A007054 -id:A007054 - cp's OEIS frontend

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.

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)).

A376161 Number of support Tau-tilting modules for some algebras.

Original entry on oeis.org

3, 5, 12, 33, 98, 306, 990, 3289, 11154, 38454, 134368, 474810, 1693812, 6091780, 22064130, 80410185, 294647250, 1084922190, 4012165080, 14895504030, 55496654460, 207431394300, 777601790940, 2922867908298, 11013796950228, 41596652545756, 157434454904160, 597029454416724, 2268232385053096

Offset: 0

F. Chapoton, Sep 13 2024

nonn

See Prop. A.6 in Wang's reference for the table counting Tau-tilting modules for the linear quiver modulo the relation alpha*beta = 0.

Paolo Xausa, Table of n, a(n) for n = 0..1000
Anna Rodriguez Rasmussen, Exact Borel subalgebras of quasi-hereditary monomial algebras, arXiv:2504.01706 [math.RT], 2025. See p. 38.
Qi Wang, Tau-tilting finite simply connected algebras, arXiv:1910.01937 [math.RT], 2019-2022.

Cf. A005807, A007054, A000108, A016789, A023999, A329533.

a := n -> -(3*n + 2)*(-4)^(n + 1)*binomial(3/2, n + 2):
seq(a(n), n = 0..28)  # Peter Luschny, Sep 13 2024

A376161[n_] := CatalanNumber[n]*(9*n + 6)/(n + 2);
Array[A376161, 30, 0] (* Paolo Xausa, Sep 14 2024 *)

def a(n):
    return 3*(3*n+2)*binomial(2*n+4,n+2)/4/(2*n+1)/(2*n+3)

a(n) = 3*(3*n+2)*binomial(2*n+4,n+2)/(4*(2*n+1)*(2*n+3)).
a(n) = A329533(n)/(n + 1).
From Peter Luschny, Sep 13 2024: (Start)
a(n) = (3*n + 2) * [x^n] ((1 - 4*x)^(3/2) + 12*x - 2)/(4*x^2).
a(n) = A016789(n)*(3/2)*(2*n)! * [x^(2*n)] hypergeom([], [3], x^2).
a(n) = CatalanNumber(n)*(9*n + 6)/(n + 2).
a(n) = -(3*n + 2)*(-4)^(n + 1)*binomial(3/2, n + 2).
a(n) = 2^n*(9*n + 6)*(2*n - 1)!! / (n + 2)!.
a(n) = A007054(n) * (3*n + 2) / 2.
a(n) = 6*A023999(n + 1)/(n + 2)!. (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

A182534 Array read by antidiagonals: coefficient of the Euler-Mascheroni constant in below expression.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 5, 4, 2, 6, 14, 10, 3, 4, 10, 42, 28, 6, 6, 5, 20, 132, 84, 14, 12, 6, 10, 35, 429, 264, 36, 28, 10, 12, 14, 70, 1430, 858, 99, 72, 20, 20, 14, 28, 126, 4862, 2860, 286, 198, 45, 40, 20, 28, 42, 252

Offset: 1

John M. Campbell, May 05 2012

The (i,j)-entry of the array is the coefficient of the Euler-Mascheroni constant in: -2^(i+2j-1)/Pi*int(log(x)*cos(x)^i*sin(x)^(2j-1)/x, x=0..infinity); see Mathematica code below.
First row:     A000108.
Second row:   -A002420.
Third row:     A007054.
Fourth row:    A002421.
Fifth row:     A007272.
Sixth row:    -A002422.
Eighth row:    A002423.
First column:  A001405.
Second column: A089408.
Odd entries on main diagonal: A126596.

			Evaluate: -256/Pi*int(cos(x)^3*log(x)*sin(x)^5/x, x=0..infinity) = 3*eulergamma-log(9/8). Thus the (3,3) entry of the array is 3, the coefficient of the Euler-Mascheroni constant in this expression.
The array begins as:
| 1   1   2   5   14  42  132 429  ... |
| 2   2   4   10  28  84  264 858  ... |
| 3   2   3   6   14  36  99  286  ... |
| 6   4   6   12  28  72  198 572  ... |
| 10  5   6   10  20  45  110 286  ... |
| 20  10  12  20  40  90  220 572  ... |
| 35  14  14  20  35  70  154 364  ... |
| 70  28  28  40  70  140 308 728  ... |
| ... ... ... ... ... ... ... ...  ... |

Cf. A000108, A002420, A007054, A002421, A007272, A002422, A002423, A001405, A089408, A126596.

A[a_, b_] :=
  A[a, b] =
   Array[Coefficient[
      Integrate[
        Log[x]*Cos[x]^#1*Sin[x]^(2 #2 - 1)/x, {x, 0,
         Infinity}] (2^(#1 + 2 #2 - 1))/(-\[Pi]), EulerGamma] &, {a, b}];
A[11, 11];
Print[A[11, 11] // MatrixForm];
Table2 = {};
k = 1;
While[k < 11, Table1 = {};
  i = 1;
  j = k;
  While[0 < j,
    AppendTo[Table1,
    First[Take[First[Take[A[11, 11], {i, i}]], {j, j}]]];
    j = j - 1;
    i = i + 1];
    AppendTo[Table2, Table1];
    k++];
Print[Flatten[Table2]]

A337841 Triangle read by rows, T(n, k) = binomial(2n-1, 2k-1) * binomial(2n-2k, n-k) * (k+1) / binomial(n+k+1, n-k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 2, 0, 3, 3, 0, 6, 10, 4, 0, 14, 30, 21, 5, 0, 36, 90, 84, 36, 6, 0, 99, 275, 308, 180, 55, 7, 0, 286, 858, 1092, 780, 330, 78, 8, 0, 858, 2730, 3822, 3150, 1650, 546, 105, 9, 0, 2652, 8840, 13328, 12240, 7480, 3094, 840, 136, 10, 0, 8398, 29070, 46512, 46512, 31977, 15561, 5320, 1224, 171, 11

Offset: 0

Werner Schulte, Oct 30 2020

			The triangle T(n,k) for 0 <= k <= n starts:
  n\k:  0     1      2      3      4      5      6     7     8    9  10
=======================================================================
   0 :  1
   1 :  0     2
   2 :  0     3      3
   3 :  0     6     10      4
   4 :  0    14     30     21      5
   5 :  0    36     90     84     36      6
   6 :  0    99    275    308    180     55      7
   7 :  0   286    858   1092    780    330     78     8
   8 :  0   858   2730   3822   3150   1650    546   105     9
   9 :  0  2652   8840  13328  12240   7480   3094   840   136   10
  10 :  0  8398  29070  46512  46512  31977  15561  5320  1224  171  11
etc.

Cf. Row sums: A000984, main diagonal: A000027, 1st subdiagonal: A014105, 2nd subdiagonal: A055112, column 0: A000007, column 1: A007054.

T := proc(n, k) option remember; if k = n then n+1 else
T(n-1, k)*(2*n-2)*(2*n-1)/((n-1)*(n+2)-(k-1)*(k+2)) fi end:
for n from 0 to 9 do seq(T(n, k), k=0..n) od; # Peter Luschny, Nov 02 2020

T(n,k) = binomial(2*n-1,n-k) * k * (2*k+1) * (2*k+2) / ((n+k)*(n+k+1)) for 1 <= k <= n, and T(n,0) = 0^n for n >= 0.
T(n,n) = n+1 for n >= 0; T(n,n-1) = (n-1) * (2*n-1) for n > 0; T(n,n-2) = (n-1) * (n-2) * (2*n-3) for n > 1.
T(n,k) = T(n-1,k) * (2*n-2) * (2*n-1) / ((n-1) * (n+2) - (k-1) * (k+2)) for 0 <= k < n with initial values T(n,n) = n+1 for n >= 0.
Row sums are A000984(n) for n >= 0.
Alternating row sums are 0 for n > 1.
Sum_{k=0..n} (-1)^k * T(n,k) * (k*(k+1)/2)^m = 0 for 0 <= m <= n-2.
T(n,1) = 12 * binomial(2*n-1,n-1)/((n+1)*(n+2)) = A007054(n) for n > 0.
T(n,k) = T(n,1)*(k*(k+1)*(2*k+1)/6)*binomial(n-1,k-1)/binomial(n+1+k,k-1) for 1 <= k <= n.
From Werner Schulte, Nov 09 2020: (Start)
T(n,k) = A128899(n,k) * (k+1) * (2*k+1) / (n+k+1) for 0 <= k <= n.
T(n,0) + Sum_{k=1..n} T(n,k) / (k*(k+1)) = A000108(n) for n >= 0. (End)

A372949 a(n) = 2f(2n)/(f(n)*f(n+2)) where f = A003266.

Original entry on oeis.org

1, 2, 8, 91, 2618, 199716, 39690618, 20689636692, 28215085220016, 100763710906257557, 942012688139052139766, 23056957423045790791793932, 1477460537993359748548214768630, 247860656992078740305125996374953260, 108861324945456389643061592667638024842480

Offset: 1

Kendra Killpatrick, May 17 2024

nonn

Fibonacci analog of the super ballot numbers.

a(n) is also the generalized FiboCatalan number for r=1. Proof that the formula always gives a positive integer can be found in a recent paper of K. Killpatrick. The sequence is the Fibonacci analog of the super ballot numbers given by Gessel (A007054). The sequence is also the Fibonacci analog of the generalized Catalan numbers, J_r*(2n)!/(n!*(n+r+1)!) where J_r=(2r+1)!/r!, for r=1. Gessel defined the generalized Catalan numbers and proved they are integers.

			a(5) = 2*f(10)/(f(5)*f(7)) = 2*122522400/(30*3120) = 2618, where f=A003266.

Ira M. Gessel, Super ballot numbers, J. Symb. Comput. 14 (1992), 179-194.
Kendra Killpatrick, Super FiboCatalan Numbers and their Lucas Analogues, arXiv:2308.13457 [math.CO], (2023).

Cf. A003266, A007054.

a[n_]:=2Fibonorial[2n]/(Fibonorial[n]Fibonorial[n+2]); Array[a,15] (* Stefano Spezia, May 23 2024 *)

a(n) ~ 10 * phi^((n-3)*(n+1)) / A062073, where phi = A001622 is the golden ratio. - Vaclav Kotesovec, May 29 2024

cp's OEIS Frontend

A337994 T(n, k) = (k*(2*k+2)*(2*k+1)*(2*n-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.

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A376161 Number of support Tau-tilting modules for some algebras.

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

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A182534 Array read by antidiagonals: coefficient of the Euler-Mascheroni constant in below expression.

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A337841 Triangle read by rows, T(n, k) = binomial(2*n-1, 2*k-1) * binomial(2*n-2*k, n-k) * (k+1) / binomial(n+k+1, n-k) for 0 <= k <= n.

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A372949 a(n) = 2*f(2*n)/(f(n)*f(n+2)) where f = A003266.

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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.

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

A337841 Triangle read by rows, T(n, k) = binomial(2n-1, 2k-1) * binomial(2n-2k, n-k) * (k+1) / binomial(n+k+1, n-k) for 0 <= k <= n.

A372949 a(n) = 2f(2n)/(f(n)*f(n+2)) where f = A003266.