A006480 -id:A006480 - cp's OEIS frontend

A120666 Triangle read by rows: T(n, k) = (n*k)!/(n!)^k.

1, 1, 6, 1, 20, 1680, 1, 70, 34650, 63063000, 1, 252, 756756, 11732745024, 623360743125120, 1, 924, 17153136, 2308743493056, 1370874167589326400, 2670177736637149247308800, 1, 3432, 399072960, 472518347558400, 3177459078523411968000, 85722533226982363751829504000, 7363615666157189603982585462030336000

Offset: 1

Roger L. Bagula, Aug 11 2006

T(m,n) is the number of ways to distribute n*m different toys among m different kids so that each kid gets exactly n toys. For example, with n=3 and m=2, the 6 different toys, t1, t2, t3, t4, t5 and t6, can be distributed in exactly 20 ways among the 2 kids, k1 and k2, since there are C(6,3)=20 ways to choose the three toys for k1, with the other three toys going to k2. The proof for the general case is based on the identity C(n*m,n)*C(n*m-n,n)*...*C(n*m-n*(m-1),n) = (n*m)!/(n!)^m. - Dennis P. Walsh, Apr 12 2018

			Triangle begins:
  1;
  1,   6;
  1,  20,   1680;
  1,  70,  34650,    63063000;
  1, 252, 756756, 11732745024, 623360743125120;

Seiichi Manyama, Rows n = 1..26, flattened
Eric Weisstein's World of Mathematics, Macdonald's Constant-Term Conjecture

Cf. A000984, A006480, A034841, A089759, A187783.

[Factorial(n*k)/(Factorial(n))^k: k in [1..n], n in [1..10]]; // G. C. Greubel, Dec 26 2022

T:= (m, n)-> (n*m)!/(m!)^n:
seq(seq(T(m, n), n=1..m), m=1..7);  # Alois P. Heinz, Apr 12 2018

Table[(n*k)!/(n!)^k, {n,10}, {k,n}]//Flatten

def A120666(n,k): return gamma(n*k+1)/(factorial(n))^k
flatten([[A120666(n,k) for k in range(1,n+1)] for n in range(1,11)]) # G. C. Greubel, Dec 26 2022

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

Edited by N. J. A. Sloane, Jun 17 2007
Offset corrected by Alois P. Heinz, Apr 12 2018
New name using formula by Joerg Arndt, Apr 15 2018

A161581 a(n) = (3n)!/(n!(n+1)!(n+2)!).

Original entry on oeis.org

21, 231, 3003, 43758, 692835, 11685817, 207157665, 3823000545, 72931087320, 1430571328200, 28734046963560, 589047962752980, 12292044987448215, 260543149635912165, 5599392250947235125, 121830987186399315825

Offset: 3

Alexander Adamchuk, Jun 14 2009

nonn

3-d analog of the Catalan numbers A000108.

Winston de Greef, Table of n, a(n) for n = 3..704
Eric Weisstein's World of Mathematics, Binomial Sums.
Eric Weisstein's World of Mathematics, Central Binomial Coefficient.
Eric Weisstein's World of Mathematics, Catalan Number.

Cf. A000108, A005789, A006480.

A161581 := proc(n) (3*n)!/n!/(n+1)!/(n+2)! ; end: seq(A161581(n),n=3..40) ; # R. J. Mathar, Jun 16 2009
a := proc (n) options operator, arrow: factorial(3*n)/(factorial(n)*factorial(n+1)*factorial(n+2)) end proc: seq(a(n), n = 3 .. 20); # Emeric Deutsch, Jun 14 2009

a(n) = A006480(n)/((n+1)^2*(n+2)).
a(n) ~ 3^(3*n + 1/2) / (2*Pi*n^4). - Vaclav Kotesovec, Feb 21 2023
a(n) = (1/2)*A005789(n) for n >= 3. - Peter Bala, Mar 01 2023
D-finite with recurrence (n+2)*(n+1)*a(n) -3*(3*n-1)*(3*n-2)*a(n-1)=0. - R. J. Mathar, Nov 22 2024

Repetitions of information contained in other sequences removed by R. J. Mathar, Jun 16 2009

More terms from Emeric Deutsch, Jun 14 2009

A178819 Pascal's prism (3-dimensional array) read by folded antidiagonal cross-sections: (h+i; h, i-j, j), h >= 0, i >= 0, 0 <= j <= i.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 3, 3, 1, 3, 6, 3, 3, 3, 1, 1, 4, 4, 6, 12, 6, 4, 12, 12, 4, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 5, 20, 30, 20, 5, 10, 30, 30, 10, 10, 20, 10, 5, 5, 1, 1, 6, 6, 15, 30, 15, 20, 60, 60, 20, 15, 60, 90, 60, 15, 6, 30, 60, 60, 30, 6, 1, 6, 15, 20, 15, 6, 1

Offset: 0

Harlan J. Brothers, Jun 16 2010

P_h = level h of Pascal's prism where P_1 = Pascal's triangle (A007318) and P_2 = denominators of Leibniz harmonic triangle (A003506). A sequence of length k through P is defined by P for n = {1, 2, 3, ..., k}.

			Prism begins (levels 1-4):
1
1 1
1 2 1
1 3 3 1
1
2 2
3 6 3
4 12 12 4
1
3 3
6 12 6
10 30 30 10
1
4 4
10 20 10
20 60 60 20

H. J. Brothers, Pascal's prism, The Mathematical Gazette, 96 (July 2012), 213-220.
H. J. Brothers, Pascal's Prism: Supplementary Material

Level 1 = A007318.
Level 2 = A003506.
Level 3 = A094305.
Level 4 = A178820.
Level 5 = A178821.
Level 6 = A178822.
Sums of shallow diagonals for each level correspond to rows of square A037027.
Contains A109649 and A046816.
P = A000984.
P = A006480.
P = A000897.
P<3n-2, 3n-2, n> = A113424.

end = 5; Column/@Table[Multinomial[h, i-j, j], {h, 0, end}, {i, 0, end}, {j, 0, i}]

a_(h, i, j) = (h+i-2; h-1, i-j, j-1), h >= 1, i >= 1, 1 <= j <= i.
Recurrence:
For P_h, element a is given by: a_(1, 1) = 1; a_(i, j) = ((i+h-2)/(i-1)) (a_(i-1, j) + a_(i-1, j-1)).

Keyword tabf by Michel Marcus, Oct 22 2017

A273630 a(n) = Sum_{k = 0..n} (-1)^kk^3binomial(n,k)^3.

Original entry on oeis.org

0, -1, 0, 162, 0, -11250, 0, 576240, 0, -25259850, 0, 1007242236, 0, -37685439792, 0, 1346871240000, 0, -46504059326010, 0, 1562983866658500, 0, -51407781284599740, 0, 1661123953798807680, 0, -52886433789393750000, 0, 1662782404368229351200

Offset: 0

Peter Bala, Jul 17 2016

Let d(n) = Sum_{k = 0..n} (-1)^k*binomial(n,k)^3. Clearly, by symmetry of the binomial coefficients we have d(2*n + 1) = 0. Dixon's identity is the result d(2*n) = (-1)^n*(3*n)!/n!^3. A generalization is: for r a nonnegative integer there holds Sum_{k = 0..n} (-1)^k*binomial(k,r)^3*binomial(n,k)^3 = (-1)^r*binomial(n,r)^3*d(n - r). This is the case r = 1. See A273631 (case r = 2) and A245086 (case r = 0).

Peter Bala, A generalization of Dixon's identity
J. Ward, 100 Years of Dixon's Identity, Irish Mathematical Society Bulletin 27, 46-54, 1991
Wikipedia, Dixon's identity

Cf. A006480, A245086, A273631.

[&+[(-1)^k*k^3 *Binomial(n, k)^3: k in [0..n]]: n in [0..70]]; // Vincenzo Librandi, Jul 23 2016

seq(add((-1)^k*k^3*binomial(n,k)^3, k = 0..n), n = 0..30);

Table[Sum[(-1)^k*k^3 Binomial[n, k]^3, {k, 0, n}], {n, 0, 27}] (* Michael De Vlieger, Jul 22 2016 *)

a(n) = sum(k=0, n, (-1)^k*k^3*binomial(n, k)^3) \\ Felix Fröhlich, Jul 22 2016

from math import factorial
def A273630(n): return (1 if (m:=n>>1)&1 else -1)*n**3*factorial(3*m)//factorial(m)**3 if n&1 else 0 # Chai Wah Wu, Oct 04 2022

a(2*n) = 0; a(2*n + 1) = (-1)^(n+1)*(2*n + 1)^3*(3*n)!/n!^3.
a(2*n + 1) = -(2*n + 1)^3*A245086(2*n) = (-1)^(n+1)* (2*n + 1)^3*A006480(n).
a(n) = Sum_{k = 1..n} (-1)^k*multinomial(n, 1, k - 1, n - k)^3.
Recurrence: a(n) = -3*n^3*(3*n - 5)*(3*n - 7)/((n - 1)^2*(n - 2)^3) * a(n-2).

A295870 a(n) = binomial(3n,n)*CQC(n), where CQC(n) = A005721(n) = A005190(2n) is a central quadrinomial coefficient.

Original entry on oeis.org

1, 12, 660, 48720, 4005540, 349260912, 31626298704, 2940502593600, 278788387440420, 26831860080682800, 2613367831568654160, 257012469788428710720, 25479526081439438845200, 2543092744417831625342400, 255292245777771431285140800, 25755871314484468746363582720

Offset: 0

Bradley Klee, Feb 23 2018

nonn

Compare with EllipticK A002894 and the Ramanujan period-energy functions A113424, A006480, A000897. The series expansion "T(x) = 2*Pi*Sum_{n>=0} a_n*x^n" determines the real period T of elliptic curves in the family "x=p^2+q^2-4*(q^2-p^2)*q, 0 < x < 1/108". This sequence serves as a counterexample to the naive idea that elliptic integrals will always evaluate to a hypergeometric function such as 2F1(a,b;c;x).
A300058 is the complex period-energy function, after scaling energy and time dimensions such that all a(n) are integers and a(0)=1. The Picard-Fuchs equation is "(12-288*x+9216*x^2)*T(x) + (-1+232*x-8160*x^2+82944*x^3)*T'(x) + (-x+164*x^2-6432*x^3+41472*x^4)*T''(x)".
Although the sequence is not generated by a hypergeometric function, it can be formulated in terms of Hypergeometric numbers, specifically the binomial coefficients. Then Zeilberger's algorithm outputs a second order recurrence with polynomial coefficients.
The contour plot is nice to look at, with reflection symmetry, three critical points, and two separatrices dividing the phase plane into eight distinct regions.
Hyperbolic Critical points are located at (q,p) locations (1/6,0) and (-1/4,sqrt(5)/4) and (-1/4,-sqrt(5)/4). Is it possible to use chord-and-tangent addition rules to produce an exponentially-convergent Diophantine approximation to sqrt(5) that moves along the upper separatrix x=1/8?
Does there exist a period-preserving transformation that takes any one of the curves with 0 < x < 1/108 into a particular Weierstrass curve from the L-function and Modular Forms Database?

D. Husemöller, Elliptic Curves, 2nd ed., New York: Springer, 2004.
J. H. Silverman, The Arithmetic of Elliptic Curves, 2nd ed., New York: Springer, 2009.

J. Cremona, Elliptic Curves over Q, LMFDB 2017.
B. Klee, The Virtues of X_{n+1} = (4+3*X_{n})/(3+2*X_{n}), seqfans mailing list, 2017.
B. Klee, Geometric G.F. for Ramanujan Periods, seqfans mailing list, 2017.
Brad Klee, Deriving Hypergeometric Picard-Fuchs Equations, Wolfram Demonstrations Project (2018).
Bradley Klee, Phase Plane Geometry.
M. Kontsevich and D. Zagier, Periods, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22.
P. Paule and M. Schorn, FastZeil: the Paule/Schorn implementation of Gosper's and Zeilberger's algorithm, RISC 2017; Local copy, pdf file only, no active links
D. Zeilberger, The Method of Creative Telescoping, Journal of Symbolic Computation, 11.3 (1991), 195-204.

Cf. A000897, A002894, A006480, A113424.
Factors: A005190, A005809, A005721.
Complex Period: A300058.

b[NN_]:=Total/@Table[((-1)^k)*Binomial[3*n,n]*Binomial[2*n,k]*Binomial[5*n-4*k-1,3*n-4*k],{n,0,NN},{k,0,Floor[3*n/4]}];
c1=8*(-30+201*n-319*n^2+145*n^3);c2=-8640*(n-5/3)*(n-4/3)*(n-1/5);c3=10*(n-6/5)*n^2;a[0]=1;a[1]=12;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],{n->n0}];
({#,SameQ[a/@Range[0, 15],#]}&@b[15])[[1]]

a(n) = A005809(n)*A005721(n).
a(n) = Sum_{k=0..floor(3n/4)} ((-1)^k)*binomial(3*n,n)*binomial(2 *n, k)*binomial(5*n - 4*k - 1, 3*n - 4*k).
c1 = 8 *(-30 + 201*n - 319*n^2 + 145*n^3); c2 = -8640*(n - 5/3)*(n - 4/3)*(n - 1/5); c3 = 10*(n - 6/5)*n^2; a(0)=1; a(1)=12; a(n) = (c1/c3)*a(n-1) + (c2/c3)*a(n-2).

A300058 a(n) = binomial(3n,n)/(2Pi)Integral_{x=0..2Pi} (12cos^2(x)sin(x) + 20sin^3(x))^(2n) dx.

Original entry on oeis.org

1, 492, 707220, 1298204880, 2654173160100, 5765723073622512, 13021894087331233104, 30217387890886676251200, 71532102917478013611243300, 171944976047709681477985038000, 418347201888204996027087975427920

Offset: 0

Bradley Klee, Feb 23 2018

nonn

Compare with A295870. The series expansion "T(x)=2*Pi*sqrt(3/5)*Sum_{n>=0} a_n*(x/25)^n" determines the period T of anharmonic oscillation along a contour of the Hamiltonian energy surface "x=2H=(5/3)*p^2+q^2+4*(p^2+q^2)*q,0

The period-energy function T(x) satisfies the Picard-Fuchs equation "(2460+28512*x+2239488*x^2)*T(x)-(125-24840*x-1423008*x^2-20155392*x^3)*T'(x)+(-125*x+1620*x^2+1189728*x^3+10077696 x^4)*T''(x)", also the P.F.Eq. of A295870 under transformation x->x'=1/108-x.

A300057 has a similar definition to A005721, with a couple of extra integers appearing in the integrand. This makes a nice analogy between real and complex periods A295870, A300058. Second-order recurrences with polynomial coefficients define both sequences.

Bradley Klee, Phase Plane Geometry.
Brad Klee, Deriving Hypergeometric Picard-Fuchs Equations, Wolfram Demonstrations Project (2018).
M. Kontsevich and D. Zagier, Periods, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22.

Cf. A002894, A113424, A006480, A000897. Factors: A005809, A300057. Real Period: A295870.

a := n -> 36^n*(3*n)!/n!^3*hypergeom([-2*n, n+1/2], [n+1], -2/3):
seq(simplify(a(n)), n=0..10); # Peter Luschny, Apr 19 2018

c1=12*(-230+2259*n-3933*n^2+1863*n^3);c2=5248800*(n-5/3)*(n-4/3)*(n-1/9);c3=9*n^2*(n-10/9);a[0]=1;a[1]=492;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],n->n0];
b[NN_]:=Expand[Total[Flatten[#]]&/@Table[Binomial[3*n,n]*Binomial[2*n,k2]*Binomial[2*n,k1]*Binomial[2*n,3*n-k1-k2]*((4+Sqrt[15])^(2*n-k1))*((4-Sqrt[15])^(2*n-k2)),{n,0,NN},{k1,0,2*n},{k2,0,2*n}]]; ({#,SameQ[#,a/@Range[0,10]]}&@b[10])[[1]]
Table[Binomial[3*n, n] * SeriesCoefficient[(1 + 9*z + 9*z^2 + z^3)^(2*n), {z, 0, 3*n}], {n, 0, 15}] (* Vaclav Kotesovec, Apr 18 2018 *)

a(n) = A005809(n)*A300057(n).
a(n) = Sum_{k1=0..2n} Sum_{k2=0..2n} binomial(3*n,n)*binomial(2*n,k2)*binomial(2*n,k1)*binomial(2*n,3*n-k1-k2)*((4+sqrt(15))^(2*n-k1))*((4-sqrt(15))^(2*n-k2))
a(0) = 1; a(1) = 492; a(n):=(c1/c3)*a(n-1)+(c2/c3)*a(n-2); with
  c1 = 12*(-230+2259*n-3933*n^2+1863*n^3);
  c2 = 5248800*(n-5/3)*(n-4/3)*(n-1/9);
  c3 = 9*n^2*(n-10/9);
a(n) ~ 2^(2*n - 1) * 3^(3*n - 1/2) * 5^(2*n + 1/2) / (Pi*n). - Vaclav Kotesovec, Apr 18 2018

A308835 The nome q=exp(T_C/T_R)=Sum_{n>=0} a(n)(x/27)^n follows from the series solutions of 2T-d/dx(9(1-x)x*dT/dx)=0.

Original entry on oeis.org

0, 1, 15, 279, 5729, 124554, 2810718, 65114402, 1538182398, 36887880105, 895303119303, 21943398532563, 542209373589501, 13489931811324550, 337599511395854298, 8491805574767197650, 214548940430198454054, 5441921826542937659088, 138512110164878076019560

Offset: 0

Bradley Klee, Jun 27 2019

nonn

Also appears in Ramanujan's theory of elliptic functions, signature 3 (cf. A006480). Almkvist et al. give a real and complex Ansatz for the second-order, ordinary differential equation: T_R = 1 + x*{Z[[x]]}, T_C = T_R*log(x) + x*{Z[[x]]}.

B. C. Berndt, "Ramanujan's Notebooks Part II", Springer, 2012, pages 80-82.

G. Almkvist et al., Generalizations of Clausen's Formula and Algebraic Transformations of Calabi-Yau Differential Equations, Proceedings of the Edinburgh Mathematical Society, 54 (2011), p. 275. [The article is on pages 273-295.]

Cf. A005797, A308836, A308837.

G[nMax_]:=Dot[RecurrenceTable[{Dot[{(3*n-5)^2 (3*n-4)^2 (9*n-4), -18(n - 1)(40 - 197*n + 351*n^2 - 279*n^3 + 81*n^4),81(n - 1)*n^3*(9*n - 13)}, a[n-#] & /@ Reverse[Range[0, 2]]] == 0, a[0] == 0, a[1] == 5/9}, a, {n, 0, nMax}], x^Range[0, nMax]];
qSer[nMax_] := Expand[Times[x, Normal[Series[Exp[ Divide[G[nMax], Hypergeometric2F1[1/3, 2/3, 1, x]]], {x, 0, nMax}]]]];
CoefficientList[(1/k)*qSer[20]/.{x->k*x},x]/.{k->27}

A364512 a(n) = (6n)!^2/((5n)! * (3n)!^2 n!).

Original entry on oeis.org

1, 120, 60984, 39673920, 28734361656, 22105177305120, 17676475936257600, 14521297485225136320, 12168600808728479801400, 10353699767677668805341120, 8916443122582617618026013984, 7754263877699070505609688536320, 6798445963232542402250454047721024

Offset: 0

Peter Bala, Jul 29 2023

Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L) where c_1 + ... + c_K = d_1 + ... + d_L  we can define the factorial ratio sequence u_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence.
The present sequence is an integral factorial ratio sequence of height 2.
It is the particular case a = 2, b = 3 of the 2-parameter family of factorial ratio sequences defined by u(n) = (2*n)!/n! * ((a+b+1)*n)! * (2*a*n)! * (2*b*n)! /( ((a+b)*n)! * ((a+1)*n)! * ((b+1)*n)! * (a*n)! * (b*n)! ). These sequences are shown to be integral by the identity u(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i) * binomial(2*a*n, a*n+i) * binomial(2*b*n, b*n+i).
For other cases see A006480 (a = b = 1), A364507 (a = b = 2), A364508 (a = b = 3), A364510 (a = 1, b = 2) and A364511 (a = 1, b = 3).

Paolo Xausa, Table of n, a(n) for n = 0..300
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2012), arXiv:1111.3057 [math.NT], (2011).
Wikipedia, Dixon's identity

Cf. A006480, A364507, A364508, A364510, A364511.

seq( (6*n)!^2/((5*n)! * (3*n)!^2 * n!), n = 0..15);

A364512[n_]:=(6n)!^2/((5n)!(3n)!^2n!);Array[A364512,15,0] (* Paolo Xausa, Oct 05 2023 *)

a(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i) * binomial(4*n, 2*n+i) * binomial(6*n, 3*n+i), showing that the sequence is integral. Compare with Dixon's identity Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^3 = (3*n)!/n!^3.
a(n) = (-1)^n * (6*n)!/((3*n)!*(2*n)!*n!) * hypergeom([-2*n, -3*n, -4*n], [n + 1, 2*n + 1], 1).
P-recursive: a(n) = (576/5)*(2*n-1)^2*(6*n-1)^2*(6*n-5)^2/((5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)*n^2) * a(n-1) with a(0) = 1.
a(n) ~ c^n * 1/(sqrt(5)*Pi*n), where c = (2^12)*(3^6)/(5^5).
a(n) = [x^n] G(x)^(24*n), where the power series G(x) = 1 + 5*x + 683*x^2 + 205020*x^3 + 81906321*x^4 + 38109640996*x^5 + 19499018805299*x^6 + 10646310099966919*x^7 + 6093917580539621690*x^8 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^24, where the power series F(x) = 1 + 5*x + 1283*x^2 + 557400*x^3 + 302894393*x^4 + 186417421346*x^5 + 124214055930695*x^6 + 87454455447781703*x^7 + 64116544959085589954*x^8 + ... appears to have integer coefficients.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3^r)) hold for all primes p >= 5 and all positive integers n and r [added Oct 11 2024: follows from Meštrović, Section 6, equation 39, since a(n) = binomial(6*n, 3*n)*binomial(6*n, n)].

A024487 a(n) = (1/(4n+2))*M(3n; n,n,n).

Original entry on oeis.org

1, 9, 120, 1925, 34398, 659736, 13302432, 278397405, 5996669250, 132166590270, 2967978162240, 67694635250424, 1564409223571600, 36561597688116000, 862822254602816640, 20535537339485077005, 492426552811873991850, 11886753074132473787250, 288645723487776840570000

Offset: 1

Clark Kimberling

Number of standard Young tableaux of shape (n,n,{1}^n). - Alois P. Heinz, Apr 05 2013

			G.f. = x + 9*x^2 + 120*x^3 + 1925*x^4 + 34398*x^5 + 659736*x^6 + ...

Alois P. Heinz, Table of n, a(n) for n = 1..250

Cf. A006480.

with(combinat):
a:= n-> multinomial(3*n, n$3)/(4*n+2):
seq(a(n), n=1..30);  # Alois P. Heinz, Apr 05 2013

a[ n_] := If[ n < 1, 0, (3 n)! / (n!^3 (4 n + 2))]; (* Michael Somos, Oct 25 2014 *)

{a(n) = if( n<1, 0, (3*n)! / (n!^3 * (4*n + 2)))}; /* Michael Somos, Oct 25 2014 */

a(n) ~ 3^(3*n+1/2) / (8*Pi*n^2). - Vaclav Kotesovec, Sep 06 2014
a(n) = A006480(n) / (4*n + 2) if n>0. - Michael Somos, Oct 25 2014
D-finite with recurrence: n^2*(2*n+1)*a(n) -3*(3*n-1)*(2*n-1)*(3*n-2)*a(n-1)=0. - R. J. Mathar, Apr 27 2020

Corrected and extended by Alois P. Heinz, Apr 05 2013

A071800 Number of lattice paths in the lattice [0..2n] X [0..2n] X [0..2n] which do not pass through the point (n,n,n). Number of paths through a lattice containing a "hole".

Original entry on oeis.org

54, 26550, 14330736, 8264889270, 4978317147804, 3090501687886704, 1961313438507566400, 1265749551338006549430, 827693426702217868006500, 547017649101008848332870300, 364691794467483796757326646400, 244920478151771001164678945670000

Offset: 1

T. D. Noe, Jun 06 2002

Eric Weisstein's World of Mathematics, Lattice Path.

Cf. A006480, A071801, A071803.

Table[Factorial[6n]/Factorial[2n]^3-(Factorial[3n]/Factorial[n]^3)^2, {n, 1, 10}]

a(n) = s(3,2n)-s(3,n)^2.

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A120666 Triangle read by rows: T(n, k) = (n*k)!/(n!)^k.

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A161581 a(n) = (3n)!/(n!(n+1)!(n+2)!).

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A178819 Pascal's prism (3-dimensional array) read by folded antidiagonal cross-sections: (h+i; h, i-j, j), h >= 0, i >= 0, 0 <= j <= i.

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A273630 a(n) = Sum_{k = 0..n} (-1)^k*k^3*binomial(n,k)^3.

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A295870 a(n) = binomial(3n,n)*CQC(n), where CQC(n) = A005721(n) = A005190(2n) is a central quadrinomial coefficient.

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A300058 a(n) = binomial(3*n,n)/(2*Pi)*Integral_{x=0..2*Pi} (12*cos^2(x)*sin(x) + 20*sin^3(x))^(2*n) dx.

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A308835 The nome q=exp(T_C/T_R)=Sum_{n>=0} a(n)*(x/27)^n follows from the series solutions of 2*T-d/dx(9*(1-x)*x*dT/dx)=0.

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A273630 a(n) = Sum_{k = 0..n} (-1)^kk^3binomial(n,k)^3.

A300058 a(n) = binomial(3n,n)/(2Pi)Integral_{x=0..2Pi} (12cos^2(x)sin(x) + 20sin^3(x))^(2n) dx.

A308835 The nome q=exp(T_C/T_R)=Sum_{n>=0} a(n)(x/27)^n follows from the series solutions of 2T-d/dx(9(1-x)x*dT/dx)=0.

A364512 a(n) = (6n)!^2/((5n)! * (3n)!^2 n!).