A135573 -id:A135573 - cp's OEIS frontend

A348898 a(n) = 15120(2n)!/(n!*(n + 5)!); super ballot numbers, row 4 of A135573.

126, 42, 36, 45, 70, 126, 252, 546, 1260, 3060, 7752, 20349, 55062, 152950, 434700, 1260630, 3721860, 11165580, 33982200, 104778450, 326908764, 1031019948, 3283989464, 10555680420, 34214964120, 111768882792, 367755678864, 1218190686237, 4060635620790, 13615072375590

Offset: 0

Peter Luschny, Nov 03 2021

Cf. A000108, A001622, A135573.

a := n -> 15120*(2*n)!/(n!*(n + 5)!): seq(a(n), n = 0..29);

a[n_] := 4^(n + 5) Hypergeometric2F1[11/2, 1/2 - n, 13/2, 1] / (11 Pi);
Table[a[n], {n, 0, 29}]

a(n)=126*binomial(2*n,n)/binomial(n+5,5) \\ Charles R Greathouse IV, Oct 23 2023

def A348898(n): return -(-4)^(5 + n)*binomial(9/2, 5 + n)/2
print([A348898(n) for n in range(30)])  # Peter Luschny, Nov 04 2021

a(n) = (1/(2*Pi))*Integral_{x=0..4} x^n*(4 - x)^(9/2)*x^(-1/2). This is the integral representation of the n-th moment of a positive function on [0, 4]. This representation is unique.
E.g.f.: x^(-4)*exp(2*x)*((-256*x^4 - 160*x^3 - 108*x^2 - 60*x - 24)*BesselI(1, 2*x) + (256*x^4 + 96*x^3 + 60*x^2 + 24*x)*BesselI(0, 2*x)).
O.g.f.: ((512*x^4 - 325*x^3 + 110*x^2 - 17*x + 1)*sqrt(1 - 4*x) - 748*x^4 + 515*x^3 - 142*x^2 + 19*x - 1)/(sqrt(1 - 4*x)*(1 + sqrt(1 - 4*x))*x^4).
a(n) = 4^(n + 5)*hypergeom([11/2, 1/2 - n], [13/2], 1) / (11*Pi).
a(n) = -(-4)^(5 + n)*binomial(9/2, 5 + n)/2. - Peter Luschny, Nov 04 2021
From Peter Bala, Mar 11 2023: (Start)
P-recursive: (n + 5)*a(n) = 2*(2*n - 1)*a(n-1) with a(0) = 126.
a(n) = Sum_{k = 0..4} (-1)^k*4^(4-k)*binomial(4,k)*Catalan(n+k) = 256*Catalan(n) - 256*Catalan(n+1) + 96*Catalan(n+2) - 16*Catalan(n+3) + Catalan(n+4), where Catalan(n) = A000108(n). Thus a(n) is an integer for all n.
a(n) is odd if n = 2^k - 5, k >= 3, otherwise a(n) is even. (End)
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = 47/630 + 44*Pi/(2187*sqrt(3)).
Sum_{n>=0} (-1)^(n+1)/a(n) = 13/218750 + 88*log(phi)/(15625*sqrt(5)), where phi is the golden ratio (A001622). (End)

A348899 a(n) = 3326404^nGamma(n + 1/2)/(sqrt(Pi)*Gamma(n + 7)); super ballot numbers, row 5 of A135573.

Original entry on oeis.org

462, 132, 99, 110, 154, 252, 462, 924, 1980, 4488, 10659, 26334, 67298, 177100, 478170, 1320660, 3721860, 10680120, 31150350, 92205036, 276615108, 840090328, 2580277436, 8007757560, 25090973688, 79319852304, 252832029219, 812127124158, 2627470107570, 8558045493228

Offset: 0

Peter Luschny, Nov 02 2021

nonn

Row 5 of array A135573.

Cf. A000108, A000984, A001622, A007054, A348893.

a := n -> 332640*4^n*GAMMA(n + 1/2)/(sqrt(Pi)*GAMMA(n + 7));
seq(a(n), n = 0..29);

a[n_] := 4^(n + 6) Hypergeometric2F1[13/2, 1/2 - n, 15/2, 1] / (13 Pi);
Table[a[n], {n, 0, 29}]
Array[332640*4^#*Gamma[# + 1/2]/(Sqrt[Pi]*Gamma[# + 7]) &, 30, 0] (* Michael De Vlieger, Nov 02 2021 *)

Let A[c, k](n) = c*4^n*Gamma(n + 1/2)/(sqrt(Pi)*Gamma(n + k)). Then
  A[1, 1](n) = A000984(n).
  A[3!, 3](n) = A007054(n).
  A[5!*7, 5](n) = A348893(n).
  A[7!*66, 7](n) = a(n).
  A[c, k](n) ~ -c*2^(2*n - 1)*(k^2 - k - 2*n + 1/4)/(n^(k + 1/2)*sqrt(Pi)).
O.g.f.: ((2048*x^5 - 1686*x^4 + 765*x^3 - 178*x^2 + 21*x - 1)*sqrt(1 - 4*x) - 3496*x^5 + 2934*x^4 - 1083*x^3 + 218*x^2 - 23*x + 1)/(sqrt(1 - 4*x)*(1 + sqrt(1 - 4*x))*x^5).
E.g.f.: 1024*exp(2*x)*((-x^5 - 3/4*x^4 - 41/64*x^3 - 123/256*x^2 - 9/32*x - 15/128)*BesselI(1, 2*x) + BesselI(0, 2*x)*x*(x^4 + 1/2*x^3 + 27/64*x^2 + 9/32*x + 15/128))/x^5.
a(n) = Integral_{x=0..4} x^n*(4-x)^(11/2)/(2*Pi*sqrt(x)). This is the integral representation as the n-th moment of a positive function on [0, 4]. The representation is unique.
a(n) = 4^(n + 6)*hypergeom([13/2, 1/2 - n], [15/2], 1) / (13*Pi).
D-finite with recurrence (n+6)*a(n) +2*(-2*n+1)*a(n-1)=0. - R. J. Mathar, Jul 27 2022
From Peter Bala, Mar 11 2023: (Start)
a(n) = 332640*(2*n)!/(n!*(n + 6)!).
a(n) = Sum_{k = 0..5} (-1)^k*4^(5-k)*binomial(n,k)*Catalan(n+k), where Catalan(n) = A000108(n). Thus a(n) is an integer for all n.
a(n) is odd if n = 2^k - 6, k >= 3, otherwise a(n) is even. (End)
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = 101/3465 + 52*Pi/(6561*sqrt(3)).
Sum_{n>=0} (-1)^(n+1)/a(n) = 8573/54140625 + 104*log(phi)/(78125*sqrt(5)), where phi is the golden ratio (A001622). (End)

A007272 Super ballot numbers: 60(2n)!/(n!(n+3)!).

Original entry on oeis.org

10, 5, 6, 10, 20, 45, 110, 286, 780, 2210, 6460, 19380, 59432, 185725, 589950, 1900950, 6203100, 20470230, 68234100, 229514700, 778354200, 2659376850, 9148256364, 31667041260, 110248217720, 385868762020, 1357193576760, 4795417304552, 17015996887120, 60619488910365

Offset: 0

N. J. A. Sloane, Simon Plouffe, Ira M. Gessel

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Matthew House, Table of n, a(n) for n = 0..1677
Emily Allen and Irina Gheorghiciuc, A Weighted Interpretation for the Super Catalan Numbers, J. Int. Seq. 17 (2014) # 14.10.7.
David Callan, A combinatorial interpretation for a super-Catalan recurrence, arXiv:math/0408117 [math.CO], 2004.
Ira M. Gessel, Super ballot numbers, J. Symbolic Comp., 14 (1992), 179-194
Ira M. Gessel and Guoce Xin, A Combinatorial Interpretation of the Numbers 6(2n)!/n!(n+2)!, Journal of Integer Sequences, Vol. 8 (2005), Article 05.2.3.
Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 96.

Row 2 of the array A135573.

Cf. A001622, A002422, A007054, A348893.

seq(10*(2*n)!/(n!)^2/binomial(n+3,n), n=0..26); # Zerinvary Lajos, Jun 28 2007

Table[60(2n)!/(n!(n+3)!), {n, 0, 30}] (* Jean-François Alcover, Jun 02 2019 *)

a(n)=if(n<0, 0, 60*(2*n)!/n!/(n+3)!) /* Michael Somos, Feb 19 2006 */

{a(n)=if(n<0, 0, n*=2; n!*polcoeff( 10*besseli(3,2*x+x*O(x^n)), n))} /* Michael Somos, Feb 19 2006 */

def A007272(n): return -(-4)^(3 + n)*binomial(5/2, 3 + n)/2
print([A007272(n) for n in range(30)])  # Peter Luschny, Nov 04 2021

G.f.: (11-32*x+9*sqrt(1-4*x))/(1-3*x+(1-x)*sqrt(1-4*x)).
E.g.f.: Sum_{n>=0} a(n)*x^(2n)/(2n)! = 60*BesselI(3, 2x)/x^3.
E.g.f.: (BesselI(0, 2*x)*(2*x+16*x^2)-BesselI(1, 2*x)*(2+6*x+16*x^2))*exp(2*x)/x^2.
Integral representation as the n-th moment of a positive function on [0, 4]: a(n) = Integral_{x=0..4} x^n*(4-x)^(5/2)/(2*Pi*x^(1/2)) dx. This representation is unique. - Karol A. Penson, Dec 04 2001
a(n) = 10*(2*n)!*[x^(2*n)](hypergeometric([],[4],x^2)). - Peter Luschny, Feb 01 2015
(n+3)*a(n) +2*(-2*n+1)*a(n-1)=0. - R. J. Mathar, Mar 06 2018
a(n) = -(-4)^(3+n)*binomial(5/2, 3+n)/2. - Peter Luschny, Nov 04 2021
From Amiram Eldar, Mar 24 2022: (Start)
Sum_{n>=0} 1/a(n) = 4/9 + 28*Pi/(3^5*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 38/1875 - 56*log(phi)/(5^4*sqrt(5)), where phi is the golden ratio (A001622). (End)
From Peter Bala, Mar 11 2023: (Start)
a(n) = Sum_{k = 0..2} (-1)^k*4^(2-k)*binomial(n,k)*Catalan(n+k) = 16*Catalan(n) - 8*Catalan(n+1) + Catalan(n+2), where Catalan(n) = A000108(n). Thus a(n) is an integer for all n.
a(n) is odd if n = 2^k - 3, k >= 2, else a(n) is even. (End)

A348893 a(n) = 840(2n)!/((n + 4)!*n!).

Original entry on oeis.org

35, 14, 14, 20, 35, 70, 154, 364, 910, 2380, 6460, 18088, 52003, 152950, 458850, 1400700, 4342170, 13646820, 43421700, 139704600, 454039950, 1489251036, 4925984196, 16419947320, 55124108860, 186281471320, 633357002488, 2165672331088, 7444498638115, 25717358931670, 89254363351090

Offset: 0

Karol A. Penson, Nov 02 2021

Row 3 of array A135573.

Cf. A000108, A001622, A007054, A007272, A348898, A348899.

Maple
```
seq(840*(2*n)!/((n + 4)!*n!),n=0..30)
```

a[n_] := 4^(n + 4) Hypergeometric2F1[9/2, 1/2 - n, 11/2, 1] / (9 Pi);
Table[a[n], {n, 0, 30}] (* Peter Luschny, Nov 03 2021 *)

a(n)=35*binomial(2*n,n)/binomial(n+4,4) \\ Charles R Greathouse IV, Oct 23 2023

def A348893(n): return (-4)^(4 + n)*binomial(7/2, 4 + n)/2
print([A348893(n) for n in range(31)])  # Peter Luschny, Nov 04 2021

O.g.f: (140*z^3 - 70*z^2 + 14*z - 1 + (1 - 4*z)^(7/2))/(2*z^4).
E.g.f: 64*exp(2*z)*((-z^3 - 1/2*z^2 - 1/4*z - 3/32)*BesselI(1,2*z) + BesselI(0,2*z)*z*(z^2 + 1/4*z + 3/32))/z^3.
O.g.f. g(z) satisfies z^4*g(z)^2 + (-140*z^3 + 70*z^2 - 14*z + 1)*g(z) + 4096*z^3 - 2268*z^2 + 476*z - 35 = 0;
a(n) = Integral_{x=0..4} x^n*64*(1 - x/4)^(7/2)/(Pi*sqrt(x)). This is the integral representation as the n-th moment of a positive function on [0, 4]. The representation is unique.
Remark: this sequence is not monotonically growing with n, as a(0) > a(1) = a(2) < a(3) < a(4)... .
From Peter Luschny, Nov 03 2021: (Start)
a(n) = 14*A007272(n)/(n + 4).
a(n) ~ 105*4^n*(8*n - 81)/(n^(11/2)*sqrt(Pi)).
a(n) = 4^(n + 4)*hypergeom([9/2, 1/2 - n], [11/2], 1) / (9*Pi). (End)
a(n) = (-4)^(4 + n)*binomial(7/2, 4 + n)/2. - Peter Luschny, Nov 04 2021
From Peter Bala, Mar 10 2023: (Start)
a(n) = 35*binomial(2*n, n) - 56*binomial(2*n, n + 1) + 28*binomial(2*n, n + 2) - 8*binomial(2*n, n + 3) + binomial(2*n, n + 4). Thus this sequence is integral.
7 divides a(n) except when n == 3 (mod 7).
P-recursive: (n + 4)*a(n) = 2*(2*n - 1)*a(n-1) with a(0) = 35.
D-finite: the o.g.f. A(x) satisfies the differential equation (1 - 4*x)*A'(x) + (4 - 2*x)*A(x) - 140 = 0, with A(0) = 35. (End)
From Peter Bala, Mar 11 2023: (Start)
a(n) = Sum_{k = 0..3} (-1)^k*4^(3-k)*binomial(3,k)*Catalan(n+k) = 64*Catalan(n) - 48*Catalan(n+1) + 12*Catalan(n+2) - Catalan(n+3), where Catalan(n) = A000108(n).
a(n) is odd if n = 2^k - 4, k >= 2, otherwise a(n) is even. (End)
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = 13/70 + 4*Pi/(81*sqrt(3)).
Sum_{n>=0} (-1)^(n+1)/a(n) = 72*log(phi)/(3125*sqrt(5)) - 103/43750, where phi is the golden ratio (A001622). (End)

A361027 Table of generalized de Bruijn's numbers (A006480) read by ascending antidiagonals.

Original entry on oeis.org

2, 30, 3, 560, 20, 20, 11550, 210, 75, 210, 252252, 2772, 504, 504, 2772, 5717712, 42042, 4620, 2352, 4620, 42042, 133024320, 700128, 51480, 15840, 15840, 51480, 700128, 3155170590, 12471030, 656370, 135135, 81675, 135135, 656370, 12471030, 75957810500, 233716340, 9237800

Offset: 0

Peter Bala, Feb 28 2023

The Catalan numbers A000108 are given by the formula Catalan(n) = (2*n)!/(n!*(n + 1)!). Gessel (1992) considered generalized Catalan numbers defined by Catalan(r,n) = J(r) * (2*n)!/(n!*(n + r + 1)!), where J(r) = (2*r + 2)!/(2*(r + 1)!) = (2^r)*Product_{j = 0..r} (2*j + 1) is chosen so that these numbers are always integers. Gessel's generalized Catalan numbers are particular cases of super ballot numbers. See A135573 for a table of these generalized Catalan numbers.
For this table we carry out an analogous construction using the de Bruijn numbers B(n) = (3*n)!/n!^3 = A006480(n) in place of the central binomial numbers. We define the generalized de Bruijn number B(r,n), r = 0, 1, 2, ..., by B(r,n) = F(r) * (3*n)!/(n!*(n + r + 1)!^2), where choosing F(r) = (3*r + 3)!/(3*(r + 1)!) = (3^r)*Product_{j = 0..r} (3*j + 1)*(3*j + 2) appears to produce integer values for these quantities. We have verified this for rows 0, 1, 2 and 3 of the table.
An alternative expression for the generalized de Bruijn numbers is B(r,n) =  G(r,n) * B(n+r+1), where G(r) = (1/3)*Product_{j = 0..r} ( (3*j + 1)*(3*j + 2)/((3*n + 3*j + 1)*(3*n + 3*j + 2)) ).
The rows of the square array below are the sequences of generalized de Bruijn numbers {B(0,k)}, {B(1,k)}, {B(2,k)}, ....

			The square array with rows n >= 0 and columns k >= 0 begins:
  n\k|       0       1       2        3        4         5         6 ...
  ----------------------------------------------------------------------
   0 |       2       3      20      210     2772     42042    700128 ...
   1 |      30      20      75      504     4620     51480    656370 ...
   2 |     560     210     504     2352    15840    135135   1361360 ...
   3 |   11550    2772    4620    15840    81675    550550   4492488 ...
   4 |  252252   42042   51480   135135   550550   3006003  20271888 ...
   5 | 5717712  700128  656370  1361360  4492488  20271888  ...
  ...
As a triangle:
 Row
  0 |        2
  1 |       30       3
  2 |      560      20     20
  3 |    11550     210     75    210
  4 |   252252    2772    504    504   2772
  5 |  5717712   42042   4620   2352   4620   42042
  ...

N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.

Ira M. Gessel, Super ballot numbers, J. Symbolic Comp., 14 (1992), 179-194.

A208881 (column 1), A361028(row 0), A361029(row 1), A361030(row 2), A361031(row 3).

Cf. A006480, A024488, A135573.

# as a square array
T := proc (n,k) (1/3)*27^(n+k+1)*binomial(n+1/3, n+k+1)*binomial(n+2/3,
n+k+1); end proc:
for n from 0 to 10 do seq(T(n,k), k = 0..10); end do;
# as a triangle
T := proc (n,k) (1/3)*27^(n+k+1)*binomial(n+1/3, n+k+1)*binomial(n+2/3,
n+k+1); end proc:
for n from 0 to 10 do seq(T(n-k,k), k = 0..n); end do;

T(n,k) = (3*n + 3)!/(3*(n + 1)!) * (3*k)!/(k!*(k + n + 1)!^2), n, k >= 0.
T(n,k) = (1/3)*27^(n+1+k)*binomial(n+1/3, n+1+k)*binomial(n+2/3, n+1+k).
T(n,k) = (1/(2*Pi))^2 * 1/27^(n+k+1) * Integral_{x = 0..27} (27 - x)^(n+2/3)*x^(k-2/3) dx * Integral_{x = 0..27} (27 - x)^(n+1/3)*x^(k-1/3) dx.
P-recursive: (n + k + 1)^2*T(n,k) = 3*(3*k - 1)*(3*k - 2)*T(n,k-1) with T(n,0) = 1/(n+1)!^2 *  (3*n + 3)!/(3*(n + 1)!).
(n + k + 1)^2*T(n,k) = 3*(3*n + 1)*(3*n + 2)*T(n-1,k) with T(0,k) = 2*(k + 1)*(3*k)!/(k + 1)!^3.
T(n,0) = A208881(n+1).

A361032 Square array read by ascending antidiagonals: T(n,k) = F(n) * (4k)!/(k!(k + n + 1)!^3), where F(n) = (1/8)(4n + 4)!/(n + 1)!; n, k >= 0.

Original entry on oeis.org

3, 315, 9, 46200, 280, 280, 7882875, 17325, 3675, 17325, 1466593128, 1513512, 116424, 116424, 1513512, 288592936632, 162954792, 5885880, 2134440, 5885880, 162954792, 59064793444800, 20193091776, 399072960, 67953600, 67953600, 399072960, 20193091776, 12445136556298875

Offset: 0

Peter Bala, Mar 04 2023

The Catalan numbers A000108 are given by the formula Catalan(k) = (2*k)!/(k!*(k + 1)!). Gessel (1992) considered generalized Catalan numbers defined by Catalan(n,k) = J(n) * (2*k)!/(k!*(k + n + 1)!), where J(n) = (2*n + 2)!/(2*(n + 1)!) = (2^n)*Product_{j = 0..n} (2*j + 1) is chosen so that these numbers are always integers. Gessel's generalized Catalan numbers are particular cases of super ballot numbers. See A135573 for a table of these generalized Catalan numbers.
For this table we carry out an analogous construction using the numbers B(k) = (4*k)!/k!^4 = A008977(k) in place of the central binomial numbers (2*k)!/k!^2. We define sequences {B(n,k) : k >= 0}, n = 0, 1, 2, ..., by B(n,k) = F(n) * (4*k)!/(k!*(k + n + 1)!^3), where choosing F(n) = (4*n + 4)!/(8*(n + 1)!) = (1/2)*(4^n)*Product_{j = 0..n} (4*j + 1)*(4*j + 2)*(4*j + 3) appears to produce integer values for these quantities. The rows of the square array below are the sequences  {B(0,k)}, {B(1,k)}, {B(2,k)}, ....
An alternative expression is B(n,k) =  G(n,k) * B(n+k+1), where G(n,k) = (1/8)*Product_{j = 0..n} ( (4*j + 1)*(4*j + 2)*(4*j + 3)/((4*k + 4*j + 1)*(4*k + 4*j + 2)*(4*k + 4*j + 3)) ).

			The square array with rows n >= 0 and columns k >= 0 begins:
  n\k|          0          1          2           3           4 ...
  ----------------------------------------------------------------------
   0 |          3          9        280       17325     1513512 ...
   1 |        315        280       3675      116424     5885880 ...
   2 |      46200      17325     116424     2134440    67953600 ...
   3 |    7882875    1513512    5885880    67953600  1449322875 ...
   4 | 1466593128  162954792  399072960  3086579925 46235189000 ...
   5 |  ...
  ...
As a triangle:
 Row
  0 |             3
  1 |           315          9
  2 |         46200        280      280
  3 |       7882875      17325     3675    17325
  4 |    1466593128    1513512   116424   116424  1513512
  5 |  288592936632  162954792  5885880  2134440  5885880  162954792

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (first 51 antidiagonals)
Ira M. Gessel, Super ballot numbers, J. Symbolic Comp., 14 (1992), 179-194.

A361033 (row 0), A361034 (row 2), A361035 (row 3).

Cf. A008977, A135573, A361027.

# as a square array
T := proc (n, k) (-1)^k*(1/8)*256^(n+1+k)*binomial(n+1/4, n+1+k)*binomial(n+2/4, n+1+k)* binomial(n+3/4, n+1+k); end proc:
for n from 0 to 10 do seq(T(n, k), k = 0..10); end do;
# as a triangle
T := proc (n, k) (-1)^k*(1/8)*256^(n+1+k)*binomial(n+1/4, n+1+k)*binomial(n+2/4, n+1+k)* binomial(n+3/4, n+1+k); end proc:
for n from 0 to 10 do seq(T(n-k, k), k = 0..n); end do;

T(n,k) = (-1)^k*(1/8)*256^(n+k+1)*binomial(n+1/4, n+k+1)*binomial(n+1/2, n+k+1)* binomial(n+3/4, n+k+1) \\ Andrew Howroyd, Jan 05 2024

T(n,k) = (-1)^k*(1/8)*256^(n+k+1)*binomial(n+1/4, n+k+1)*binomial(n+1/2, n+k+1)* binomial(n+3/4, n+k+1).
P-recursive: (n + k + 1)^3*T(n,k) = 4*(4*k - 1)*(4*k - 2)*(4*k - 3)*T(n,k-1) with T(n,0) = (1/8)*(4*n + 4)!/(n + 1)!^4 = (1/8)*A008977(n+1).
(n + k + 1)^3*T(n,k) = 4*(4*n + 1)*(4*n + 2)*(4*n + 3)*T(n-1,k) with T(0,k) = 3*(4*k)!/(k!*(k+1)!^3) = A361033(k).
T(n,k) = (1/2) * (1/(2*Pi))^3 * 256^(n+k+1) * Integral_{x = 0..1} (1 - x)^(n+1/4)*x^(k-1/4) dx * Integral_{x = 0..1} (1 - x)^(n+1/2)*x^(k-1/2) dx * Integral_{x = 0..1} (1 - x)^(n+3/4)*x^(k-3/4) dx.

A361037 a(n) = 20(3n)!/((2n)!(n+2)!).

Original entry on oeis.org

10, 10, 25, 84, 330, 1430, 6630, 32300, 163438, 852150, 4552275, 24812400, 137547000, 773564328, 4405019090, 25357898940, 147375745990, 863805209750, 5101386767295, 30332569967700, 181465130121450, 1091677288630950

Offset: 0

Peter Bala, Mar 04 2023

Gessel (1992) introduced sequences {b(r,n): n >= 0} of super ballot numbers defined by b(r,n) = J(r) * (2*n)!/(n!*(n + r + 1)!), r = 0,1,2,..., where J(r) = (2*r + 2)!/(2*(r + 1)!) = (2^r)*Product_{j = 0..r} (2*j + 1) is chosen so that these numbers are always integers. The sequence {b(1,n) : n >= 0} is A000108, the sequence of Catalan numbers. See A135573 for a table of these generalized Catalan numbers.

We carry out an analogous construction using the numbers B(n) = A005809(n) = binomial(3*n,n) = (3*n)!/((2*n)!*n!) in place of the central binomial numbers. We define B(r,n), r = 0,1,2, ..., by B(r,n) = F(r) * (3*n)!/((2*n)!*(n + r + 1)!), where F(r) is the minimal choice to produce integer values for these quantities for all n. This sequence is the case r = 1. See A007226 (r = 0), A361038 (r = 2) and A361039 (r = 3).

Ira M. Gessel, Super ballot numbers, J. Symbolic Comp., 14 (1992), 179-194.

Cf. A000139, A001764, A005809, A007226, A361038, A361039.

seq( 20*(3*n)!/((2*n)!*(n+2)!), n = 0..20);

Table[20 (3n)!/((2n)!(n+2)!),{n,0,30}] (* Harvey P. Dale, Aug 05 2024 *)

a(n) = 10*binomial(3*n,n) - 7*binomial(3*n,n+1) + binomial(3*n,n+2). Thus a(n) is an integer.
P-recursive: 2*(n + 2)(2*n - 1)*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 10.
a(n) ~ (27/4)^n * 10*sqrt(3/Pi)/n^(5/2).
The o.g.f. satisfies the differential equation
x^2*(27*x - 4)*A''(x) + 2*x*(27*x - 5)*A'(x) + 2*(3*x + 2)*A(x) - 40 = 0, with A(0) = 10 and A'(0) = 10.

A182073 Square array read by antidiagonals: T(n,k) = n!k! / (floor(n/2)! floor(k/2)! * floor((n+k)/2)!).

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 6, 2, 2, 6, 6, 3, 2, 3, 6, 30, 6, 6, 6, 6, 30, 20, 10, 4, 6, 4, 10, 20, 140, 20, 20, 12, 12, 20, 20, 140, 70, 35, 10, 15, 6, 15, 10, 35, 70, 630, 70, 70, 30, 30, 30, 30, 70, 70, 630, 252, 126, 28, 42, 12, 30, 12, 42, 28, 126, 252

Offset: 0

Peter Bala, Apr 10 2012

Compare with A068555 whose entries are given by (2*n)!*(2*k)!/(n!*k!*(n+k)!). See also A211226.

			As a square array
.n\k.|...0....1....2....3....4....5....6...
= = = = = = = = = = = = = = = = = = = = = =
..0..|...1....1....2....6....6...30...20...
..1..|...1....1....2....3....6...10...20...
..2..|...2....2....2....6....4...20...10...
..3..|...6....3....6....6...12...15...30...
..4..|...6....6....4...12....6...30...12...
..5..|..30...10...20...15...30...30...60...
..6..|..20...20...10...30...12...60...20...
...
Formatted as a triangle
.n\k.|...0....1....2....3....4....5....6
= = = = = = = = = = = = = = = = = = = = =
..0..|...1
..1..|...1....1
..2..|...2....1....2
..3..|...6....2....2....6
..4..|...6....3....2....3....6
..5..|..30....6....6....6....6...30
..6..|..20...10....4....6....4...10...20
...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A068555, A135573, A211226.

T[n_, k_] := n!*k!/(Floor[n/2]!*Floor[k/2]!*Floor[(n + k)/2]!);
TableForm[Table[T[n, k], {n, 0, 5}, {k, 0, 10}]]
Table[T[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* G. C. Greubel, Aug 20 2017 *)

That T(n,k) is an integer follows from the formulas:
T(2*n,2*k) = (2*n)!*(2*k)!/(n!*k!*(n+k)!) = A068555(n,k);
T(2*n,2*k+1) = (2*n)!*(2*k+1)!/(n!*k!*(n+k)!) = (2*k+1)*A068555(n,k);
T(2*n+1,2*k) = (2*n+1)!*(2*k)!/(n!*k!*(n+k)!) = (2*n+1)*A068555(n,k);
T(2*n+1,2*k+1) = (2*n+1)!*(2*k+1)!/(n!*k!*(n+k+1)!) = (2*k+1)*A135573(n,k).

A361040 a(n) = 420(3n)!/(n!(2n + 3)!).

Original entry on oeis.org

70, 21, 30, 70, 210, 735, 2856, 11970, 53130, 246675, 1187550, 5890248, 29954680, 155602020, 823184880, 4424618730, 24116031162, 133072694475, 742405558650, 4182821562150, 23776769743650, 136248095712855, 786482994679200

Offset: 0

Peter Bala, Mar 04 2023

The Catalan numbers A000108 are defined by the formula Catalan(n) = (2*n)!/(n!*(n+1)!). Gessel (1992) considered generalized Catalan numbers defined by Catalan(r,n) = J(r)*(2*n)!/(n!*(n+r+1)!), where J(r) = (2^r)*Product_{j = 0..r} (2*j + 1) is chosen so that these numbers are always integers. Gessel's generalized Catalan numbers are particular cases of super ballot numbers. See A135573 for a table of these generalized Catalan numbers.

For r = 0,1,2,..., it appears that there is an integer C(r) such the sequence {C(r)*(3*n)!/(n!*(2*n + r)!) : n >= 0} is integral. This is the case r = 3. For other cases see A005809 (r = 0, C(0) = 1), A001764 (r = 1, C(1) = 1), A000139 (r = 2, C(2) = 4) and A361041 (r = 4, C(4) = 1680).

Ira M. Gessel, Super ballot numbers, J. Symbolic Comp., 14 (1992), 179-194.

Cf. A000139, A001764, A005809, A007272, A135573, A361038, A361041.

seq( 420*(3*n)!/(n!*(2*n + 3)!), n = 0..20)

a(n) = 70*binomial(3*n,2*n) - 189*binomial(3*n,2*n+1) + 114*binomial(3*n,2*n+2) -32*binomial(3*n,2*n+3). Thus a(n) is an integer.
P-recursive: 2*(n + 1)*(2*n + 3)*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 70.
a(n) ~ (27/4)^n * 105*sqrt(3/(16*Pi))/n^(7/2).
The o.g.f. A(x) satisfies the differential equation
x^2*(4 - 27*x^4)*A''(x) + 2*x*(7 - 27*x)*A'(x) + (6 - 6*x)*A(x) - 420 = 0, with A(0) = 70 and A'(0) = 21.

A364318 Irregular table T read by rows: T(n,k) gives for permutations of [n] = {1, 2, ..., n}, n >= 1, the number of cycles corresponding to the k-th partition of n without part 1 (in Abramowitz-Stegun order).

Original entry on oeis.org

0, 1, 2, 6, 3, 24, 20, 120, 90, 40, 15, 720, 504, 420, 210, 5040, 3360, 2688, 1260, 1260, 1120, 105, 40320, 25920, 20160, 18144, 9072, 15120, 2240, 2520, 362880, 226800, 172800, 151200, 72576, 75600, 120960, 56700, 50400, 18900, 25200, 945

Offset: 1

Wolfdieter Lang, Aug 14 2023

The length of row n is 1 for n = 1, and A002865(n), for n >= 2.
This table selects in row n the entries of the partition array (multinomial M2 numbers) corresponding to partitions without part 1.
The characteristic array for partitions of n without part 1 is given in A145573, and the complete M_2 multinomial table is found in A036039 (see the W. Lang link there for n = 1, 2, ..., 10).
In table A008306 the entries of the present table belonging to the same length of the cycles (that is, the same number of parts of the corresponding partition) are summed.
The row sums give the subfactorial (derangement) numbers A000166. (See the W. Lang comment from Jun 01 2010 there on the M_2 multinomial array for the partitions without part 1.)

			The table T begins:
n\k     1     2     3     4    5     6    7    8 ...  Row sums A000166
----------------------------------------------------------------------
1:      0                                                        0
2:      1                                                        1
3:      2                                                        2
4:      6     3                                                  9
5:     24    20                                                 44
6:    120    90    40    15                                    265
7:    720   504   420   210                                   1854
8:   5040  3360  2688  1260 1260  1120  105                  14833
9:  40320 25920 20160 18144 9072 15120 2240 2520            133496
...
n = 10: 362880 226800 172800 151200 72576 75600 120960 56700 50400 18900 25200 945, with sum 1334961.
n = 11: 3628800 2217600 1663200 1425600 1330560 712800 1108800 997920 443520 415800 166320 415800 123200 34650, with sum 14684570.
...
T(8, 6) corresponds to the partition (2,3^2) of n = 8, hence its M_2 multinomial number is 8!/((2^1*1!)*(3^2*2!)) = 1120.
With the number of cycle calculation T(8, 6) = #Cy(8;2,1)*#Cy(8-2;3,2) with #Cy(N;j,ej) = (j-1)!^ej*Product_{q=0..ej-1} binomial(N - q*j, j)/ej!, Hence #Cy(8;2,1) = 1!*binomial(8, 2)/1! = 28, and #Cy(6;3,2) = 2!^2*binomial(6, 3)*binomial(6-3, 3)/2! = 40, and T(8, 6) = 28*40 = 1120.

Cf. A000166, A002865, A008306, A135573.

Column k = 1 is A000142(n-1) = (n-1)!.

T(n, k) is the table M_2 = A036039 without the entries corresponding to the partitions of n with at least one part 1 (the Abromowitz-Stegun order of partitions is used).

cp's OEIS Frontend

A348898 a(n) = 15120*(2*n)!/(n!*(n + 5)!); super ballot numbers, row 4 of A135573.

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A348899 a(n) = 332640*4^n*Gamma(n + 1/2)/(sqrt(Pi)*Gamma(n + 7)); super ballot numbers, row 5 of A135573.

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A007272 Super ballot numbers: 60*(2n)!/(n!*(n+3)!).

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A348893 a(n) = 840*(2*n)!/((n + 4)!*n!).

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A361027 Table of generalized de Bruijn's numbers (A006480) read by ascending antidiagonals.

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A361032 Square array read by ascending antidiagonals: T(n,k) = F(n) * (4*k)!/(k!*(k + n + 1)!^3), where F(n) = (1/8)*(4*n + 4)!/(n + 1)!; n, k >= 0.

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A361037 a(n) = 20*(3*n)!/((2*n)!*(n+2)!).

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A348898 a(n) = 15120(2n)!/(n!*(n + 5)!); super ballot numbers, row 4 of A135573.

A348899 a(n) = 3326404^nGamma(n + 1/2)/(sqrt(Pi)*Gamma(n + 7)); super ballot numbers, row 5 of A135573.

A007272 Super ballot numbers: 60(2n)!/(n!(n+3)!).

A348893 a(n) = 840(2n)!/((n + 4)!*n!).

A361032 Square array read by ascending antidiagonals: T(n,k) = F(n) * (4k)!/(k!(k + n + 1)!^3), where F(n) = (1/8)(4n + 4)!/(n + 1)!; n, k >= 0.

A361037 a(n) = 20(3n)!/((2n)!(n+2)!).

A182073 Square array read by antidiagonals: T(n,k) = n!k! / (floor(n/2)! floor(k/2)! * floor((n+k)/2)!).

A361040 a(n) = 420(3n)!/(n!(2n + 3)!).