493641, 14808497812637121, 199138509770855489, 2690461489090104009

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 8^C(n+1,2). - Philippe Deléham, Oct 29 2007
Shifts left when INVERT transform applied eight 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, A007564, A059231.

a:=n->Sum([0..n],k->8^k*(1/n)*Binomial(n,k)*Binomial(n,k+1));;
Concatenation([1],List([1..18],n->a(n))); # Muniru A Asiru, Feb 10 2018

Q:=Rationals(); R:=PowerSeriesRing(Q, 40); Coefficients(R!((1+7*x-Sqrt(49*x^2-18*x+1))/(16*x))) // G. C. Greubel, Feb 05 2018

A082147_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]+8*add(a[j]*a[w-j-1],j=1..w-1) od;
convert(a, list) end: A082147_list(18); # Peter Luschny, May 19 2011

Table[SeriesCoefficient[(1+7*x-Sqrt[49*x^2-18*x+1])/(16*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
f[n_] := Sum[ 8^k*Binomial[n, k]*Binomial[n, k + 1]/n, {k, 0, n}]; f[0] = 1; Array[f, 21, 0] (* Robert G. Wilson v, Feb 24 2018 *)
a[n_] := Hypergeometric2F1[1 - n, -n, 2, 8];
Table[a[n], {n, 0, 18}] (* Peter Luschny, Mar 19 2018 *)

a(n)=if(n<1,1,sum(k=0,n,8^k/n*binomial(n,k)*binomial(n,k+1)))

G.f.: (1 + 7*x - sqrt(49*x^2-18*x+1))/(16*x).
a(n) = Sum_{k=0..n} A088617(n, k)*8^k*(-7)^(n-k). - Philippe Deléham, Jan 21 2004
a(n) = (9(2n-1)a(n-1) - 49(n-2)a(n-2)) / (n+1) for n >= 2, a(0) = a(1) = 1. - Philippe Deléham, Aug 19 2005
a(n) = upper left term in M^n, M = the production matrix:
  1, 1
  8, 8, 8
  1, 1, 1, 1
  8, 8, 8, 8, 8
  1, 1, 1, 1, 1, 1
  ...
- Gary W. Adamson, Jul 08 2011
a(n) ~ sqrt(16+18*sqrt(2))*(9+4*sqrt(2))^n/(16*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2012
G.f.: 1/(1 - x/(1 - 8*x/(1 - x/(1 - 8*x/(1 - x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, Apr 21 2017
a(n) = hypergeom([1 - n, -n], [2], 8). - Peter Luschny, Mar 19 2018

A000910 a(n) = 5*binomial(n, 6).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 5, 35, 140, 420, 1050, 2310, 4620, 8580, 15015, 25025, 40040, 61880, 92820, 135660, 193800, 271320, 373065, 504735, 672980, 885500, 1151150, 1480050, 1883700, 2375100, 2968875, 3681405, 4530960, 5537840, 6724520, 8115800, 9738960, 11623920, 13803405

Offset: 0

N. J. A. Sloane

Charles Jordan, Calculus of Finite Differences, Budapest, 1939, p. 449.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

A diagonal of A088617.

Cf. A080159, A000579, A210569.

Table[5Binomial[n, 6], {n, 0, 100}] (* Stefan Steinerberger, Apr 30 2006 *)

a(n)=5*binomial(n,6) \\ Charles R Greathouse IV, Oct 07 2015

[5*binomial(n,6) for n in (0..40)] # G. C. Greubel, May 22 2022

a(n) = 5*A000579(n+3) = A080159(n+3, 3).
G.f.: 5*x^6/(1-x)^7. - Colin Barker, Mar 01 2012
E.g.f.: x^6*exp(x)/144. - G. C. Greubel, May 22 2022
From Amiram Eldar, Jul 19 2022: (Start)
Sum_{n>=6} 1/a(n) = 6/25.
Sum_{n>=6} (-1)^n/a(n) = 192*log(2)/5 - 661/25. (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

A133305 a(n) = (1/n)Sum_{i=0..n-1} C(n,i)C(n,i+1)4^i5^(n-i), a(0) = 1.

Original entry on oeis.org

1, 5, 45, 505, 6345, 85405, 1204245, 17558705, 262577745, 4005148405, 62070886845, 974612606505, 15471084667545, 247876665109005, 4003225107031845, 65101209768055905, 1065128963164067745, 17520376884067071205, 289572455530026439245, 4806489064223483202905

Offset: 0

Philippe Deléham, Oct 18 2007

nonn

Fifth column of array A103209.

The Hankel transform of this sequence is 20^C(n+1,2). - Philippe Deléham, Oct 28 2007

G. C. Greubel, Table of n, a(n) for n = 0..795
Samuele Giraudo, Operads from posets and Koszul duality, arXiv:1504.04529 [math.CO], 2015-2016.

Cf. A000108, A060693, A103209, A103210, A103211.

Q:=Rationals(); R:=PowerSeriesRing(Q, 40); Coefficients(R!((1-x-Sqrt(x^2-18*x+1))/(8*x))) // G. C. Greubel, Feb 10 2018

a[n_] := Hypergeometric2F1[-n, n + 1, 2, -4];
Table[a[n], {n, 0, 16}] (* Peter Luschny, Jan 08 2018 *)
CoefficientList[Series[(1-x-Sqrt[x^2-18*x+1])/(8*x), {x, 0, 50}], x] (* G. C. Greubel, Feb 10 2018 *)

x='x+O('x^30); Vec((1-x-sqrt(x^2-18*x+1))/(8*x)) \\ G. C. Greubel, Feb 10 2018

G.f.: (1-z-sqrt(z^2-18*z+1))/(8*z).
a(n) = Sum_{k=0..n} A088617(n,k)*4^k.
a(n) = Sum_{k=0..n} A060693(n,k)*4^(n-k).
a(n) = Sum_{k=0..n} C(n+k, 2k)*4^k*C(k), C(n) given by A000108.
a(0) = 1, a(n) = a(n-1) + 4*Sum_{k=0..n-1} a(k)*a(n-1-k). - Philippe Deléham, Oct 23 2007
Conjecture: (n+1)*a(n) + 9*(-2*n+1)*a(n-1) + (n-2)*a(n-2) = 0. - R. J. Mathar, May 23 2014
G.f.: 1/(1 - 5*x/(1 - 4*x/(1 - 5*x/(1 - 4*x/(1 - 5*x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, May 10 2017
a(n) = hypergeom([-n, n + 1], [2], -4). - Peter Luschny, Jan 08 2018
a(n) ~ 5^(1/4) * phi^(6*n + 3) / (2^(5/2) * sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Nov 21 2021

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A107131 A Motzkin related triangle.

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A007004 a(n) = (3*n)! / ((n+1)*(n!)^3).

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A078018 a(n) = Sum_{k=0..n} 6^k*N(n,k), with a(0)=1, where N(n,k) = C(n,k) * C(n,k+1)/n are the Narayana numbers (A001263).

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A008550 Table T(n,k), n>=0 and k>=0, read by antidiagonals: the k-th column given by the k-th Narayana polynomial.

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A082147 a(0)=1; for n >= 1, a(n) = Sum_{k=0..n} 8^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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A000910 a(n) = 5*binomial(n, 6).

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A007004 a(n) = (3n)! / ((n+1)(n!)^3).

A078018 a(n) = Sum_{k=0..n} 6^kN(n,k), with a(0)=1, where N(n,k) = C(n,k) C(n,k+1)/n are the Narayana numbers (A001263).

A082147 a(0)=1; for n >= 1, a(n) = Sum_{k=0..n} 8^kN(n,k) where N(n,k) = (1/n)C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

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).

A133305 a(n) = (1/n)Sum_{i=0..n-1} C(n,i)C(n,i+1)4^i5^(n-i), a(0) = 1.