A078820 -id:A078820 - cp's OEIS frontend

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

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).

A078819 a(n) = 140*C(2n,n)/(n+4).

Original entry on oeis.org

35, 56, 140, 400, 1225, 3920, 12936, 43680, 150150, 523600, 1847560, 6584032, 23661365, 85652000, 312018000, 1142971200, 4207562730, 15557374800, 57750861000, 215145084000, 804104751450, 3014244096864, 11329763650800, 42691863032000, 161238018415500, 610258100044320

Offset: 0

Henry Bottomley, Dec 07 2002

			a(5)=140*C(10,5)/9=3920

Cf. A000108, A000984, A001622, A038629, A078817, A078818, A078820.

Table[140*Binomial[2*n, n]/(n + 4), {n, 0, 30}] (* Amiram Eldar, Feb 16 2023 *)

D-finite with recurrence a(n) = a(n-1)*(4n^2+10n-6)/(n^2+4n) = A078817(n, 3) = 7*A078820(n)/(2n+1) = 140*A000984(n)/(n+4).
From Amiram Eldar, Feb 16 2023: (Start)
Sum_{n>=0} 1/a(n) = Pi/(126*sqrt(3)) + 3/70.
Sum_{n>=0} (-1)^n/a(n) = 37/1750 - 3*log(phi)/(125*sqrt(5)), where phi is the golden ratio (A001622). (End)

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

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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A078819 a(n) = 140*C(2n,n)/(n+4).

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

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cp's OEIS Frontend

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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Author

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Examples

Links

Crossrefs

Programs

Maple

Formula

A078819 a(n) = 140*C(2n,n)/(n+4).

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

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

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