A008977 -id:A008977 - cp's OEIS frontend

A001451 a(n) = (5n)!/((3n)!n!n!).

1, 20, 1260, 100100, 8817900, 823727520, 79919739900, 7962100660800, 808906548235500, 83426304143982800, 8707404737345073760, 917663774856743842200, 97491279924241456098300, 10427604345391237790688000, 1121786259855036145008408000

Offset: 0

N. J. A. Sloane

From Peter Bala, Jul 15 2016: (Start)
This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = (-1)^m*a(m) for n = 2*m. The sum vanishes for n odd. Cf. A273628 and A273629.
Note the similar results:
Sum_{k = 0..n} (-1)^k*binomial(n,k)* binomial(2*n + k,n)*binomial(3*n + k,n) = (-1)^n*(3*n)!/n!^3 = (-1)^n*A006480(n);
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n)^2 = A002894(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n) = A000984(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n). (End)
Choose three noncollinear step vectors to satisfy the zero sum, 3*v_1 + v_2 + v_3 = 0. Then a(n) is the number of loop plane walks of length 5*n which depart from and return to the origin. Equivalently, a(n) counts distinct permutations of a (5*n)-digit integer with digits 1,2,3 of multiplicity 3*n,n,n respectively. - Bradley Klee, Aug 12 2018

			G.f. = 1 + 20*x + 1260*x^2 + 100100*x^3 + 8817900*x^4 + 823727520*x^5 + ... - _Michael Somos_, Aug 12 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000897, A002894, A002897, A006480, A008977, A186420, A188662, A273628, A273629.

List([0..15],n->Factorial(5*n)/(Factorial(3*n)*Fact0rial(n)*Factorial(n))); # Muniru A Asiru, Aug 12 2018

[Factorial(5*n)/(Factorial(3*n)*Factorial(n)*Factorial(n)): n in [0..30]]; // Vincenzo Librandi, May 22 2011

Maple
```
f := n->(5*n)!/((3*n)!*n!*n!);
```

Table[(5*n)!/((3*n)!*n!*n!), {n, 0, 20}] (* Vincenzo Librandi, Sep 04 2012 *)

a(n) = binomial(4*n,n)*binomial(5*n,n) = ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) = [x^n](F(x)^(20*n)), where F(x) = 1 + x + 12*x^2 + 390*x^3 + 16984*x^4 + 867042*x^5 + 48848541*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A186420 and A188662. - Peter Bala, Jul 14 2016
a(n) ~ 3^(-3*n-1/2)*5^(5*n+1/2)/(2*Pi*n). - Ilya Gutkovskiy, Jul 13 2016
G.f.: G(x) = 4F3(1/5,2/5,3/5,4/5;1/3,2/3,1;(5^5/3^3)*x). Let G^(n)(x) = d^n/dx^n G(x), and c = {120, 15000*x-6, 45000*x^2-114*x, 25000*x^3-135*x^2, 3125*x^4-27*x^3}, then Sum_{n=0..4} c_n*G^(n)(x) = 0. - Bradley Klee, Aug 12 2018
From Peter Bala, Mar 20 2022: (Start)
Right-hand side of the following identities valid for n >= 1:
Sum_{k = 0..3*n} 2*n*(2*n+k-1)!/(k!*n!^2) = (5*n)!/((3*n)!*n!^2);
Sum_{k = 0..n} 4*n*(4*n+k-1)!/(k!*n!*(3*n)!) = (5*n)!/((3*n)!*n!^2). Cf. A000897 and A113424. (End)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(4*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 12 2024

A060538 Square array read by antidiagonals of number of ways of dividing n*k labeled items into n labeled boxes with k items in each box.

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 20, 90, 24, 1, 70, 1680, 2520, 120, 1, 252, 34650, 369600, 113400, 720, 1, 924, 756756, 63063000, 168168000, 7484400, 5040, 1, 3432, 17153136, 11732745024, 305540235000, 137225088000, 681080400, 40320, 1, 12870

Offset: 1

Henry Bottomley, Apr 02 2001

			       1        1        1        1
       2        6       20       70
       6       90     1680    34650
      24     2520   369600 63063000

Harry J. Smith, Table of n, a(n) for n = 1..210

Subtable of A187783.
Rows include A000012, A000984, A006480, A008977, A008978 etc.
Columns include A000142, A000680, A014606, A014608, A014609 etc.
Main diagonal is A034841.

T(n,k)=(n*k)!/k!^n;
for(n=1, 6, for(k=1, 6, print1(T(n,k), ", ")); print) \\ Harry J. Smith, Jul 06 2009

T(n, k) = (nk)!/k!^n = T(n-1, k)*binomial(nk, k) = T(n-1, k)*A060539(n, k) = A060540(n, k)*A000142(k).

A071550 a(n) = (8n)!/n!^8.

Original entry on oeis.org

1, 40320, 81729648000, 369398958888960000, 2390461829733887910000000, 18975581770994682860770223800320, 171889289584866507880743491472699801600

Offset: 0

Benoit Cloitre, May 30 2002

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..100

Sequences (k*n)!/n!^k: A000984 (k = 2), A006480 (k = 3), A008977 (k = 4), A008978 (k = 5), A008979 (k = 6), A071549 (k = 7), A071551 (k = 9), A071552 (k = 10).

[Factorial(8*n)/Factorial(n)^8: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014

Table[(8 n)!/(n)!^8, {n, 0, 20}] (* Vincenzo Librandi, Aug 13 2014 *)

A071551 a(n) = (9n)!/n!^9.

Original entry on oeis.org

1, 362880, 12504636144000, 1080491954750208000000, 140810154080474667338550000000, 23183587808948692737291767860055162880, 4439413043841128802009762476941510771390464000

Offset: 0

Benoit Cloitre, May 30 2002

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..100

Sequences (k*n)!/n!^k: A000984 (k = 2), A006480 (k = 3), A008977 (k = 4), A008978 (k = 5), A008979 (k = 6), A071549 (k = 7), A071550 (k = 8), A071552 (k = 10).

[Factorial(9*n)/Factorial(n)^9: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014

Table[(9n)!/(n)!^9, {n, 0, 20}] (* Vincenzo Librandi, Aug 13 2014 *)

A071552 a(n) = (10n)!/n!^10.

Original entry on oeis.org

1, 3628800, 2375880867360000, 4386797336285844480000000, 12868639981414579848070084500000000, 49120458506088132224064306071170476903628800

Offset: 0

Benoit Cloitre, May 30 2002

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..100

Sequences (k*n)!/n!^k: A000984 (k = 2), A006480 (k = 3), A008977 (k = 4), A008978 (k = 5), A008979 (k = 6), A071549 (k = 7), A071550 (k = 8), A071551 (k = 9).

[Factorial(10*n)/Factorial(n)^10: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014

Table[(10n)!/(n)!^10, {n, 0, 20}] (* Vincenzo Librandi, Aug 13 2014 *)

A069466 Triangle T(n, k) of numbers of square lattice walks that start and end at origin after 2*n steps and contain exactly k steps to the east, possibly touching origin at intermediate stages.

Original entry on oeis.org

1, 2, 2, 6, 24, 6, 20, 180, 180, 20, 70, 1120, 2520, 1120, 70, 252, 6300, 25200, 25200, 6300, 252, 924, 33264, 207900, 369600, 207900, 33264, 924, 3432, 168168, 1513512, 4204200, 4204200, 1513512, 168168, 3432, 12870, 823680, 10090080, 40360320, 63063000, 40360320, 10090080, 823680, 12870

Offset: 0

Martin Wohlgemuth, Mar 24 2002

A Pólya plane walk takes steps (N,E,S,W) along cardinal directions in the plane, visiting only points of Z^2 (cf. Links). T(n,k) is the number of walks departing from and returning to the origin, with exactly 2*k steps along the NS axis and 2*(n-k) steps along the EW direction. Equivalently, triangle T(n,k) is the number of distinct permutations of a 2*n-letter word with letters (N,E,S,W) in multiplicity (k,n-k,k,n-k). Moving only along either NS or EW directions, T(n,0) = T(n,n) = A000894(n). Row sums appear as Equation 4 in the original Pólya article, Sum_{k=0..n} T(n,k) = A002894(n). This identity is proven routinely using Zeilberger's algorithm. - Bradley Klee, Aug 12 2018

			Triangle begins:
    1,
    2,    2,
    6,   24,     6,
   20,  180,   180,    20,
   70, 1120,  2520,  1120,   70,
  252, 6300, 25200, 25200, 6300, 252
  ...
T(4,2) = 2520 because there are 2520 distinct lattice walks of length 2*4=8 starting and ending at the origin and containing exactly 2 steps to the east.
For T(2,k), the lattice-path words are:
T(2,0):{EEWW, WEEW, WWEE, EWWE, WEWE, EWEW}
T(2,1):{NESW, NEWS, NSEW, NSWE, NWES, NWSE, ENSW, ENWS, ESNW, ESWN, EWNS, EWSN, SNEW, SNWE, SENW, SEWN, SWNE, SWEN, WNES, WNSE, WENS, WESN, WSNE, WSEN}
T(2,2):{NNSS, SNNS, SSNN, NSSN, SNSN, NSNS}

Muniru A Asiru, Table of n, a(n) for n = 0..1325 (Rows n=0..50, flattened)
A. Bostan, Calcul Formel pour la Combinatoire des Marches, Habilitation à Diriger des Recherches, Université Paris 13, 2017, page 11.
G. Pólya, Über eine Aufgabe der Wahrscheinlichkeitsrechnung betreffend die Irrfahrt im Straßennetz, Mathematische Annalen, 84 (1921), 149-160.

T(2*n, n) = A008977(n).
Cf. A002894, A000984.
Cf. A007318 (Pascal, m=1), this sequence (m=2), A320824 (m=3).

T:=Flat(List([0..8],n->List([0..n],k->Binomial(2*n,n)*(Binomial(n,k))^2))); # Muniru A Asiru, Oct 21 2018

T:=(n,k)->binomial(2*n,n)*(binomial(n,k))^2: seq(seq(T(n,k),k=0..n),n=0..8); # Muniru A Asiru, Oct 21 2018

T[k_, r_] := Binomial[2k, k]*Binomial[k, r]^2; Table[T[k, r], {k, 0, 8}, {r, 0, k}] // Flatten (* Jean-François Alcover, Nov 21 2012, from explicit formula *)

Recurrences: T(1, 0) = T(1, 1)=2; T(k, r) = 2*k*(2*k-1)/(k-r)^2 * T(k-1, r); T(k, r) = (k+1-r)^2/r^2 * T(k, r-1).
T(n, k) = binomial(2*n, n) * binomial(n, k)^2.
Sum_{k=0..n} T(n, k) = A002894(n).
From Bradley Klee, Aug 12 2018: (Start)
T(n,k) = (2*n)!/((n-k)!*k!)^2.
T(n,k) = C(2*n,2*k)*C(2*(n-k),n-k)*C(2*k,k).
Sum_{k=0..n} T(n,k) = Sum_{k=0..n} C(2*n,2*k)*C(2*(n-k),n-k)*C(2*k,k) = C(2*n,n)^2.
Sum_{k=0..n} T(n,k) = Sum_{k=0..n} (2n)!/(k!(n-k)!)^2 = C(2*n,n)^2.
(End)

A089659 a(n) = S1(n,2), where S1(n, t) = Sum_{k=0..n} (k^t * Sum_{j=0..k} binomial(n,j)).

Original entry on oeis.org

0, 2, 19, 104, 440, 1600, 5264, 16128, 46848, 130560, 352000, 923648, 2369536, 5963776, 14766080, 36044800, 86900736, 207224832, 489357312, 1145569280, 2660761600, 6136266752, 14060355584, 32027705344, 72561459200, 163577856000, 367068708864, 820204535808

Offset: 0

N. J. A. Sloane, Jan 04 2004

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jun Wang and Zhizheng Zhang, On extensions of Calkin's binomial identities, Discrete Math., 274 (2004), 331-342.
Index entries for linear recurrences with constant coefficients, signature (8,-24,32,-16).

Columns : A000012, A000012, A000984, A006480, A008977, A008978.

Sequences of S1(n, t): A001792 (t=0), A089658 (t=1), this sequence (t=2), A089660 (t=3), A089661 (t=4), A089662 (t=5), A089663 (t=6).

I:=[0,2,19,104]; [n le 4 select I[n] else 8*Self(n-1)-24*Self(n-2)+32*Self(n-3)-16*Self(n-4): n in [1..41]]; // Vincenzo Librandi, Jun 22 2016

LinearRecurrence[{8,-24,32,-16}, {0,2,19,104}, 40] (* Vincenzo Librandi, Jun 22 2016 *)

[2^(n-3)*n*(7*n^2 + 12*n + 5)/3 for n in (0..40)] # G. C. Greubel, May 24 2022

a(n) = 2^(n-3)*n*(7*n^2 + 12*n + 5)/3. (see Wang and Zhang p. 333)
From Chai Wah Wu, Jun 21 2016: (Start)
a(n) = 8*a(n-1) - 24*a(n-2) + 32*a(n-3) - 16*a(n-4) for n > 3.
G.f.: x*(2 + 3*x)/(1 - 2*x)^4. (End)
E.g.f.: x*(12 + 33*x + 14*x^2)*exp(2*x)/6. - Ilya Gutkovskiy, Jun 21 2016

A184896 a(n) = C(2n,n) * (7^n/n!^2) * Product_{k=0..n-1} (7k+1)*(7k+6).

Original entry on oeis.org

1, 84, 45864, 35672000, 32445913500, 32247604076688, 33935228690034672, 37165308416775931392, 41919854708375196052500, 48365506771435816732770000, 56812832722107710740048677120, 67715433011522917282547695380480

Offset: 0

Paul D. Hanna, Jan 25 2011

nonn

			G.f.: A(x) = 1 + 84*x + 45864*x^2 + 35672000*x^3 +...
A(x)^(1/2) = 1 + 42*x + 22050*x^2 + 16909900*x^3 +...+ A184895(n)*x^n +...

Cf. A184895; variants: A184423, A008977, A184892, A001421, A184898.

{a(n)=(2*n)!/n!^2*(7^n/n!^2)*prod(k=0,n-1,(7*k+1)*(7*k+6))}

Self-convolution of A184895, where A184895(n) = (7^n/n!^2) * Product_{k=0..n-1} (14k+1)*(14k+6).

a(n) ~ sin(Pi/7) * 2^(2*n) * 7^(3*n) / (Pi*n)^(3/2). - Vaclav Kotesovec, Oct 23 2020

A139541 There are 4n players who wish to play bridge at n tables. Each player must have another player as partner and each pair of partners must have another pair as opponents. The choice of partners and opponents can be made in exactly a(n)=(4n)!/(n!*8^n) different ways.

Original entry on oeis.org

1, 3, 315, 155925, 212837625, 618718975875, 3287253918823875, 28845653137679503125, 388983632561608099640625, 7637693625347175036443671875, 209402646126143497974176151796875, 7752714167528210725497923667975703125, 377130780679409810741846496828678078515625

Offset: 0

Reinhard Zumkeller, Apr 25 2008

nonn

From Karol A. Penson, Oct 05 2009: (Start)
Integral representation as n-th moment of a positive function on a positive semi-axis (solution of the Stieltjes moment problem), in Maple notation:
a(n)=int(x^n*((1/4)*sqrt(2)*(Pi^(3/2)*2^(1/4)*hypergeom([], [1/2, 3/4], -(1/32)*x)*sqrt(x)-2*Pi*hypergeom([], [3/4, 5/4], -(1/32)*x)*GAMMA(3/4)*x^(3/4)+sqrt(Pi)*GAMMA(3/4)^2*2^(1/4)*hypergeom([], [5/4, 3/2],-(1/32)*x)*x)/(Pi^(3/2)*GAMMA(3/4)*x^(5/4))), x=0..infinity), n=0,1... .
This solution may not be unique. (End)

G. Pólya and G. Szegő, Problems and Theorems in Analysis II (Springer 1924, reprinted 1976), Appendix: Problem 203.1, p164.

Andrew Howroyd, Table of n, a(n) for n = 0..100
Eric Weisstein's World of Mathematics, Tournament
Index entries for sequences related to tornaments.

Cf. A008299, A000142, A100733, A001018.

a(n)={(4*n)!/(n!*8^n)} \\ Andrew Howroyd, Jan 07 2020

a(n) = (4*n)!/(n!*8^n).
a(n) = A001147(n)*A001147(2*n).
a(n) = A008977(n)*(A049606(n)/A001316(n))^3. - Reinhard Zumkeller, Apr 28 2008

Terms a(11) and beyond from Andrew Howroyd, Jan 07 2020

A184892 a(n) = C(2n,n) * (5^n/n!^2) * Product_{k=0..n-1} (5k+1)*(5k+4).

Original entry on oeis.org

1, 40, 8100, 2310000, 768075000, 278719056000, 107022956040000, 42753018765600000, 17585519046944062500, 7397979398239787500000, 3168258657090171394750000, 1376657183877933677265000000

Offset: 0

Paul D. Hanna, Jan 25 2011

nonn

			G.f.: A(x) = 1 + 40*x + 8100*x^2 + 2310000*x^3 +...
A(x)^(1/2) = 1 + 20*x + 3850*x^2 + 1078000*x^3 +...+ A184891(n)*x^n +...

Cf. A184891, A184890, A184423, A008977, A001421, A184896, A184898.

Table[Binomial[2*n, n] * 5^n / n!^2 * Product[(5*k + 1)*(5*k + 4), {k, 0, n - 1}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 07 2020 *)

{a(n)=(2*n)!/n!^2*(5^n/n!^2)*prod(k=0,n-1,(5*k+1)*(5*k+4))}

Self-convolution of A184891, where
. A184891(n) = (5^n/n!^2) * Product_{k=0..n-1} (10k+1)*(10k+4).
a(n) ~ sqrt(5 - sqrt(5)) * 2^(2*n - 3/2) * 5^(3*n) / (Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Oct 07 2020

cp's OEIS Frontend

A001451 a(n) = (5*n)!/((3*n)!*n!*n!).

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A060538 Square array read by antidiagonals of number of ways of dividing n*k labeled items into n labeled boxes with k items in each box.

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A071550 a(n) = (8n)!/n!^8.

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Magma

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A071551 a(n) = (9n)!/n!^9.

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Magma

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A071552 a(n) = (10n)!/n!^10.

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Magma

Mathematica

A069466 Triangle T(n, k) of numbers of square lattice walks that start and end at origin after 2*n steps and contain exactly k steps to the east, possibly touching origin at intermediate stages.

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A089659 a(n) = S1(n,2), where S1(n, t) = Sum_{k=0..n} (k^t * Sum_{j=0..k} binomial(n,j)).

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A184896 a(n) = C(2n,n) * (7^n/n!^2) * Product_{k=0..n-1} (7k+1)*(7k+6).

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A001451 a(n) = (5n)!/((3n)!n!n!).

A139541 There are 4n players who wish to play bridge at n tables. Each player must have another player as partner and each pair of partners must have another pair as opponents. The choice of partners and opponents can be made in exactly a(n)=(4n)!/(n!*8^n) different ways.