A007946 -id:A007946 - cp's OEIS frontend

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

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

A078817 Table by antidiagonals giving variants on Catalan sequence: T(n,k)=C(2n,n)C(2k,k)(2k+1)/(n+k+1).

Original entry on oeis.org

1, 3, 1, 10, 4, 2, 35, 15, 9, 5, 126, 56, 36, 24, 14, 462, 210, 140, 100, 70, 42, 1716, 792, 540, 400, 300, 216, 132, 6435, 3003, 2079, 1575, 1225, 945, 693, 429, 24310, 11440, 8008, 6160, 4900, 3920, 3080, 2288, 1430, 92378, 43758, 30888, 24024, 19404

Offset: 0

Henry Bottomley, Dec 07 2002

			Rows start:
     1,     3,    10,    35,   126,   462,  1716,
     1,     4,    15,    56,   210,   792,  3003,
     2,     9,    36,   140,   540,  2079,  8008,
     5,    24,   100,   400,  1575,  6160, 24024,
    14,    70,   300,  1225,  4900, 19404, 76440,
    42,   216,   945,  3920, 15876, 63504,252252,
   132,   693,  3080, 12936, 52920,213444,853776,
etc.

Ira Gessel, Super ballot numbers, J. Symbolic Computation 14 (1992), 179-194.
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
Jovan Mikic, A Note on the Gessel Numbers, arXiv:2203.12931 [math.CO], 2022.

Columns include A000108 (catalan), A038629, A078818 and A078819. Rows include A001700, A001791, A007946 and A078820. Diagonals include A002894 and A060150.

Essentially a reflected version of A033820.

A078817 := proc(n,k)
    binomial(2*n,n)*binomial(2*k,k)*(2*k+1)/(n+k+1) ;
end proc: # R. J. Mathar, Dec 06 2018

T(n, k) = A000984(n)*A002457(k)/(n+k+1) = T(k, n)*(2k+1)/(2n+1).

A078818 a(n) = 30*binomial(2n,n)/(n+3).

Original entry on oeis.org

10, 15, 36, 100, 300, 945, 3080, 10296, 35100, 121550, 426360, 1511640, 5408312, 19501125, 70794000, 258529200, 949074300, 3500409330, 12964479000, 48198087000, 179799820200, 672822343050, 2524918756464, 9500112378000, 35830670759000, 135439935469020

Offset: 0

Henry Bottomley, Dec 07 2002

			a(5) = 30*binomial(10,5)/8 = 945.

Muniru A Asiru, Table of n, a(n) for n = 0..300

Cf. A000108, A000984, A001622, A007946, A038629, A078817, A078819.

List([0..30],n->30*Binomial(2*n,n)/(n+3)); # Muniru A Asiru, Aug 09 2018

[30*Binomial(2*n,n)/(n+3): n in [0..30]]; // Vincenzo Librandi, Aug 11 2018

Table[(30 Binomial[2 n, n] / (n + 3)), {n, 0, 30}] (* Vincenzo Librandi, Aug 11 2018 *)

D-finite with recurrence a(n) = a(n-1)*(4n^2+6n-4)/(n^2+3n) = A078817(n, 2) = 5*A007946(n)/(2n+1) = 30*A000984(n)/(n+3).
From Amiram Eldar, Feb 16 2023: (Start)
Sum_{n>=0} 1/a(n) = 4*Pi/(135*sqrt(3)) + 7/45.
Sum_{n>=0} (-1)^n/a(n) = 9/125 - 32*log(phi)/(375*sqrt(5)), where phi is the golden ratio (A001622). (End)

A078820 a(n) = 20C(2n,n)(2n+1)/(n+4).

Original entry on oeis.org

5, 24, 100, 400, 1575, 6160, 24024, 93600, 364650, 1421200, 5542680, 21633248, 84504875, 330372000, 1292646000, 5061729600, 19835652870, 77786874000, 305254551000, 1198665468000, 4709756401350, 18516070880736, 72834194898000, 286645366072000, 1128666128908500

Offset: 0

Henry Bottomley, Dec 07 2002

			a(5)=20*C(10,5)*11/9=6160.

Cf. A001622, A001700, A001791, A002457, A007946, A078817, A078819.

Table[20Binomial[2n,n] (2n+1)/(n+4),{n,0,30}] (* Harvey P. Dale, Nov 02 2011 *)

D-finite with recurrence a(n) = a(n-1)*(4n^2+14n+6)/(n^2+4n) = A078817(3, n) = (2n+1)*A078819(n)/7 = 20*A002457(n)/(n+4).
From Amiram Eldar, Feb 16 2023: (Start)
Sum_{n>=0} 1/a(n) = 11*Pi/(90*sqrt(3)) + 1/30.
Sum_{n>=0} (-1)^n/a(n) = 17*log(phi)/(25*sqrt(5)) + 1/50, where phi is the golden ratio (A001622). (End)

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

A242986 a(n) = 6(n+1)!/((3+floor(n/2))(floor(n/2)!)^2).

Original entry on oeis.org

2, 4, 9, 36, 36, 216, 140, 1120, 540, 5400, 2079, 24948, 8008, 112112, 30888, 494208, 119340, 2148120, 461890, 9237800, 1790712, 39395664, 6953544, 166885056, 27041560, 703080560, 105306075, 2948570100, 410605200, 12318156000, 1602881040, 51292193280, 6263890380

Offset: 0

Peter Luschny, Aug 25 2014

nonn

Cf. A007946, A056040.

[6*Factorial(n+1)/((3+Floor(n/2))*(Factorial(Floor(n/2)))^2) : n in [0..30]]; // Wesley Ivan Hurt, Aug 26 2014

A056040 := n -> n!/iquo(n,2)!^2;
A242986 := n -> (6*(n+1)/(3+iquo(n,2)))*A056040(n);
seq(A242986(n), n=0..32);

Table[6(n + 1)!/((3 + Floor[n/2])*(Floor[n/2]!)^2), {n, 0, 30}] (* Wesley Ivan Hurt, Aug 26 2014 *)

@CachedFunction
def A242986(n):
    if n == 0: return 2
    h = (n+1)*A242986(n-1)
    if 2.divides(n):
        h *= (4*(n+4))/(n^2*(n+6))
    return h
[A242986(n) for n in range(33)]

a(n) = (6*(n+1)/(3+floor(n/2)))*A056040(n).
a(2*n) = A007946(n).
Recurrence: a(n) = a(n-1)*(n+1)*(4*(n+4))/(n^2*(n+6)) if n mod 2 = 0 else a(n-1)*(n+1) for n>0, a(0) = 2.
Asymptotic: a(x) ~ exp(x*log(2) - log(Pi)/2 - cos(Pi*x)*(log(x/2) + 1/(2*x))/2 + log(6*(x+1)) - log(3+floor(x/2))) for x>=1.
G.f.: (4*x-1)/(2*x^6) + (-16*x^7+16*x^6-48*x^5+12*x^4+48*x^3-12*x^2-8*x+2)/(4*(1-4*x^2)^(3/2)*x^6). - Robert Israel, Aug 25 2014
Sum_{n>=0} 1/a(n) = Pi^2/54 + 19*Pi/(54*sqrt(3)) + 1/9. - Amiram Eldar, Feb 17 2023

A246507 a(n) = 70(n+1)binomial(2*n+1,n+1)/(n+5).

Original entry on oeis.org

14, 70, 300, 1225, 4900, 19404, 76440, 300300, 1178100, 4618900, 18106088, 70984095, 278369000, 1092063000, 4286142000, 16830250920, 66118842900, 259878874500, 1021939149000, 4020523757250, 15824781508536, 62313700079400, 245478212434000, 967428110493000, 3814113125277000

Offset: 0

Karol A. Penson, Aug 27 2014

nonn

4*a(n+1) is the number of annular noncrossing permutations of parameter 4, see the references.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Benoît Collins, James A. Mingo, Piotr Śniady, and Roland Speicher, Second order freeness and fluctuations of random matrices. III: Higher order freeness and free cumulants, Documenta Mathematica 12 (2007), 1-70.
James A. Mingo and Alexandru Nica, Annular noncrossing permutations and partitions, and second-order asymptotics for random matrices, International Mathematics Research Notices, Vol. 2004, No. 28 (2004), pp. 1413-1460; arXiv preprint, arXiv:math/0303312 [math.OA], 2003.

Cf. A001622, A001791, A007946, A078820.

[70*(n+1)*Binomial(2*n+1,n+1)/(n+5): n in [0..30]]; // Vincenzo Librandi, Aug 29 2014

Table[70 (n+1) Binomial[2 n + 1, n + 1]/(n + 5), {n, 0, 30}] (* Vincenzo Librandi, Aug 29 2014 *)

for(n=0,25, print1(70*(n+1)*binomial(2*n+1,n+1)/(n+5), ", ")) \\ G. C. Greubel, Apr 06 2017

O.g.f.: 2*(1-sqrt(1-4*z)-2*z-2*z^2-4*z^3-10*z^4)/(sqrt(1-4*z) *4*z^5).
Representation as the n-th moment of a signed function w(x) = 2*sqrt(x)*(x^4-2*x^3-2*x^2-4*x-10)/(4*Pi*sqrt(4-x)) on the segment x = (0,4): a(n) = Integral_{x=0..4} x^n*w(x) dx. The function w(x) -> 0 for x -> 0, and w(x) -> infinity for x->4.
a(n) ~ (35/65536)*4^n*(-755913243+151182552*n - 30236416*n^2 + 6047744*n^3 - 1212416*n^4 + 262144*n^5)/(n^(11/2)*sqrt(Pi)).
Another asymptotic series starts: a(n) ~ exp(n*log(4) + log((70*(2*n+1))/(n+5)) - log(Pi*n)/2 - 1/(8*n)). - Peter Luschny, Aug 28 2014
n*(n+5)*a(n) -2*(n+4)*(2*n+1)*a(n-1) = 0. - R. J. Mathar, Jun 14 2016
From Amiram Eldar, Feb 16 2023: (Start)
Sum_{n>=0} 1/a(n) = 2*Pi/(45*sqrt(3)) + 1/105.
Sum_{n>=0} (-1)^n/a(n) = 44*log(phi)/(175*sqrt(5)) + 1/175, where phi is the golden ratio (A001622). (End)

cp's OEIS Frontend

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

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A078817 Table by antidiagonals giving variants on Catalan sequence: T(n,k)=C(2n,n)*C(2k,k)*(2k+1)/(n+k+1).

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A078818 a(n) = 30*binomial(2n,n)/(n+3).

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A078820 a(n) = 20*C(2n,n)*(2n+1)/(n+4).

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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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Mathematica

Sage

Formula

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

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Maple

Mathematica

Sage

Formula

A242986 a(n) = 6*(n+1)!/((3+floor(n/2))*(floor(n/2)!)^2).

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Maple

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Sage

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A246507 a(n) = 70*(n+1)*binomial(2*n+1,n+1)/(n+5).

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

A078817 Table by antidiagonals giving variants on Catalan sequence: T(n,k)=C(2n,n)C(2k,k)(2k+1)/(n+k+1).

A078820 a(n) = 20C(2n,n)(2n+1)/(n+4).

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

A242986 a(n) = 6(n+1)!/((3+floor(n/2))(floor(n/2)!)^2).

A246507 a(n) = 70(n+1)binomial(2*n+1,n+1)/(n+5).