A008977 -id:A008977 - cp's OEIS frontend

A187783 De Bruijn's triangle, T(m,n) = (m*n)!/(n!^m) read by downward antidiagonals.

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 6, 6, 1, 1, 1, 20, 90, 24, 1, 1, 1, 70, 1680, 2520, 120, 1, 1, 1, 252, 34650, 369600, 113400, 720, 1, 1, 1, 924, 756756, 63063000, 168168000, 7484400, 5040, 1

Offset: 0

Robert G. Wilson v, Jan 05 2013

From Tilman Piesk, Oct 28 2014: (Start)
Number of permutations of a multiset that contains m different elements n times. These multisets have the signatures A249543(m,n-1) for m>=1 and n>=2.
In an m-dimensional Pascal tensor (the generalization of a symmetric Pascal matrix) P(x1,...,xn) = (x1+...+xn)!/(x1!*...*xn!), so the main diagonal of an m-dimensional Pascal tensor is D(n) = (m*n)!/(n!^m). These diagonals are the rows of this array (with m>0), which begins like this:
  m\n:0   1       2            3                4                    5
  0:  1   1       1            1                1                    1 ... A000012;
  1:  1   1       1            1                1                    1 ... A000012;
  2:  1   2       6           20               70                  252 ... A000984;
  3:  1   6      90         1680            34650               756756 ... A006480;
  4:  1  24    2520       369600         63063000          11732745024 ... A008977;
  5:  1 120  113400    168168000     305540235000      623360743125120 ... A008978;
  6:  1 720 7484400 137225088000 3246670537110000 88832646059788350720 ... A008979;
with columns: A000142 (n=1), A000680 (n=2), A014606 (n=3), A014608 (n=4), A014609 (n=5).
A089759 is the transpose of this matrix. A034841 is its diagonal. A141906 is its lower triangle. A120666 is the upper triangle of this matrix with indices starting from 1. A248827 are the diagonal sums (or the row sums of the triangle).
(End)

			T(3,5) = (3*5)!/(5!^3) = 756756 = A014609(3) = A006480(5) is the number of permutations of a multiset that contains 3 different elements 5 times, e.g., {1,1,1,1,1,2,2,2,2,2,3,3,3,3,3}.

Tilman Piesk, First 54 rows of the triangle, flattened
T. Chappell, A. Lascoux, S. Ole Warnaar, and W. Zudilin, Logarithmic and complex constant term identities, arXiv:1112.3130 [math.CO], 2012.
Tilman Piesk, Array for indices 0..16
Tilman Piesk, PHP code used to create the b-file
Tilman Piesk, Illustration of the multisets for m,n=0..4
Wikipedia, Permutations of multisets, Pascal matrix and simplex

Cf. A089759 (transposed), A141906 (subtriangle), A120666 (subtriangle transposed), A060538 (1st row/column removed).
Rows: A000012, A000984, A006480, A006480, A008978, A008979, A071549, A071550, A071551, A071552.
Columns: A000012, A000142, A000680, A014606, A014608, A014609, A248814, A172603, A172609, A172613.
Main diagonal gives: A034841.
Row sums of the triangle: A248827.

[Factorial(k*(n-k))/(Factorial(n-k))^k: k in [0..n], n in [0..10]]; // G. C. Greubel, Dec 26 2022

T[n_, k_]:= (k*n)!/(n!)^k; Table[T[n, k-n], {k, 9}, {n, 0, k-1}]//Flatten

def A187783(n,k): return gamma(k*(n-k)+1)/(factorial(n-k))^k
flatten([[A187783(n,k) for k in range(n+1)] for n in range(11)]) # G. C. Greubel, Dec 26 2022

T(m,n) = (m*n)!/(n!)^m.

A060540(m,n) = T(m,n)/m! . - R. J. Mathar, Jun 21 2023

Row m=0 prepended by Tilman Piesk, Oct 28 2014

A188662 Binomial coefficients: a(n) = binomial(3*n,n)^2.

Original entry on oeis.org

1, 9, 225, 7056, 245025, 9018009, 344622096, 13521038400, 540917591841, 21966328580625, 902702926350225, 37456461988358400, 1566697064677290000, 65973795093338597136, 2794203818390077646400, 118933541228935777741056, 5084343623375056062840609

Offset: 0

Emanuele Munarini, Apr 07 2011

Even-order terms in the diagonal of rational function 1/(1 - (x^2 + y^2 + z^2 - x*y - y*z - x*z)). - Gheorghe Coserea, Aug 09 2018

Seiichi Manyama, Table of n, a(n) for n = 0..604

Cf. A005809, A000897, A002894, A002897, A006480, A008977, A186420.

[Binomial(3*n,n)^2: n in [0..100]]; // Vincenzo Librandi, Apr 08 2011

Mathematica
```
Table[Binomial[3 n, n]^2, {n, 0, 16}]
```
Maxima
```
makelist(binomial(3*n,n)^2,n,0,16);
```

a(n) = binomial(3*n,n)^2; \\ Michel Marcus, Nov 01 2016

from math import comb
def A188662(n): return comb(3*n,n)**2 # Chai Wah Wu, Mar 15 2023

a(n) = A005809(n)^2.
a(n) = binomial(3*n,n)^2 = ( [x^n](1 + x)^(3*n) )^2 = [x^n](F(x)^(9*n)), where F(x) = 1 + x + 4*x^2 + 49*x^3 + 795*x^4 + 15180*x^5 + 320422*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977 and A186420. - Peter Bala, Jul 12 2016
a(n) ~ 3^(6*n+1)*4^(-2*n-1)/(Pi*n). - Ilya Gutkovskiy, Jul 13 2016
a(n) = Sum_{k=0..n} binomial(n, k)^2*binomial(3*n+k, 2*n). - Seiichi Manyama, Jan 09 2017

A008979 a(n) = (6n)!/(n!)^6.

Original entry on oeis.org

1, 720, 7484400, 137225088000, 3246670537110000, 88832646059788350720, 2670177736637149247308800, 85722533226982363751829504000, 2889253496242619386328267523990000, 101097362223624462291180422369532000000, 3644153415887633116359073848179365185734400, 134567406165969006655507763343147223231094784000

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..100
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Cf. A000897, A000984, A002894, A002897, A006480, A008977, A008978, A071549, A071550, A071551, A071552, A186420, A188662.

[Factorial(6*n)/Factorial(n)^6: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014

seq( (6*n)!/(n!)^6, n=0..20); # G. C. Greubel, Feb 17 2020

Table[(6 n)!/(n)!^6, {n, 0, 20}] (* Vincenzo Librandi, Aug 13 2014 *)

vector(21, n, my(m=n-1); (6*m)!/(m!)^6 ) \\ G. C. Greubel, Feb 17 2020

[factorial(6*n)/factorial(n)^6 for n in (0..20)] # G. C. Greubel, Feb 17 2020

From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n)*
  binomial(5*n,n)*binomial(6*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) * ( [x^n](1 + x)^(6*n) ) = [x^n](F(x)^(720*n)), where F(x) = 1 + x + 4478*x^2 + 53085611*x^3 + 926072057094*x^4 + 19977558181209910*x^5 + 493286693783478576177*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A008978, A186420 and A188662. (End)
a(n) ~ 3^(6*n+1/2)*4^(3*n-1)/(Pi*n)^(5/2). - Ilya Gutkovskiy, Jul 12 2016
From Peter Bala, Feb 14 2020: (Start)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z*u*v)^n] (1 + x + y + z + u + v)^(6*n). (End)

A023979 Sum of exponents in prime-power factorization of multinomial coefficient M(4n; n,n,n,n).

Original entry on oeis.org

0, 4, 7, 11, 12, 16, 17, 23, 21, 23, 24, 28, 27, 33, 36, 37, 33, 38, 37, 42, 41, 44, 46, 52, 48, 51, 52, 51, 50, 55, 55, 60, 55, 57, 61, 62, 58, 64, 66, 70, 65, 72, 71, 78, 78, 76, 78, 84, 79, 81, 80, 82, 82, 88, 84, 86, 82, 85, 88, 94, 90, 98, 102, 102, 93, 95, 94, 100, 100, 104, 103, 107, 102

Offset: 0

Clark Kimberling

nonn

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A001222, A008977, A022559.

Cf. A023978, A023980, A023981, A023982, A023983, A023984, A023985, A023986, A023987, A023990, A023991, A023992, A023993.

a[n_] := PrimeOmega[Multinomial[n, n, n, n]]; Array[a, 100, 0] (* Amiram Eldar, Jun 11 2025 *)

a(n) = bigomega((4*n)! / n!^4); \\ Amiram Eldar, Jun 11 2025

From Amiram Eldar, Jun 11 2025: (Start)
a(n) = A001222(A008977(n)).
a(n) = A022559(4*n) - 4*A022559(n). (End)

Offset changed to 0 and a(0) prepended by Amiram Eldar, Jun 11 2025

A082488 a(n) = Sum_{k = 0..n} C(n,k) * C(n+k,k) * C(n+2k,k) C(n+3*k,k).

Original entry on oeis.org

1, 25, 2641, 392641, 67982041, 12838867105, 2564949195985, 533008982952625, 114035552691160585, 24950692835328410305, 5557138347370070346601, 1255741805437716400557625, 287180884347761929741524361, 66343186345544102086872515761

Offset: 0

Emanuele Munarini, Apr 28 2003

Diagonal of the rational function 1/(1-(x + y + z + w + x*y*z*w)). - Gheorghe Coserea, Jul 15 2016

			G.f.: A(x) = 1 + 25*x + 2641*x^2 + 392641*x^3 + 67982041*x^4 + 12838867105*x^5 +...
where
A(x) = 1/(1-x) + (4!/1!^4)*x/(1-x)^5 + (8!/2!^4)*x^2/(1-x)^9 + (12!/3!^4)*x^3/(1-x)^13 + (16!/4!^4)*x^4/(1-x)^17 + (20!/5!^4)*x^5/(1-x)^21 +... [Hanna]
Equivalently,
A(x) = 1/(1-x) + 24*x/(1-x)^5 + 2520*x^2/(1-x)^9 + 369600*x^3/(1-x)^13 + 63063000*x^4/(1-x)^17 + 11732745024*x^5/(1-x)^21 +...+ A008977(n)*x^n/(1-x)^(4*n+1) +...

Alois P. Heinz, Table of n, a(n) for n = 0..400
A. Bostan, S. Boukraa, J.-M. Maillard, J.-A. Weil, Diagonals of rational functions and selected differential Galois groups, arXiv preprint arXiv:1507.03227 [math-ph], 2015.
Jacques-Arthur Weil, Supplementary Material for the Paper "Diagonals of rational functions and selected differential Galois groups"

Cf. A081798.
Column k = 4 of A229142.
Cf. A008977, A001850, A082489, A229049, A229050.
Related to diagonal of rational functions: A268545-A268555.

End

[&+[Binomial(n, k)*Binomial(n+k, k)*Binomial(n+2*k, k)*Binomial(n+3*k, k): k in [0..n]]: n in [0..20]]; // Vincenzo Librandi, Oct 16 2018

with(combinat):
a:= n-> add(multinomial(n+3*k, n-k, k$4), k=0..n):
seq(a(n), n=0..15);  # Alois P. Heinz, Sep 23 2013

Table[Sum[Binomial[n,k]*Binomial[n+k,k]*Binomial[n+2*k,k]* Binomial[n+3*k,k], {k, 0, n}],{n,0,20}] (* Vaclav Kotesovec, Sep 23 2013 *)

{a(n)=polcoeff(sum(m=0, n, (4*m)!/m!^4*x^m/(1-x+x*O(x^n))^(4*m+1)), n)}
for(n=0, 15, print1(a(n), ", "))

{a(n)=sum(k=0, n, binomial(n, k)*binomial(n+k, k)*binomial(n+2*k, k)*binomial(n+3*k, k))}
for(n=0, 15, print1(a(n), ", "))

G.f.: Sum_{n>=0} (4*n)!/n!^4 * x^n / (1-x)^(4*n+1). - Paul D. Hanna, Sep 22 2013
Recurrence: n^3*(2*n-3)*(4*n-9)*(4*n-5)*a(n) = (4*n-9)*(4*n-3)*(520*n^4 - 1820*n^3 + 2109*n^2 - 905*n + 121)*a(n-1) - (192*n^6 - 1536*n^5 + 4748*n^4 - 7050*n^3 + 5065*n^2 - 1563*n + 171)*a(n-2) + (4*n-1)*(32*n^5 - 296*n^4 + 1040*n^3 - 1689*n^2 + 1209*n - 279)*a(n-3) - (n-3)^3*(2*n-1)*(4*n-5)*(4*n-1)*a(n-4). - Vaclav Kotesovec, Sep 23 2013
a(n) ~ c*d^n/(Pi^(3/2)*n^(3/2)), where d = 65 + 46*sqrt(2) + 2*sqrt(2*(1055 + 746*sqrt(2))) = 259.976980158726979... is the maximal positive root of the equation 1 - 4*d + 6*d^2 - 260*d^3 + d^4 = 0 and c = sqrt(8 + 5*sqrt(2) + sqrt(14*(11 + 8*sqrt(2))))/8 = 0.71529801573844067904424114047445568721... - Vaclav Kotesovec, Sep 23 2013, updated Jul 16 2016
G.f.: hypergeom([1/8, 3/8],[1],256*x/(1-x)^4)^2/(1-x). - Mark van Hoeij, Sep 23 2013
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 13*x^2 + 893*x^3 + 99125*x^4 + 13706093*x^5 + ... appears to have integer coefficients. - Peter Bala, Jan 13 2016
0 = x^2*(3*x+1)^2*(1-260*x+6*x^2-4*x^3+x^4)*y''' + 3*x*(3*x+1)*(1-390*x-378*x^2+8*x^3-15*x^4+6*x^5)*y'' + (1-836*x+133*x^2+768*x^3-69*x^4-60*x^5+63*x^6)*y' + (-25+397*x-378*x^2-6*x^3+3*x^4+9*x^5)*y, where y is the g.f. - Gheorghe Coserea, Jul 15 2016

A001421 a(n) = (6n)!/((n!)^3(3*n)!).

Original entry on oeis.org

1, 120, 83160, 81681600, 93699005400, 117386113965120, 155667030019300800, 214804163196079142400, 305240072216678400087000, 443655767845074392936328000, 656486312795713480715743268160, 985646873056680684690542988249600, 1497786250388951255453847206769124800

Offset: 0

N. J. A. Sloane, Glenn K Painter (KUPK78A(AT)prodigy.com)

nonn

Self-convolution of A092870, where A092870(n) = (12^n/n!^2) * Product_{k=0..n-1} (12k+1)*(12k+5). - Paul D. Hanna, Jan 25 2011

			G.f.: A(x) = 1 + 120*x + 83160*x^2 + 81681600*x^3 + ...
A(x)^(1/2) = 1 + 60*x + 39780*x^2 + 38454000*x^3 + ... + A092870(n)*x^n + ...

Seiichi Manyama, Table of n, a(n) for n = 0..310 (terms 0..75 from Vincenzo Librandi)
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
M. Kaneko and D. Zagier, Supersingular j-invariants, hypergeometric series and Atkin's orthogonal polynomials, pp. 97-126 of D. A. Buell and J. T. Teitelbaum, eds., Computational Perspectives on Number Theory, Amer. Math. Soc., 1998 (See Eq. 31.)
R. S. Maier, Nonlinear differential equations satisfied by certain classical modular forms, arXiv:0807.1081 [math.NT], 2008_2010, p. 34 equation (7.29b).
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Cf. A092870; variants: A184423, A008977, A184892, A184896, A184898. - Paul D. Hanna, Jan 25 2011
Cf. A289292.
Cf. A145492, A145493, A145494, A008979.

[Factorial(6*n)/(Factorial(n)^3*Factorial(3*n)): n in [0..15]]; // Vincenzo Librandi, Oct 26 2011

Maple
```
f := n->(6*n)!/( (n!)^3*(3*n)!);
```

Factorial[6 n]/(Factorial[3n] Factorial[n]^3) (* Jacob Lewis (jacobml(AT)uw.edu), Jul 28 2009 *)
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/6, 1/2, 5/6}, {1, 1}, 1728 x], {x, 0, n}] (* Michael Somos, Jul 11 2011 *)

{a(n)=(2*n)!/n!^2*(12^n/n!^2)*prod(k=0, n-1, (6*k+1)*(6*k+5))} \\ Paul D. Hanna, Jan 25 2011

O.g.f.: Hypergeometric2F1(5/12, 1/12; 1; 1728x)^2. - Jacob Lewis (jacobml(AT)uw.edu), Jul 28 2009
a(n) = binomial(2n,n) * (12^n/n!^2) * Product_{k=0..n-1} (6k+1)*(6k+5). - Paul D. Hanna, Jan 25 2011
G.f.: F(1/6, 1/2, 5/6; 1, 1; 1728*x), a hypergeometric series. - Michael Somos, Feb 28 2011
0 = y^3*z^3 - 360*y^4*z^2 + 43200*y^5*z - 1728000*y^6 - 16632*x*y^2*z^3 + 7691328*x*y^3*z^2 - 1738520064*x*y^4*z + 176027074560*x*y^5 + 92207808*x^2*y*z^3 - 69176553984*x^2*y^2*z^2 + 23624298528768*x^2*y^3*z - 2853152143441920*x^2*y^4 - 170400029184*x^3*z^3 + 224945232150528*x^3*y*z^2 - 92759146352345088*x^3*y^2*z + 11686511179538104320*x^3*y^3 where x = a(n), y = a(n+1), z = a(n+2) for all n in z. - Michael Somos, Sep 21 2014
a(n) ~ 2^(6*n - 1) * 3^(3*n) / (Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Apr 07 2018
From Peter Bala, Feb 14 2020: (Start)
a(n) = binomial(6*n,n)*binomial(5*n,n)*binomial(4*n,n) = ( [x^n](1 + x)^(6*n) ) * ( [x^n](1 + x)^(5*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(120*n)), where F(x) = 1 + x + 227*x^2 + 123980*x^3 + 92940839*x^4 + 82527556542*x^5 + 81459995686401*x^6 + ...
appears to have integer coefficients. For similar results see A008979.
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z)^n] (1 + x + y + z)^(6*n). (End)
a(n) = (8^n/n!^3)*Product_{k = 0..3*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
a(n) = 24*(6*n - 1)*(2*n - 1)*(6*n - 5)*a(n-1)/n^3. - Neven Sajko, Jul 19 2023
From Karol A. Penson, Jan 21 2025: (Start)
a(n) = Integral_{x=0..1728} x^n*W(x), with W(x) =  W1(x) + W2(x) + W3(x), where
  W1(x) = hypergeometric([1/6, 1/6, 1/6], [1/3, 2/3], x/1728)/(6*sqrt(Pi)*x^(5/6)*Gamma(5/6)^3),
  W2(x) = - hypergeometric([1/2, 1/2, 1/2], [2/3, 4/3], x/1728)/(24*Pi^2*sqrt(x)), and
  W3(x) = hypergeometric([5/6, 5/6, 5/6], [4/3, 5/3], x/1728)*Gamma(5/6)^3/(1536*Pi^(7/2)*x^(1/6)). This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem on x = (0, 1728). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with singularity x^(-1/6), and for x > 0 is monotonically decreasing to zero at x = 1728. (End)

A082368 a(n) = (4n-1)! / (n! n! * n! * (n-1)! * 3!).

Original entry on oeis.org

1, 105, 15400, 2627625, 488864376, 96197645544, 19688264481600, 4148378852099625, 893864677761055000, 196056702961398759480, 43627992869961630486720, 9825387560922608865863400, 2235197406895366368301560000, 512889830640524227455318600000

Offset: 1

John A. Trono (jtrono(AT)smcvt.edu), May 10 2003

Number of combinations that are possible when placing teams ranked #1 to #4*N in a single elimination tournament where there are four columns of N teams (as in the NCAA Men's Division-1 basketball tournament that is played in March) and each column is a separate regional tournament that produces one of the four semi-finalists. (The teams in the columns appear in sorted order and the relative positions of the four columns is irrelevant.)

Number of ways of dividing 4n labeled items into 4 unlabeled boxes with n items in each box. - Dan Parrish, Apr 09 2015

			8 ranked teams (n=2) in a four region, single elimination tournament generates 105 different possible tournament orderings, where the teams in each region are ordered from best to worst. (Teams would be matched up from top to bottom and continue towards the middle two for other matchups, when more than two teams are listed in each column.) 105 tournaments is too many to list here. As this formula applies to single elimination tournaments, this enumeration formula really only makes sense when n is even.

Vincenzo Librandi, Table of n, a(n) for n = 1..300

Row 4 of A060540.

Cf. A000984, A005809, A005810, A008977.

[Factorial(4*n-1) / (Factorial(n)*Factorial(n)* Factorial(n)*Factorial(n-1)*6): n in [1..15]]; // Vincenzo Librandi, Jun 16 2017

[seq(binomial(4*n,n)*binomial(3*n,n)*binomial(2*n,n)/24,n=1..17)]; # Zerinvary Lajos, Jun 25 2006

Table[(4 n)! / (4! n!^4), {n, 30}] (* Vincenzo Librandi, Jun 16 2017 *)

a(n)=(4*n)!/(4!*n!^4) \\ Charles R Greathouse IV, Apr 09 2015

a(n) = binomial(4*n,n)*binomial(3*n,n)*binomial(2*n,n)/24. - Zerinvary Lajos, Jun 25 2006
a(n) = (4n)!/(4!*n!^4). - Dan Parrish, Apr 09 2015
From Robert Israel, Apr 09 2015: (Start)
a(n) = Gamma(2*n+1/2)*Gamma(n+1/2)*64^n/(24*Pi*(n!)^3).
a(n+1) = 8*(2*n+1)*(4*n+1)*(4*n+3)*a(n)/(n+1)^3.
G.f.: g(x) = x*hypergeom([1,5/4,3/2,7/4],[2,2,2],256*x) satisfies
x^4*(256*x-1)*g''''(x) + 5*x^3*(384*x-1)*g'''(x) + 4*x^2*(780*x-1)*g''(x) + 840*x^2*g'(x) = 0. (End)
From Karol A. Penson, Dec 31 2023: (Start)
a(n) = Integral_{x=0..256} x^n*W(x) dx, n>=0, where W(x) = x^(1/4)*hypergeometric3F2([1/4, 1/4, 1/4], [1/2, 3/4], x/256)/(96*Gamma(3/4)^4) - sqrt(x)*hypergeometric3F2([1/2, 1/2, 1/2], [3/4, 5/4], x/256)/(96*Pi^2) + Gamma(3/4)^4*x^(3/4)*hypergeometric3F2([3/4, 3/4, 3/4], [5/4, 3/2], x/256)/(768*Pi^4) is positive and unimodal on x = [0, 256]. It has a single maximum at approximately x = 31, and it goes to zero with W'(x) diverging, at both x = 0 and x = 256. This integral representation as the n-th power moment of the positive function W(x) on the interval [0, 256] is unique, as W(x) is the solution of the Hausdorff moment problem. (End)
a(n) = A008977(n)/24. - Vaclav Kotesovec, Feb 14 2024

More terms from Zerinvary Lajos, Jun 25 2006

A186420 a(n) = binomial(2n,n)^4.

Original entry on oeis.org

1, 16, 1296, 160000, 24010000, 4032758016, 728933458176, 138735983333376, 27435582641610000, 5588044012339360000, 1165183173971324375296, 247639903129149250277376, 53472066459540320483696896, 11701285507234585729600000000, 2589980371199606611713600000000

Offset: 0

Emanuele Munarini, Feb 21 2011

			G.f.: 4F3({1/2,1/2,1/2,1/2},{1,1,1},256x) where 4F3 is a hypergeometric series.

Seiichi Manyama, Table of n, a(n) for n = 0..417

Cf. binomial(2n,n)^k: A000984 (k=1), A002894 (k=2), A002897 (k=3), this sequence (k=4).

Cf. A000108, A000888, A186414, A186415, A186416, A186418, A186419, A000897, A006480, A008977, A188662.

Table[Binomial[2n,n]^4,{n,0,20}]
Table[Coefficient[Series[HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {1, 1, 1}, 256 x], {x, 0, n}], x, n], {n, 0, 14}] (* Michael De Vlieger, Jul 13 2016 *)

Maxima
```
makelist(binomial(2*n,n)^4,n,0,40);
```

a(n) = A000984(n)^4 = A002894(n)^2.
a(n) = binomial(2*n,n)^4 = ( [x^n](1 + x)^(2*n) )^4 = [x^n](F(x)^(16*n)), where F(x) = 1 + x + 25*x^2 + 1798*x^3 + 183442*x^4 + 22623769*x^5 + 3142959012*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977 and A188662. - Peter Bala, Jul 14 2016
a(n) ~ 256^n/(Pi*n)^2. - Ilya Gutkovskiy, Jul 13 2016

A071549 a(n) = (7n)!/n!^7.

Original entry on oeis.org

1, 5040, 681080400, 182509367040000, 66475579247327250000, 28837919555681211870935040, 14007180988362844601443040716800, 7363615666157189603982585462030336000, 4104167472585675600759440022842715359250000, 2392741010223442438553822446842770682716580000000

Offset: 0

Benoit Cloitre, May 30 2002

nonn

Number of closed paths of length 7n whose steps are 7th roots of unity. - Andrew Howroyd, Nov 01 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Gilbert Labelle and Annie Lacasse, Closed paths whose steps are roots of unity, in FPSAC 2011, Reykjavik, Iceland DMTCS proc. AO, 2011, 599-610.
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Row n=7 of A187783, column k=7 of A089759.

Sequences (k*n)!/n!^k: A000984 (k = 2), A006480 (k =3), A008977 (k = 4), A008978 (k = 5), A008979 (k = 6), A071550 (k = 8), A071551 (k = 9), A071552 (k = 10).

[Factorial(7*n)/Factorial(n)^7: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014

Table[(7 n)!/(n)!^7, {n, 0, 20}] (* Vincenzo Librandi, Aug 13 2014 *)

a(n) = (7*n)!/(n!^7); \\ Andrew Howroyd, Nov 01 2018

From Peter Bala, Feb 14 2020: (Start)
a(n) = C(7*n,n)*C(6*n,n)*C(5*n,n)*C(4*n,n)*C(3*n,n)*C(2*n,n).
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply  Mestrovic, Equation 39, p. 12.
a(n) = [x^n](F(x)^(5040*n)), where F(x) = 1 + x + 62528*x^2 + 11087269661*x^3 + 3021437267047869*x^4 + 1045823730475703710735*x^5 + ...
appears to have integer coefficients. For similar results see A008979.
a(n) = [(x*y*z*u*v*w)^n] (1 + x + y + z + u + v + w)^(7*n). (End)

a(8)-a(9) added by Andrew Howroyd, Nov 01 2018

A361637 Constant term in the expansion of (1 + x + y + z + 1/(xyz))^n.

Original entry on oeis.org

1, 1, 1, 1, 25, 121, 361, 841, 4201, 25705, 118441, 423721, 1628881, 8065201, 41225185, 184416961, 768211081, 3420474121, 16620237001, 79922011465, 364149052705, 1638806098945, 7655390077105, 36739991161105, 174363209490625, 811840219629121, 3790118889635521

Offset: 0

Seiichi Manyama, Mar 19 2023

nonn

Diagonal of the rational function 1/(1 - (x^4 + y^4 + z^4 + w^4 + x*y*z*w)). - Seiichi Manyama, Jul 04 2025

Winston de Greef, Table of n, a(n) for n = 0..1429

Cf. A002426, A344560, A361703.
Cf. A008977, A328725.
Cf. A082488, A274783, A274785.

a(n) = n!*sum(k=0, n\4, 1/(k!^4*(n-4*k)!));

a(n) = n! * Sum_{k=0..floor(n/4)} 1/(k!^4 * (n-4*k)!).
G.f.: Sum_{k>=0} (4*k)!/k!^4 * x^(4*k)/(1-x)^(4*k+1).
From Vaclav Kotesovec, Mar 20 2023: (Start)
Recurrence: n^3*a(n) = (2*n - 1)*(2*n^2 - 2*n + 1)*a(n-1) - (n-1)*(6*n^2 - 12*n + 7)*a(n-2) + 2*(n-2)*(n-1)*(2*n - 3)*a(n-3) + 255*(n-3)*(n-2)*(n-1)*a(n-4).
a(n) ~ 5^(n + 3/2) / (2^(5/2) * Pi^(3/2) * n^(3/2)). (End)

cp's OEIS Frontend

A187783 De Bruijn's triangle, T(m,n) = (m*n)!/(n!^m) read by downward antidiagonals.

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A188662 Binomial coefficients: a(n) = binomial(3*n,n)^2.

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A008979 a(n) = (6n)!/(n!)^6.

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A023979 Sum of exponents in prime-power factorization of multinomial coefficient M(4n; n,n,n,n).

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A082488 a(n) = Sum_{k = 0..n} C(n,k) * C(n+k,k) * C(n+2*k,k) * C(n+3*k,k).

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A001421 a(n) = (6*n)!/((n!)^3*(3*n)!).

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A082488 a(n) = Sum_{k = 0..n} C(n,k) * C(n+k,k) * C(n+2k,k) C(n+3*k,k).

A001421 a(n) = (6n)!/((n!)^3(3*n)!).

A082368 a(n) = (4n-1)! / (n! n! * n! * (n-1)! * 3!).

A361637 Constant term in the expansion of (1 + x + y + z + 1/(xyz))^n.