A002740 -id:A002740 - cp's OEIS frontend

A217213 2*A002740(n).

0, 0, 0, 4, 30, 168, 840, 3960, 18018, 80080, 350064, 1511640, 6466460, 27457584, 115892400, 486748080, 2035917450, 8485840800, 35263382880, 146157442200, 604404010980, 2494365759600, 10275832148400, 42264944401680, 173588164506900, 712027089322848, 2917167840871200, 11938779496898800

Offset: 0

N. J. A. Sloane, Oct 03 2012

nonn

Luca Ferrari and Emanuele Munarini, Enumeration of saturated chains in Dyck lattices, arXiv preprint arXiv:1203.6807, 2012

Cf. A002740.

A253180 Number T(n,k) of 2n-length strings of balanced parentheses of exactly k different types that are introduced in ascending order; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 2, 2, 0, 5, 15, 5, 0, 14, 98, 84, 14, 0, 42, 630, 1050, 420, 42, 0, 132, 4092, 11880, 8580, 1980, 132, 0, 429, 27027, 129129, 150150, 60060, 9009, 429, 0, 1430, 181610, 1381380, 2432430, 1501500, 380380, 40040, 1430, 0, 4862, 1239810, 14707550, 37777740, 33795762, 12864852, 2246244, 175032, 4862

Offset: 0

Alois P. Heinz, Mar 23 2015

In general, column k>0 is asymptotic to (4*k)^n / (k!*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jun 01 2015

			T(3,1) = 5: ()()(), ()(()), (())(), (()()), ((())).
T(3,2) = 15: ()()[], ()[](), ()[][], ()([]), ()[()], ()[[]], (())[], ([])(), ([])[], (()[]), ([]()), ([][]), (([])), ([()]), ([[]]).
T(3,3) = 5: ()[]{}, ()[{}], ([]){}, ([]{}), ([{}]).
Triangle T(n,k) begins:
  1;
  0,   1;
  0,   2,     2;
  0,   5,    15,      5;
  0,  14,    98,     84,     14;
  0,  42,   630,   1050,    420,    42;
  0, 132,  4092,  11880,   8580,  1980,  132;
  0, 429, 27027, 129129, 150150, 60060, 9009, 429;
  ...

Alois P. Heinz, Rows n = 0..140, flattened

Columns k=0-10 give: A000007, A000108 (for n>0), A258390, A258391, A258392, A258393, A258394, A258395, A258396, A258397, A258398.
Main diagonal gives A000108.
First lower diagonal gives A002740(n+2).
T(2n,n) gives A258399.
Row sums give A064299.
Cf. A000142, A256061.

ctln:= proc(n) option remember; binomial(2*n, n)/(n+1) end:
A:= proc(n, k) option remember; k^n*ctln(n) end:
T:= (n, k)-> add(A(n, k-i)*(-1)^i/((k-i)!*i!), i=0..k):
seq(seq(T(n, k), k=0..n), n=0..10);

A[n_, k_] := A[n, k] = k^n*CatalanNumber[n]; T[0, 0] = 1; T[n_, k_] := Sum[A[n, k-i]*(-1)^i/((k-i)!*i!), {i, 0, k}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 11 2017, adapted from Maple *)

T(n,k) = A256061(n,k)/k! = Sum_{i=0..k} (-1)^i * C(k,i) * (k-i)^n * A000108(n) / A000142(n).

A129175 Triangle read by rows: MacMahon's q-analog of the Catalan numbers.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 1, 0, 1, 1, 0, 1, 1, 2, 2, 3, 2, 4, 3, 4, 3, 4, 2, 3, 2, 2, 1, 1, 0, 1, 1, 0, 1, 1, 2, 2, 4, 3, 5, 5, 7, 6, 9, 7, 9, 8, 9, 7, 9, 6, 7, 5, 5, 3, 4, 2, 2, 1, 1, 0, 1, 1, 0, 1, 1, 2, 2, 4, 4, 6, 6, 9, 9, 13, 12, 16, 16, 19, 18, 22, 20, 23, 21, 23

Offset: 0

Emeric Deutsch, Apr 20 2007

Previous name: T(n,k) is the number of Dyck words of length 2n having major index k (n >= 0, k >= 0). A Dyck word of length 2n is a word of n 0's and n 1's for which no initial segment contains more 1's than 0's. The major index of a Dyck word is the sum of the positions of those 1's that are followed by a 0.
Representing a Dyck word p of length 2n as a Dyck path p', the major index of p is equal to the sum of the abscissae of the valleys of p'.
Row n has 1+n*(n-1) terms.
Row sums are the Catalan numbers (A000108).
T(n,k) = T(n,n^2-n-k) (i.e., rows are palindromic).
Alternating row sums are the central binomial coefficients binomial(n, floor(n/2)) = A001405(n).
Sum_{k=0..n*(n-1)} k*T(n,k) = A002740(n+1).
T(n,k) = A129174(n,n+k) (i.e., except for the initial 0's, rows of A129174 and A129175 are the same).
For another q-analog of the Catalan numbers, due to Carlitz and Riordan, that enumerates Dyck paths by an area statistic see A227543. - Peter Bala, Feb 28 2023

			T(4,8)=2 because we have 01001101 (with 10's starting at positions 2 and 6) and 00101011 (with 10's starting at positions 3 and 5).
Triangle starts:
  1;
  1;
  1,0,1;
  1,0,1,1,1,0,1;
  1,0,1,1,2,1,2,1,2,1,1,0,1;
  1,0,1,1,2,2,3,2,4,3,4,3,4,2,3,2,2,1,1,0,1;

G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976.
P. A. MacMahon, Combinatory analysis, Two volumes (bound as one), Chelsea Publishing Co., New York, 1960 (see p. 214).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see p. 236, Exercise 6.34 b. [From Emeric Deutsch, Nov 05 2008]

Alois P. Heinz, Rows n = 0..32, flattened
FindStat - Combinatorial Statistic Finder, The major index of a Dyck path.
J. Furlinger and J. Hofbauer, q-Catalan numbers, J. Comb. Theory, A, 40, 248-264, 1985.
M. Shattuck, Parity theorems for statistics on permutations and Catalan words, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 5, Paper A07, 2005.

Cf. A000108, A001405, A002740, A129174, A227543.

br:=n->sum(q^i,i=0..n-1): f:=n->product(br(j),j=1..n): cbr:=(n,k)->f(n)/f(k)/f(n-k): P:=n->sort(expand(simplify(cbr(2*n,n)/br(n+1)))): for n from 0 to 7 do seq(coeff(P(n),q,k),k=0..n*(n-1)) od; # yields sequence in triangular form
# second Maple program:
b:= proc(x, y, t) option remember; `if`(y<0 or y>x, 0, `if`(x=0, 1,
      expand(b(x-1, y-1, true)+b(x-1, y+1, false)*`if`(t, z^x, 1))))
    end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(2*n, 0, false)):
seq(T(n), n=0..8);  # Alois P. Heinz, Sep 15 2014

b[x_, y_, t_] := b[x, y, t] = If[y<0 || y>x, 0, If[x == 0, 1, Expand[b[x-1, y-1, True] + b[x-1, y+1, False]*If[t, z^x, 1]]]]; T[n_] := Function[{p}, Table[ Coefficient[p, z, i], {i, 0, Exponent[p, z]}]][b[2*n, 0, False]]; Table[T[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, May 26 2015, after Alois P. Heinz *)
p[n_] := QBinomial[2n,n,q]/QBinomial[n+1,1,q]; Table[CoefficientList[p[n] // FunctionExpand, q], {n,0,9}] // Flatten (* Peter Luschny, Jul 22 2016 *)

from sage.combinat.q_analogues import q_catalan_number
def T(n): return list(q_catalan_number(n))
for n in (0..6): print(T(n)) # Peter Luschny, Jul 19 2016

The generating polynomial for row n is P[n](t) = binomial[2n,n]/[n+1], where [n+1]=1+t+t^2+...+t^n and binomial[2n,n] is a Gaussian polynomial (due to MacMahon).

New name from Peter Luschny, Jul 24 2016

A342987 Triangle read by rows: T(n,k) is the number of tree-rooted planar maps with n edges, k faces and no isthmuses, n >= 0, k = 1..n+1.

Original entry on oeis.org

1, 0, 1, 0, 2, 2, 0, 3, 15, 5, 0, 4, 60, 84, 14, 0, 5, 175, 650, 420, 42, 0, 6, 420, 3324, 5352, 1980, 132, 0, 7, 882, 13020, 42469, 37681, 9009, 429, 0, 8, 1680, 42240, 246540, 429120, 239752, 40040, 1430, 0, 9, 2970, 118998, 1142622, 3462354, 3711027, 1421226, 175032, 4862

Offset: 0

Andrew Howroyd, Apr 03 2021

The number of vertices is n + 2 - k.

For k >= 2, column k is a polynomial of degree 4*(k-2)+1.

			Triangle begins:
  1;
  0, 1;
  0, 2,    2;
  0, 3,   15,     5;
  0, 4,   60,    84,     14;
  0, 5,  175,   650,    420,     42;
  0, 6,  420,  3324,   5352,   1980,    132;
  0, 7,  882, 13020,  42469,  37681,   9009,   429;
  0, 8, 1680, 42240, 246540, 429120, 239752, 40040, 1430;
  ...

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)
T. R. S. Walsh and A. B. Lehman, Counting rooted maps by genus. III: Nonseparable maps, J. Combinatorial Theory Ser. B 18 (1975), 222-259, Table VIIIb.

Columns k=1..4 are A000007, A000027, A006470, A006471.
Diagonals are A000108, A002740, A006432, A006433.
Row sums are A342988.
Cf. A342981, A342982, A342984, A342985.

\\ here G(n,y) is A342984 as g.f.
F(n,y)={sum(n=0, n, x^n*sum(i=0, n, my(j=n-i); y^i*(2*i+2*j)!/(i!*(i+1)!*j!*(j+1)!))) + O(x*x^n)}
G(n,y)={my(g=F(n,y)); subst(g, x, serreverse(x*g^2))}
H(n)={my(g=G(n,y)-x, v=Vec(sqrt(serreverse(x/g^2)/x))); [Vecrev(t) | t<-v]}
{ my(T=H(8)); for(n=1, #T, print(T[n])) }

G.f.: A(x,y) satisfies A(x,y) = G(x*A(x,y)^2,y) where G(x,y) + x is the g.f. of A342984.

A097070 Consider all compositions (ordered partitions) of n into n parts, allowing zeros. E.g., for n = 3 we get 300, 030, 003, 210, 120, 201, 102, 021, 012, 111. Then a(n) is the total number of 1's.

Original entry on oeis.org

1, 2, 9, 40, 175, 756, 3234, 13728, 57915, 243100, 1016158, 4232592, 17577014, 72804200, 300874500, 1240940160, 5109183315, 21002455980, 86213785350, 353452638000, 1447388552610, 5920836618840, 24197138082780, 98801168731200, 403095046038750, 1643337883690776, 6694900194799404

Offset: 1

Amy J. Kolan, Sep 15 2004

Number of compositions of n into n parts, allowing zeros = binomial(2*n-1,n) = A088218 = essentially A001700.

			The compositions for n=2 are 20, 02, 11. There are two 1's in these so a(2) = 2.
From _Robert G. Wilson v_, Sep 16 2004: (Start)
The case n = 5:
A. There are 5 combinations associated with the numbers 50000: 50000, 05000, 00500, 00050, 00005.
B. There are 20 combinations associated with the numbers 41000.
C. There are 20 combinations associated with 32000.
D. There are 30 combinations associated with 31100.
E. There are 30 combinations associated with 22100.
F. There are 20 combinations associated with 21110.
G. There is one combinations associated with 11111.
The number of 1's associated with A is 0, with B 20, with C 0, with D 60, with E 30, with F 60 and with G 5. 0 + 20 + 0 + 60 + 30 + 60 + 5 = 175.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
G.-S. Cheon, H. Kim, and L. W. Shapiro, Mutation effects in ordered trees, arXiv preprint arXiv:1410.1249 [math.CO], 2014.

Cf. A037965.
Cf. A000984, A001622, A088218.
cf. A005430, A002740, A002457, A000917, A110609, A097613, A001791, A110609.

List([1..30], n-> n*Binomial(2*n-3, n-1)); # G. C. Greubel, Jul 27 2019

[n*Binomial(2*n-3, n-1): n in [1..30]]; // Vincenzo Librandi, Jul 13 2019

A097070 := n -> ifelse(n=1, 1, 2^(n-2)*JacobiP(n-1, -1/2, -n+2, 3)):
seq(simplify(A097070(n)), n = 1..28);  # Peter Luschny, Jan 22 2025

Table[n*Binomial[2n-3, n-1], {n, 30}] (* Robert G. Wilson v, Sep 17 2004 *)

a(n) = n*binomial(2*n-3, n-1); \\ Joerg Arndt, Feb 17 2015

[n*binomial(2*n-3, n-1) for n in (1..30)] # G. C. Greubel, Jul 27 2019

a(n) = n*binomial(2*n-3, n-1).
More generally, total number of k's (k>=0) in all ordered partitions of n into n parts, allowing zeros, is n*binomial(2*n-k-2, n-2) if n >= k, 0 otherwise.
Total number of 0's is given by A005430.
From Vladeta Jovovic, Sep 17 2004: (Start)
a(n) = Sum_{k=0..n} k*binomial(n, k)*binomial(n-2, k-2).
G.f.: x*(1 -2*x +(1-4*x)^(3/2))/(2*(1-4*x)^(3/2)).
E.g.f.: (x/2)*(exp(2*x)*BesselI(0, 2*x)+1). (End)
a(n) = A014107(n)*A000108(n-2). - Philippe Deléham, Apr 12 2007
a(n) = n*A088218(n-1) for n > 0. - Werner Schulte, Jan 22 2017
From Bruce J. Nicholson, Jul 11 2019: (Start)
a(n) = A002740(n) + A097613(n).
a(n) = A110609(n-1) - A002457(n-2) + A097613(n).
a(n) = A005430(n-1) - A000917(n-3) for n > 1.
a(n) = A002457(n-1) - A037965(n) - A000917(n-3) for n > 1.
a(n) = A037965(n)/2.
a(n) = A001700(n-2)*n.
a(n) = A001791(n-2)*n + A000984(n-2)*n for n > 1. (End)
From Amiram Eldar, May 16 2022: (Start)
Sum_{n>=1} 1/a(n) = 4*Pi/(3*sqrt(3)) - Pi^2/9.
Sum_{n>=1} (-1)^(n+1)/a(n) = 8*log(phi)/sqrt(5) - 4*log(phi)^2, where phi is the golden ratio (A001622). (End)
a(n) = 2^(n-2)*JacobiP(n-1, -1/2, -n+2, 3) for n > 1. - Peter Luschny, Jan 22 2025

Formula, more terms and comments from Vladeta Jovovic, Sep 15 2004

A181204 T(n,k) = 0!1!2!...(k-1)! (nk)! k(k-1)n(n-1) / 2n!(n+1)!...(n+k-1)!

Original entry on oeis.org

0, 0, 0, 0, 4, 0, 0, 30, 30, 0, 0, 168, 756, 168, 0, 0, 840, 16632, 16632, 840, 0, 0, 3960, 360360, 1729728, 360360, 3960, 0, 0, 18018, 7876440, 199536480, 199536480, 7876440, 18018, 0, 0, 80080, 174594420, 25241364720, 140229804000, 25241364720

Offset: 1

R. H. Hardin Oct 10 2010

(Empricial) T(n,k)=Number of nXk matrices containing a defective permutation of 1..n*k in strictly increasing order rowwise and columnwise, with one permutation value omitted and one repeated (see example)
Formula is n*(n-1)*k*(k-1)/2 times n-th k-dimensional Catalan number
Table starts
.0.......0.............0....................0...........................0
.0.......4............30..................168.........................840
.0......30...........756................16632......................360360
.0.....168.........16632..............1729728...................199536480
.0.....840........360360............199536480................140229804000
.0....3960.......7876440..........25241364720.............118949931243000
.0...18018.....174594420........3445446284280..........117015012361447200
.0...80080....3926434512......500598983364480.......129624266420759510400
.0..350064...89492111280....76591644454765440....158211402715245473193600
.0.1511640.2064420294300.12237255920840932800.209298196564031904834960000

			Some solutions for 4X2
..2..4....1..4....1..3....1..3....2..5....1..3....1..3....2..4....1..2....1..3
..3..5....2..5....3..6....2..4....3..6....2..4....2..4....3..6....2..4....2..4
..5..7....6..7....4..7....4..6....4..7....4..5....4..7....5..7....3..5....4..7
..6..8....7..8....5..8....5..8....5..8....7..8....5..8....6..8....6..8....6..8

R. H. Hardin, Table of n, a(n) for n=1..1000

Column 2 is twice A002740(n+1)

Cf. A060854 for permutation without defect.

A091811 Array read by rows: T(n,k) = binomial(n+k-2,k-1)binomial(2n-1,n-k).

Original entry on oeis.org

1, 3, 2, 10, 15, 6, 35, 84, 70, 20, 126, 420, 540, 315, 70, 462, 1980, 3465, 3080, 1386, 252, 1716, 9009, 20020, 24024, 16380, 6006, 924, 6435, 40040, 108108, 163800, 150150, 83160, 25740, 3432, 24310, 175032, 556920, 1021020, 1178100, 875160

Offset: 1

Benoit Cloitre, Mar 18 2004

Alternating sum of elements of n-th row = 1.
If a certain event has a probability p of occurring in any given trial, the probability of its occurring at least n times in 2n-1 trials is Sum_{k=1..n} T(n,k)*(-1)^(k-1)*p^(n+k-1). For example, the probability of its occurring at least 4 out of 7 times is 35p^4 - 84p^5 + 70p^6 - 20p^7. - Matthew Vandermast, Jun 05 2004
With the row polynomial defined as R(n,x) = Sum_{k = 1..n} T(n,k)*x^k, the row polynomial is related to the regularized incomplete Beta function I_x(a,b), through the relation R(n,x) = -(-x)^{-n+1}*I_{-x}(n,n). - Leo C. Stein, Jun 06 2019

			Triangle starts:
    1,
    3,   2,
   10,  15,   6,
   35,  84,  70,  20,
  126, 420, 540, 315, 70,
  ...

Cf. A001700 (first column), A002740 (second column), A000984 (main diagonal), A033876 (second diagonal), A178792 (row sums).

[[Binomial(n+k-2,k-1)*Binomial(2*n-1,n-k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Jun 15 2015

t[n_, k_] := Binomial[n+k-2, k-1]*Binomial[2n-1, n-k]; Table[t[n, k], {n, 1, 9}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 06 2012 *)

T(x,y)=binomial(x+y-2,y-1)*binomial(2*x-1,x-y)

From Peter Bala, Apr 10 2012: (Start)
O.g.f.: x*t*(1+2*x-sqrt(1-4*t*(x+1)))/(2*(x+t)*sqrt(1-4*t*(x+1))) = x*t + (3*x+2*x^2)*t^2 + (10*x+15*x^2+6*x^3)*t^3 + ....
Sum_{k = 1..n} (-1)^(k-1)*T(n,k)*2^(n-k) = 4^(n-1).
Row polynomial R(n+1,x) = ((2*n+1)!/n!^2)*x*Integral_{y = 0..1} (y*(1+x*y))^n dy. Row sums A178792. (End)

A145885 a(n) = (n-1)^2binomial(2n,n)/(2(n+1)).

Original entry on oeis.org

0, 1, 10, 63, 336, 1650, 7722, 35035, 155584, 680238, 2939300, 12584726, 53488800, 225990180, 950094810, 3977737875, 16594533120, 69018792150, 286296636780, 1184823735810, 4893253404000, 20171905282620, 83020426503300

Offset: 1

Emeric Deutsch, Nov 06 2008

a(n) = sum of valley abscissae in all Dyck paths of semilength n minus number of valleys in all Dyck paths of semilength (n). Example: a(3)=10; indeed, the Dyck paths of semilength 3, followed by their valley abscissae are UDUDUD (2,4), UDUUDD (2), UUDDUD (4), UUDUDD (3), UUUDDD ( ); therefore a(3)=2+4+2+4+3 - 5 = 10. Instead of Dyck paths one can consider Dyck words; then sum of valley abscissae corresponds to major index and number of valleys to number of descents.

R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see p. 236, Exercise 6.34 d.

Cf. A000108, A002740, A002054, A145884.

seq((1/2)*(n-1)^2*binomial(2*n,n)/(n+1),n=1..24);

Table[CatalanNumber[n]*(n - 1)^2/2, {n, 1, 23}] (* Zerinvary Lajos, Jul 08 2009 *)

a(n) = A002740(n+1) - A002054(n-1) (n >= 2).
a(n) = Sum_{k=0..(n-1)^2} k*A145884(n,k) for n >= 1.
a(n) = (n-1)^2*Cat(n)/2, where Cat(n)=binomial(2n,n)/(n+1)=A000108(n) are the Catalan numbers.
G.f.: 4*z^2*(8*z - 1 + 3*sqrt(1-4*z))/((1 + sqrt(1-4*z))^3*(1-4*z)^(3/2)).
D-finite with recurrence (n+1)*(n-2)^2*a(n) - 2*(2*n-1)*(n-1)^2*a(n-1) = 0. - R. J. Mathar, Aug 10 2017
E.g.f.: (exp(2*x) * ((1 + 2*x) * BesselI(0,2*x) - 2 * (2 - x) * BesselI(1,2*x)) - 1) / 2. - Ilya Gutkovskiy, Nov 03 2021

A337994 T(n, k) = (k(2k+2)(2k+1)(2n-1)binomial(2(n-1),n-1))/(n(n+1)(n+2)) for n, k > 0 and T(0, 0) = 1. Triangle read by rows, for 0 <= k <= n.

Original entry on oeis.org

1, 0, 2, 0, 3, 15, 0, 6, 30, 84, 0, 14, 70, 196, 420, 0, 36, 180, 504, 1080, 1980, 0, 99, 495, 1386, 2970, 5445, 9009, 0, 286, 1430, 4004, 8580, 15730, 26026, 40040, 0, 858, 4290, 12012, 25740, 47190, 78078, 120120, 175032

Offset: 0

Peter Luschny, Nov 01 2020

T(n, k) is divisible by A099398(n) for all 0 <= k <= n.

			Triangle starts:
[0] 1
[1] 0, 2
[2] 0, 3,    15
[3] 0, 6,    30,    84
[4] 0, 14,   70,    196,   420
[5] 0, 36,   180,   504,   1080,  1980
[6] 0, 99,   495,   1386,  2970,  5445,   9009
[7] 0, 286,  1430,  4004,  8580,  15730,  26026,  40040
[8] 0, 858,  4290,  12012, 25740, 47190,  78078,  120120, 175032
[9] 0, 2652, 13260, 37128, 79560, 145860, 241332, 371280, 541008, 755820

Cf. A119578 (row sums), (-1)^n*A005430 (alternating row sums), A002740 (main diagonal), A007054 (col 1), A099398 (universal divisor), A000108 (Catalan).

T := proc(n, k) if n = 0 then 1 else
(k*(2*k+2)*(2*k+1)*(2*n-1)*binomial(2*(n-1), n-1))/(n*(n+1)*(n+2)) fi end:
# Recursive:
CatalanNumber := n -> binomial(2*n, n)/(n+1):
T := proc(n, k) option remember; if k=0 then k^n elif k=n then CatalanNumber(n+1)* binomial(n+1, 2) else (4 - 10/(n + 2))*T(n-1, k) fi end:
seq(seq(T(n, k), k=0..n), n=0..9);

T[n_, k_] := If[n == 0, 1, (k (2k + 2)(2k + 1)(2n - 1) CatalanNumber[n-1])/((n + 1) (n + 2))]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten

Let t(n) denote the triangular numbers and C(n) the Catalan numbers.
T(n, k) = k*(2*n - 1)*(t(2*k + 1)/t(n + 1))*C(n - 1) for n, k > 0.
T(n, k) = k^n if k = 0; if k = n then C(n+1)*t(n+1); else T(n-1, k)*(4-10/(n+2)).

A342311 T(n, k) = (n - k + 2)binomial(2n, n + k - 2). Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

0, 0, 2, 4, 12, 12, 30, 60, 60, 30, 168, 280, 280, 168, 56, 840, 1260, 1260, 840, 360, 90, 3960, 5544, 5544, 3960, 1980, 660, 132, 18018, 24024, 24024, 18018, 10010, 4004, 1092, 182, 80080, 102960, 102960, 80080, 48048, 21840, 7280, 1680, 240

Offset: 0

Peter Luschny, Mar 08 2021

			[0] 0
[1] 0,      2
[2] 4,      12,     12
[3] 30,     60,     60,     30
[4] 168,    280,    280,    168,    56
[5] 840,    1260,   1260,   840,    360,    90
[6] 3960,   5544,   5544,   3960,   1980,   660,    132
[7] 18018,  24024,  24024,  18018,  10010,  4004,   1092,  182
[8] 80080,  102960, 102960, 80080,  48048,  21840,  7280,  1680,  240
[9] 350064, 437580, 437580, 350064, 222768, 111384, 42840, 12240, 2448, 306

Cf. A002939 (main diagonal), A217213, A002740.

T := (n, k) -> (n - k + 2)*binomial(2*n, n + k - 2):
seq(seq(T(n, k), k=0..n), n=0..9);

cp's OEIS Frontend

A217213 2*A002740(n).

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A253180 Number T(n,k) of 2n-length strings of balanced parentheses of exactly k different types that are introduced in ascending order; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A129175 Triangle read by rows: MacMahon's q-analog of the Catalan numbers.

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A342987 Triangle read by rows: T(n,k) is the number of tree-rooted planar maps with n edges, k faces and no isthmuses, n >= 0, k = 1..n+1.

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A097070 Consider all compositions (ordered partitions) of n into n parts, allowing zeros. E.g., for n = 3 we get 300, 030, 003, 210, 120, 201, 102, 021, 012, 111. Then a(n) is the total number of 1's.

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A181204 T(n,k) = 0!*1!*2!*...*(k-1)! *(n*k)! *k*(k-1)*n*(n-1) / 2*n!*(n+1)!*...*(n+k-1)!

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A091811 Array read by rows: T(n,k) = binomial(n+k-2,k-1)*binomial(2*n-1,n-k).

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A181204 T(n,k) = 0!1!2!...(k-1)! (nk)! k(k-1)n(n-1) / 2n!(n+1)!...(n+k-1)!

A091811 Array read by rows: T(n,k) = binomial(n+k-2,k-1)binomial(2n-1,n-k).

A145885 a(n) = (n-1)^2binomial(2n,n)/(2(n+1)).

A337994 T(n, k) = (k(2k+2)(2k+1)(2n-1)binomial(2(n-1),n-1))/(n(n+1)(n+2)) for n, k > 0 and T(0, 0) = 1. Triangle read by rows, for 0 <= k <= n.

A342311 T(n, k) = (n - k + 2)binomial(2n, n + k - 2). Triangle read by rows, T(n, k) for 0 <= k <= n.