A051960 -id:A051960 - cp's OEIS frontend

A228329 a(n) = Sum_{k=0..n} (k+1)^2*T(n,k)^2 where T(n,k) is the Catalan triangle A039598.

1, 8, 98, 1320, 18590, 268736, 3952228, 58837680, 883941750, 13373883600, 203487733020, 3110407163760, 47726453450988, 734694122886080, 11341161925265480, 175489379096245984, 2721169178975361702, 42273090191785999728, 657788911222324942060, 10250564041646388681200

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(2).

G. C. Greubel, Table of n, a(n) for n = 0..825
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega2(n) = a(n-1).
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.

Cf. A039598, A000108, A024492 (h(0)), A000894 (h(1)), A000515 (h(3)), A228330 (h(4)), A228331 (h(5)), A228332 (h(6)), A228333 (h(7)).

Cf. A000142, A007318, A051960.

B:=(n,k)->binomial(2*n, n-k) - binomial(2*n, n-k-2); #A039598
Omega:=(m,n)->add((k+1)^m*B(n,k)^2,k=0..n);
h:=m->[seq(Omega(m,n),n=0..20)];
h(2);
# Second solution:
h := n -> I*HeunG(8/5,0,-1/4,1/4,3/2,1/2,16*x)/sqrt(16*x-1);
seq(coeff(series(h(x),x,n+2),x,n),n=0..19); # Peter Luschny, Nov 26 2013

a[n_] := Binomial[4n, 2n] (3n+1)/(2n+1);
Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Jul 30 2018, after Philippe Deléham *)

@CachedFunction
def A228329(n):
    return A228329(n-1)*(6*n+2)*(4*n-3)*(4*n-1)/(n*(2*n+1)*(3*n-2)) if n>0 else 1
[A228329(n) for n in (0..19)]  # Peter Luschny, Nov 26 2013

Conjecture: n*(2*n+1)*a(n) + 2*(-26*n^2+25*n-11)*a(n-1) + 20*(4*n-5)*(4*n-7)*a(n-2) = 0. - R. J. Mathar, Sep 08 2013
a(n) = ((4n)!*(3n+1))/((2n)!^2*(2n+1)) = binomial(4n,2n)*(3n+1)/(2n+1). - Philippe Deléham, Nov 25 2013
Therefore a(n) = A051960(2*n) / 2. - F. Chapoton, Jun 14 2024
From Peter Luschny, Nov 26 2013: (Start)
a(n) = 16^n*(3*n+1)*gamma(2*n+1/2)/(sqrt(Pi)*gamma(2*n+2)).
a(n) = a(n-1)*(6*n+2)*(4*n-3)*(4*n-1)/(n*(2*n+1)*(3*n-2)) if n > 0 else 1.
a(n) = [x^n] I*HeunG(8/5,0,-1/4,1/4,3/2,1/2,16*x)/sqrt(16*x-1) where [x^n] f(x) is the coefficient of x^n in f(x) and HeunG is the Heun general function. (End)

A265612 a(n) = CatalanNumber(n+1)n(1+3n)/(6+2n).

Original entry on oeis.org

0, 1, 7, 35, 156, 660, 2717, 11011, 44200, 176358, 700910, 2778446, 10994920, 43459650, 171655785, 677688675, 2674776720, 10555815270, 41656918050, 164401379610, 648887951400, 2561511781920, 10113397410402, 39937416869070, 157743149913776, 623178050662300

Offset: 0

Peter Luschny, Dec 15 2015

nonn

This is row n=7 in the array A(n,k) = (rf(k+n-2,k-1)-(k-1)*(k-2)*rf(k+n-2, k-3))/ (k-1)! if n>=3 and A(n,0)=0, A(n,1)=1, A(n,2)=n; rf(n,k) denotes the rising factorial. See the cross-references for other values of n and the table in A264357.

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

Cf. A000108, A007946, A097613, A051960, A029651, A051924, A129869, A220101, A264357, A265613.

A265612 := n -> 2*4^n*GAMMA(3/2+n)*n*(1+3*n)/(sqrt(Pi)*GAMMA(4+n)):
seq(simplify(A265612(n)), n=0..25);

Table[SeriesCoefficient[(5 x + (I (x - 1) (7 x - 2))/Sqrt[4 x - 1] - 2 - x^2)/(2 x^3), {x, 0, n}], {n, 0, 25}] (* or *)
Table[2*4^n Gamma[3/2 + n] n (1 + 3 n)/(Sqrt[Pi] Gamma[4 + n]), {n, 0, 25}] (* or *)
Table[CatalanNumber[n + 1] n ((1 + 3 n)/(6 + 2 n)), {n, 0, 25}] (* Michael De Vlieger, Dec 15 2015 *)

for(n=0,25, print1(round(2*4^n*gamma(3/2+n)*n*(1+3*n)/(sqrt(Pi)*gamma(4+n))), ", ")) \\ G. C. Greubel, Feb 06 2017

a = lambda n: catalan_number(n+1)*n*(1+3*n)/(6+2*n)
[a(n) for n in range(26)]

G.f.: (5*x+(I*(x-1)*(7*x-2))/sqrt(4*x-1)-2-x^2)/(2*x^3).
a(n) = 2*4^n*Gamma(3/2+n)*n*(1+3*n)/(sqrt(Pi)*Gamma(4+n)).
a(n) = (rf(5+n, n-1)-(n-1)*(n-2)*rf(5+n, n-3))/(n-1)! for n>=3, rf(n,k) the rising factorial.
a(n) = a(n-1)*(2*n*(1+3*n)*(1+2*n)/((n-1)*(3*n-2)*(3+n))) for n>=2.
a(n) ~  4^n*(6-(127/4)/n+(7995/64)/n^2-(223405/512)/n^3+(23501457/16384)/n^4-...) /sqrt(n*Pi).
a(n) = [x^n] x*(1 + x)/(1 - x)^(n+4). - Ilya Gutkovskiy, Oct 09 2017

A146983 a(n) = A002531(n)*A002531(n+1).

Original entry on oeis.org

1, 2, 10, 35, 133, 494, 1846, 6887, 25705, 95930, 358018, 1336139, 4986541, 18610022, 69453550, 259204175, 967363153, 3610248434, 13473630586, 50284273907, 187663465045, 700369586270, 2613814880038, 9754889933879, 36405744855481, 135868089488042

Offset: 0

Paul Barry, Nov 04 2008

a(n+1) is the Hankel transform of A051960 aerated.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,3,-1).

Cf. A001654, A182432, A211955.

a:=[1,2,10];; for n in [4..30] do a[n]:=3*a[n-1]+3*a[n-2]-a[n-3]; od; a; # G. C. Greubel, Jan 09 2020

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1-x+x^2)/((1+x)*(1-4*x+x^2)) )); // G. C. Greubel, Jan 09 2020

seq(coeff(series((1-x+x^2)/((1+x)*(1-4*x+x^2)), x, n+1), x, n), n = 0..30); # G. C. Greubel, Jan 09 2020

LinearRecurrence[{3,3,-1}, {1,2,10}, 30] (* G. C. Greubel, Jan 09 2020 *)

my(x='x+O('x^30)); Vec((1-x+x^2)/((1+x)*(1-4*x+x^2))) \\ G. C. Greubel, Jan 09 2020

def A146983_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-x+x^2)/((1+x)*(1-4*x+x^2)) ).list()
A146983_list(30) # G. C. Greubel, Jan 09 2020

From Peter Bala, May 01 2012: (Start)
a(n) = (-1)^n + 3*Sum_{k = 1..n} (-1)^(n-k)*6^(k-1)*binomial(n+k,2*k).
a(n) = (-1)^n*R(n,-3), where R(n,x) is the n-th row polynomial of A211955.
a(n) = (-1)^n*1/u*T(n,u)*T(n+1,u) with u = sqrt(-1/2) and T(n,x) denotes the Chebyshev polynomial of the first kind  Cf. A182432.
Recurrence: a(n) = 4*a(n-1) -a(n-2) +3*(-1)^n, with a(0) = 1 and a(1) = 2; a(n)*a(n-2) = a(n-1)*(a(n-1)+3*(-1)^n).
Sum_{k>=0} (-1)^k/a(k) = 1/sqrt(3). (End)
From Colin Barker, Jul 29 2013: (Start)
a(n) = 3*a(n-1) + 3*a(n-2) - a(n-3).
G.f.: (1-x+x^2)/((1+x)*(1-4*x+x^2)). (End)

More terms from Colin Barker, Jul 29 2013

A264357 Array A(r, n) of number of independent components of a symmetric traceless tensor of rank r and dimension n, written as triangle T(n, r) = A(r, n-r+2), n >= 1, r = 2..n+1.

Original entry on oeis.org

0, 2, 0, 5, 2, 0, 9, 7, 2, 0, 14, 16, 9, 2, 0, 20, 30, 25, 11, 2, 0, 27, 50, 55, 36, 13, 2, 0, 35, 77, 105, 91, 49, 15, 2, 0, 44, 112, 182, 196, 140, 64, 17, 2, 0, 54, 156, 294, 378, 336, 204, 81, 19, 2, 0

Offset: 1

Wolfdieter Lang, Dec 10 2015

A (totally) symmetric traceless tensor of rank r >= 2 and dimension n >= 1 is irreducible.
The array of the number of independent components of a rank r symmetric traceless tensor A(r, n), for r >= 2 and n >=1, is given by risefac(n,r)/r! - risefac(n,r-2)/(r-2)!, where the first term gives the number of independent components of a symmetric tensors of rank r (see a Dec 10 2015 comment under A135278) and the second term is the number of constraints from the tracelessness requirement. The tensor has to be traceless in each pair of indices.
The first rows of the array A, or the first columns (without the first r-2 zeros) of the triangle T are for r = 2..6: A000096, A005581, A005582, A005583, A005584.
Equals A115241 with the first column of positive integers removed. - Georg Fischer, Jul 26 2023

			The array A(r, n) starts:
   r\n 1 2  3   4   5    6    7     8     9    10 ...
   2:  0 2  5   9  14   20   27    35    44    54
   3:  0 2  7  16  30   50   77   112   156   210
   4:  0 2  9  25  55  105  182   294   450   660
   5:  0 2 11  36  91  196  378   672  1122  1782
   6:  0 2 13  49 140  336  714  1386  2508  4290
   7:  0 2 15  64 204  540 1254  2640  5148  9438
   8:  0 2 17  81 285  825 2079  4719  9867 19305
   9:  0 2 19 100 385 1210 3289  8008 17875 37180
  10:  0 2 21 121 506 1716 5005 13013 30888 68068
  ...
The triangle T(n, r) starts:
   n\r  2   3   4   5   6   7  8  9 10 11 ...
   1:   0
   2:   2   0
   3:   5   2   0
   4:   9   7   2   0
   5:  14  16   9   2   0
   6:  20  30  25  11   2   0
   7:  27  50  55  36  13   2  0
   8:  35  77 105  91  49  15  2  0
   9:  44 112 182 196 140  64 17  2  0
  10:  54 156 294 378 336 204 81 19  2  0
  ...
A(r, 1) = 0 , r >= 2, because a symmetric rank r tensor t of dimension one has one component t(1,1,...,1) (r 1's) and if the traces vanish then t vanishes.
A(3, 2) = 2 because a symmetric rank 3 tensor t with three indices taking values from 1 or 2 (n=2) has the four independent components t(1,1,1), t(1,1,2), t(1,2,2), t(2,2,2), and (invoking symmetry) the vanishing traces are Sum_{j=1..2} t(j,j,1) = 0 and Sum_{j=1..2} t(j,j,2) = 0. These are two constraints, which can be used to eliminate, say, t(1,1,1) and t(2,2,2), leaving 2 = A(3, 2) independent components, say, t(1,1,2) and t(1,2,2).
From _Peter Luschny_, Dec 14 2015: (Start)
The diagonals diag(n, k) start:
   k\n  0       1       2       3       4      5       6
   0:   0,      2,      9,     36,    140,   540,   2079, ... A007946
   1:   2,      7,     25,     91,    336,  1254,   4719, ... A097613
   2:   5,     16,     55,    196,    714,  2640,   9867, ... A051960
   3:   9,     30,    105,    378,   1386,  5148,  19305, ... A029651
   4:  14,     50,    182,    672,   2508,  9438,  35750, ... A051924
   5:  20,     77,    294,   1122,   4290, 16445,  63206, ... A129869
   6:  27,    112,    450,   1782,   7007, 27456, 107406, ... A220101
   7:  35,    156,    660,   2717,  11011, 44200, 176358, ... A265612
   8:  44,    210,    935,    4004, 16744, 68952, 281010, ... A265613
  A000096,A005581,A005582,A005583,A005584.
(End)

Cf. A000096, A005581, A005582, A005583, A005584, A007946, A024482, A029651, A051924, A051960, A097613, A115241, A129869, A135278, A220101, A265612, A265613.

A[r_, n_] := Pochhammer[n, r]/r! - Pochhammer[n, r-2]/(r-2)!;
T[n_, r_] := A[r, n-r+2];
Table[T[n, r], {n, 1, 10}, {r, 2, n+1}] (* Jean-François Alcover, Jun 28 2019 *)

A = lambda r, n: rising_factorial(n,r)/factorial(r) - rising_factorial(n,r-2)/factorial(r-2)
for r in (2..10): [A(r,n) for n in (1..10)] # Peter Luschny, Dec 13 2015

T(n, r) = A(r, n-r+2) with the array A(r, n) = risefac(n,r)/r! - risefac(n,r-2)/(r-2)! where the rising factorial risefac(n,k) = Product_{j=0..k-1} (n+j) and risefac(n,0) = 1.
From Peter Luschny, Dec 14 2015: (Start)
A(n+2, n+1) = A007946(n-1) = CatalanNumber(n)*3*n*(n+1)/(n+2) for n>=0.
A(n+2, n+2) = A024482(n+2) = A097613(n+2) = CatalanNumber(n+1)*(3*n+4)/2 for n>=0.
A(n+2, n+3) = A051960(n+1) = CatalanNumber(n+1)*(3*n+5) for n>=0.
A(n+2, n+4) = A029651(n+2) = CatalanNumber(n+1)*(6*n+9) for n>=0.
A(n+2, n+5) = A051924(n+3) = CatalanNumber(n+2)*(3*n+7) for n>=0.
A(n+2, n+6) = A129869(n+4) = CatalanNumber(n+2)*(3*n+8)*(2*n+5)/(n+4) for n>=0.
A(n+2, n+7) = A220101(n+4) = CatalanNumber(n+3)*(3*(n+3)^2)/(n+5) for n>=0.
A(n+2, n+8) = CatalanNumber(n+4)*(n+3)*(3*n+10)/(2*n+12) for n>=0.
Let for n>=0 and k>=0 diag(n,k) = A(k+2,n+k+1) and G(n,k) = 2^(k+2*n)*Gamma((3-(-1)^k+2*k+4*n)/4)/(sqrt(Pi)*Gamma(k+n+0^k)) then
diag(n,0) = G(n,0)*(n*3)/(n+2),
diag(n,1) = G(n,1)*(3*n+4)/((n+1)*(n+2)),
diag(n,2) = G(n,2)*(3*n+5)/(n+2),
diag(n,3) = G(n,3)*3,
diag(n,4) = G(n,4)*(3*n+7),
diag(n,5) = G(n,5)*(3*n+8),
diag(n,6) = G(n,6)*3*(3+n)^2,
diag(n,7) = G(n,7)*(3+n)*(10+3*n). (End)

A265613 a(n) = CatalanNumber(n+1)n(3n^2+5n+2)/((4+n)*(3+n)).

Original entry on oeis.org

0, 1, 8, 44, 210, 935, 4004, 16744, 68952, 281010, 1136960, 4576264, 18349630, 73370115, 292746300, 1166182800, 4639918800, 18443677230, 73261092240, 290845019400, 1154169552900, 4578702310182, 18159992594568, 72014135814704, 285542883894800, 1132125641947300

Offset: 0

Peter Luschny, Dec 15 2015

nonn

This is row n=8 in the array A(n,k) = (rf(k+n-2,k-1)-(k-1)*(k-2)*rf(k+n-2, k-3))/ (k-1)! if n>=3 and A(n,0)=0, A(n,1)=1, A(n,2)=n; rf(n,k) denotes the rising factorial. See the cross-references for other values of n and the table in A264357.

Robert Israel, Table of n, a(n) for n = 0..1630

Cf. A000108, A007946, A097613, A051960, A029651, A051924, A129869, A220101, A264357, A265612.

A265613 := n -> (4*4^n*n*(n+1)*(3*n+2)*GAMMA(n+3/2))/(sqrt(Pi)*GAMMA(n+5)):
seq(simplify(A265613(n)), n=0..25);

Table[SeriesCoefficient[I (14 x^2 + I Sqrt[4 x - 1] (4 x^2 - 7 x + 2) - 11 x + 2 (1 - x^3))/(2 x^4 Sqrt[4 x - 1]), {x, 0, n}], {n, 0, 25}]
(* or *)
Table[(4^(n + 1) n (n + 1) (3 n + 2) Gamma[n + 3/2])/(Sqrt[Pi] Gamma[n + 5]), {n, 0, 25}] (* or *)
Table[CatalanNumber(n+1) n (3 n^2 + 5 n + 2)/((4 + n) (3 + n)), {n, 0, 25}] (* Michael De Vlieger, Dec 15 2015 *)

a = lambda n: catalan_number(n+1)*n*(3*n^2+5*n+2)/((4+n)*(3+n))
[a(n) for n in range(26)]

G.f.: I*(14*x^2+I*sqrt(4*x-1)*(4*x^2-7*x+2)-11*x+2*(1-x^3))/(2*x^4*sqrt(4*x-1)).
a(n) = (4^(n+1)*n*(n+1)*(3*n+2)*Gamma(n+3/2))/(sqrt(Pi)*Gamma(n+5)).
a(n) = (rf(n+6, n-1)-(n-1)*(n-2)*rf(n+6, n-3))/(n-1)! for n>=3, rf(n,k) the rising factorial.
a(n) = a(n-1)*((2*(n+1))*(3*n+2)*(1+2*n)/((n-1)*(3*n-1)*(4+n))) for n>=2.
a(n) ~ 4^n*(12-(191/2)/n+(17595/32)/n^2-(705005/256)/n^3+(104705937/8192)/ n^4-...)/sqrt(n*Pi).
a(n) = [x^n] x*(1 + x)/(1 - x)^(n+5). - Ilya Gutkovskiy, Oct 09 2017

A028283 Central elements in 4-Pascal triangle A028275 (by row).

Original entry on oeis.org

1, 4, 10, 32, 110, 392, 1428, 5280, 19734, 74360, 281996, 1074944, 4115020, 15808912, 60917800, 235350720, 911315430, 3535767000, 13742347740, 53495534400, 208537056420, 813950932080, 3180614712600, 12441628655040

Offset: 0

Mohammad K. Azarian

Equals A000984, [1, 2, 6, 20, 70, ...] convolved with [1, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jun 05 2009

Cf. A000108, A028275, A051960.

a(n) = 2*A051960(n-1).

a(n) = 2*A000108(n)*(3n-1). - Ralf Stephan, Aug 24 2003

A274104 a(n) = Sum_{k=0..n} (3k+2)Catalan(k).

Original entry on oeis.org

2, 7, 23, 78, 274, 988, 3628, 13495, 50675, 191673, 729145, 2786655, 10691111, 41150011, 158825371, 614483086, 2382366586, 9253540456, 36001307656, 140269835866, 547245301906, 2137552658206, 8358366985726, 32715599554876, 128168506456852, 502538379368656, 1971926625140816

Offset: 0

N. J. A. Sloane, Jun 13 2016

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Moa Apagodu and Doron Zeilberger, Using the "Freshman's Dream" to Prove Combinatorial Congruences, arXiv:1606.03351 [math.CO], 2016. Also Amer. Math. Monthly. 124 (2017), 597-608.

Partial sums of A051960.

Cf. A000108, A006134, A014137.

[(&+[(3*k+2)*Catalan(k): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Jun 30 2024

CoefficientList[Series[(1 +2 x -Sqrt[1-4 x])/(2 x Sqrt[1-4 x] (1-x)), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2016 *)

[sum((3*k+2)*catalan_number(k) for k in range(n+1)) for n in range(41)] # G. C. Greubel, Jun 30 2024

D-finite with recurrence: (n+1)*a(n) - (3*n+5)*a(n-1) - 2*(3*n-8)*a(n-2) + 4*(2*n-3)*a(n-3) = 0. - R. J. Mathar, Jun 15 2016
G.f.: (1 + 2*x - sqrt(1-4*x))/(2*x*(1-x)*sqrt(1-4*x)). - Ilya Gutkovskiy, Jun 15 2016
a(n) = A014137(n+1) + (n+1)*A000108(n+1) - 1. - G. C. Greubel, Jun 30 2024
From Mélika Tebni, Sep 02 2024: (Start)
a(n) = A006134(n) + A006134(n+1)/2 - 1/2.
E.g.f.: exp(2*x)*(5*BesselI(0, 2*x)/2 + BesselI(1, 2*x)) + exp(x)/2*(3*Integral_{x=-oo..oo} BesselI(0,2*x)*exp(x) dx - 1). (End)

A383776 a(n) = (11n + 3 + 6/(n+2)) Catalan(n).

Original entry on oeis.org

6, 16, 53, 186, 672, 2472, 9207, 34606, 130988, 498576, 1906346, 7316596, 28170768, 108760560, 420889995, 1632155670, 6340808820, 24673450560, 96148670310, 375164728620, 1465589068320, 5731488987120, 22436098732710, 87905595401676, 344702077523352, 1352701532137312, 5312100899224532, 20874451526714856

Offset: 0

F. Chapoton, May 09 2025

It appears that for n >= 2 a(n-2) is the number of lattice points in the n-dimensional lattice polytope defined, in the space with coordinates (x_1,x_2,...,x_n), by the equations x_i >= 0 for every i, sum_i x_i <= n and x_1 + x_2 <= 2. For n=2, this is a triangle with 6 lattice points.

Paolo Xausa, Table of n, a(n) for n = 0..1000

Cf. A000108, A000984, A007054, A051960.

A383776[n_] := (11*n + 3 + 6/(n + 2))*CatalanNumber[n];
Array[A383776, 30, 0] (* Paolo Xausa, May 15 2025 *)

[(11*n+3+6/(n+2))*catalan_number(n) for n in range(12)]

a(n) = (11*n + 3 + 6/(n + 2))*Catalan(n).

G.f.: 2*(7 + 5*sqrt(1 - 4*x) - 6*x)/((1 + sqrt(1 - 4*x))^2*sqrt(1 - 4*x)). - Stefano Spezia, May 15 2025

A241188 Triangle T(n,s) of Dynkin type D_n read by rows (n >= 2, 0 <= s <= n).

Original entry on oeis.org

1, 2, 1, 1, 3, 5, 5, 1, 4, 9, 16, 20, 1, 5, 14, 30, 55, 77, 1, 6, 20, 50, 105, 196, 294, 1, 7, 27, 77, 182, 378, 714, 1122, 1, 8, 35, 112, 294, 672, 1386, 2640, 4290, 1, 9, 44, 156, 450, 1122, 2508, 5148, 9867, 16445

Offset: 2

N. J. A. Sloane, Apr 24 2014

			Triangle begins:
1, 2, 1,
1, 3, 5, 5,
1, 4, 9, 16, 20,
1, 5, 14, 30, 55, 77,
1, 6, 20, 50, 105, 196, 294,
1, 7, 27, 77, 182, 378, 714, 1122,
1, 8, 35, 112, 294, 672, 1386, 2640, 4290,
1, 9, 44, 156, 450, 1122, 2508, 5148, 9867, 16445,
...

M. A. A. Obaid, S. K. Nauman, W. M. Fakieh, C. M. Ringel, The numbers of support-tilting modules for a Dynkin algebra, 2014.
M. A. A. Obaid, S. K. Nauman, W. M. Fakieh, C. M. Ringel, The numbers of support-tilting modules for a Dynkin algebra, arXiv:1403.5827 [math.RT], 2014 and J. Int. Seq. 18 (2015) 15.10.6.

See A009766 for the case of type A.
See A059481 for the case of type B/C.
Diagonals give A029869, A051960, A029651, A051924. Row sums are also A051924.

f[t_, s_] := Binomial[t, s] (s + t)/t;
T[, 0] = 1; T[n, n_] := f[2 n - 2, n - 2]; T[n_, s_] := f[n + s - 2, s];
Table[T[n, s], {n, 2, 9}, {s, 0, n}] // Flatten (* Jean-François Alcover, Feb 12 2019 *)

T(n,s) = [n+s-2,s] for 0 <= s < n, T(n,n) = [2n-2,n-2], where [t,s] stands for binomial(t,s)*(s+t)/t.

A367505 Triangle read by rows: row n gives the h-vector of the n-th halohedron.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 13, 27, 13, 1, 1, 21, 76, 76, 21, 1, 1, 31, 175, 300, 175, 31, 1, 1, 43, 351, 925, 925, 351, 43, 1, 1, 57, 637, 2401, 3675, 2401, 637, 57, 1, 1, 73, 1072, 5488, 11956, 11956, 5488, 1072, 73, 1, 1, 91, 1701, 11376, 33516, 47628, 33516, 11376, 1701, 91, 1

Offset: 0

F. Chapoton, Nov 21 2023

Theorem 6.1.11 in Almeter's thesis gives the f-vector generating series. Then replacing x with x-1 gives the h-vector generating series.

			As a table:
  (1),
  (1,  1),
  (1,  3,  1),
  (1,  7,  7,  1),
  (1, 13, 27, 13,  1),
  (1, 21, 76, 76, 21,  1),
  ...

Jordan Grady Almeter, P-graph associahedra and hypercube graph associahedra, arXiv:2211.02113 [math.CO], 2022; Ph.D. thesis, North Carolina State University, 2022.
Forcey's Hedra Zoo, Halohedron.

Row sums are A051960(n-1) for n>=1.
Alternating sums form an aerated version of A110556.
Columns k=0-2 give A000012, A002061, A039623(n-1) for n>=2.

T[0,0]:=1;T[n_,k_]:= Binomial[n-1,n-k]*Binomial[n,n-k]+Binomial[n-1,n-k-1]^2;Flatten[Table[T[n,k],{n,0,10},{k,0,n}]] (* Detlef Meya, Nov 23 2023 *)

x = polygen(QQ, 'x')
t = x.parent()[['t']].0
F = (1 + (1+x) * t) / (2 * sqrt(1 - 2 * (x+1) * t + (x-1)**2 * t**2)) + 1/2
for poly in F.list(): print(poly.list())

G.f.: (1 + (1+x)*t)/(2*sqrt(1 - 2*(x+1)*t + (x-1)^2*t^2)) + 1/2.

T(0,0) = 1; T(n,k) = binomial(n-1,n-k)*binomial(n,n-k)+binomial(n-1,n-k-1)^2. - Detlef Meya, Nov 23 2023

cp's OEIS Frontend

A228329 a(n) = Sum_{k=0..n} (k+1)^2*T(n,k)^2 where T(n,k) is the Catalan triangle A039598.

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A265612 a(n) = CatalanNumber(n+1)*n*(1+3*n)/(6+2*n).

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A146983 a(n) = A002531(n)*A002531(n+1).

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A264357 Array A(r, n) of number of independent components of a symmetric traceless tensor of rank r and dimension n, written as triangle T(n, r) = A(r, n-r+2), n >= 1, r = 2..n+1.

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A265613 a(n) = CatalanNumber(n+1)*n*(3*n^2+5*n+2)/((4+n)*(3+n)).

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A028283 Central elements in 4-Pascal triangle A028275 (by row).

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A274104 a(n) = Sum_{k=0..n} (3*k+2)*Catalan(k).

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A265612 a(n) = CatalanNumber(n+1)n(1+3n)/(6+2n).

A265613 a(n) = CatalanNumber(n+1)n(3n^2+5n+2)/((4+n)*(3+n)).

A274104 a(n) = Sum_{k=0..n} (3k+2)Catalan(k).

A383776 a(n) = (11n + 3 + 6/(n+2)) Catalan(n).