A006480 -id:A006480 - cp's OEIS frontend

A248658 G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k)^3 * x^(2*k).

1, 1, 1, 2, 9, 28, 66, 153, 433, 1345, 3952, 10991, 30954, 90988, 271845, 804153, 2361457, 6979690, 20842285, 62493914, 187274712, 561448399, 1688263179, 5093148285, 15393417178, 46570446829, 141063389488, 427979185898, 1300470246165, 3956367018001, 12048354848013, 36728336040306

Offset: 0

Paul D. Hanna, Oct 10 2014

nonn

Limit_{n->oo} a(n)/a(n+1) = 1 - t = t^3 = 0.3176721961... where t = ((sqrt(93)+9)/18)^(1/3) - ((sqrt(93)-9)/18)^(1/3).

Diagonal of the rational function 1 / ((1 - x)*(1 - y)*(1 - z) - (x*y*z)^3). - Ilya Gutkovskiy, Apr 23 2025

			G.f. A(x) = 1 + x + x^2 + 2*x^3 + 9*x^4 + 28*x^5 + 66*x^6 + 153*x^7 +...
which equals the series:
A(x) = 1/(1-x-x^3) + 3!/1!^3*x^4/(1-x-x^3)^4 + 6!/2!^3*x^8/(1-x-x^3)^7 + 9!/3!^3*x^12/(1-x-x^3)^10 + 12!/4!^3*x^16/(1-x-x^3)^13 +...
The g.f. also equals the series:
A(x) = 1 +
x*(1 + x^2) +
x^2*(1 + 2^3*x^2 + x^4) +
x^3*(1 + 3^3*x^2 + 3^3*x^4 + x^6) +
x^4*(1 + 4^3*x^2 + 6^3*x^4 + 4^3*x^6 + x^8) +
x^5*(1 + 5^3*x^2 + 10^3*x^4 + 10^3*x^6 + 5^3*x^8 + x^10) +...

Paul D. Hanna, Table of n, a(n) for n = 0..500

Cf. A181545, A181543.

Table[Sum[Binomial[n-2*k,k]^3,{k,0,Floor[n/3]}],{n,0,20}] (* Vaclav Kotesovec, Oct 15 2014 *)

{a(n)=local(A=1); A=sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^3*x^(2*k)) +x*O(x^n)); polcoeff(A, n)}
for(n=0, 40, print1(a(n), ", "))

{a(n)=polcoeff(sum(m=0, n,x^(4*m)/(1-x-x^3 +x*O(x^n))^(3*m+1)*(3*m)!/(m!)^3), n)}
for(n=0, 40, print1(a(n), ", "))

{a(n)=sum(k=0, n\3, binomial(n-2*k, k)^3)}
for(n=0, 40, print1(a(n), ", "))

G.f.: A(x) = Sum_{n>=0} (3*n)!/(n!)^3 * x^(4*n) / (1-x-x^3)^(3*n+1).
a(n) = Sum_{k=0..[n/3]} C(n-2*k,k)^3.
G.f.: A(x) = G( x^4/(1-x-x^3)^3 )/(1-x-x^3) where G(x) satisfies:
* G(x^3) = G( x*(1+3*x+9*x^2)/(1+6*x)^3 )/(1+6*x) and G(x) is the g.f. of A006480.

A318107 Triangle read by rows: T(n,k) = (3n - 2k)!/((n-k)!^3*k!).

Original entry on oeis.org

1, 6, 1, 90, 24, 1, 1680, 630, 60, 1, 34650, 16800, 2520, 120, 1, 756756, 450450, 92400, 7560, 210, 1, 17153136, 12108096, 3153150, 369600, 18900, 336, 1, 399072960, 325909584, 102918816, 15765750, 1201200, 41580, 504, 1, 9465511770, 8779605120, 3259095840, 617512896, 63063000, 3363360, 83160, 720, 1

Offset: 0

Gheorghe Coserea, Sep 18 2018

Diagonal of rational function R(x,y,z,t) = 1/(1 - (x + y + z + t*x*y*z)) with respect to x,y,z, i.e., T(n,k) = [(xyz)^n*t^k] R(x,y,z,t).

Annihilating differential operator: x*(2*t*x + 1)*((t*x - 1)^3 + 27*x)*Dx^2 + (6*t^4*x^4 - 8*t^3*x^3 - 3*t*(t - 18)*x^2 + 6*(t + 9)*x - 1)*Dx + (t*x - 1)*(t*(2*t^2*x^2 + 2*t*x - 1) - 6).

			A(x;t) = 1 + (6 + t)*x + (90 + 24*t + t^2)*x^2 + (1680 + 630*t + 60*t^2 + t^3)*x^3 + ...
Triangle starts:
n\k [0]        [1]        [2]        [3]       [4]      [5]    [6]  [7]
[0] 1;
[1] 6,         1;
[2] 90,        24,        1;
[3] 1680,      630,       60,        1;
[4] 34650,     16800,     2520,      120,      1;
[5] 756756,    450450,    92400,     7560,     210,     1;
[6] 17153136,  12108096,  3153150,   369600,   18900,   336,   1;
[7] 399072960, 325909584, 102918816, 15765750, 1201200, 41580, 504, 1;
[8] ...

Gheorghe Coserea, Rows n=0..100, flattened

Cf. A000172, A006480, A081798, A124435, A318108, A318109.

T(n,k) = (3*n - 2*k)!/((n-k)!^3*k!);
concat(vector(10, n, vector(n, k, T(n-1, k-1))))
/* test:
P(n, v='t) = subst(Polrev(vector(n+1, k, T(n, k-1)), 't), 't, v);
diag(expr, N=22, var=variables(expr)) = {
  my(a = vector(N));
  for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
  for (n = 1, N, a[n] = expr;
    for (k = 1, #var, a[n] = polcoef(a[n], n-1)));
  return(a);
};
apply_diffop(p, s) = { \\ apply diffop p (encoded as Pol in Dx) to Ser s
  s=intformal(s);
  sum(n=0, poldegree(p, 'Dx), s=s'; polcoef(p, n, 'Dx) * s);
};
\\ diagonal property:
x='x; y='y; z='z; t='t;
diag(1/(1 - (x+y+z + t*x*y*z)), 11, [x,y,z]) == vector(11, n, P(n-1))
\\ annihilating diffop:
y = Ser(vector(101, n, P(n-1)), 'x);
p=x*(2*t*x + 1)*((t*x - 1)^3 + 27*x)*Dx^2 + (6*t^4*x^4 - 8*t^3*x^3 - 3*t*(t - 18)*x^2 + 6*(t + 9)*x  - 1)*Dx + (t*x - 1)*(t*(2*t^2*x^2 + 2*t*x - 1) - 6);
0 == apply_diffop(p, y)
*/

Let P_n(t) = Sum_{k=0..n} T(n,k)*t^k. Then A000172(n) = P_n(-4), A318108(n) = P_n(-3),  A318109(n) = P_n(-2), A124435(n) = P_n(-1), A006480(n) = P_n(0), A081798(n) = P_n(1).
G.f. y = Sum_{n>=0} P_n(t)*x^n satisfies:
0 = x*(2*t*x + 1)*((t*x - 1)^3 + 27*x)*y'' + (6*t^4*x^4 - 8*t^3*x^3 - 3*t*(t - 18)*x^2 + 6*(t + 9)*x  - 1)*y' + (t*x - 1)*(t*(2*t^2*x^2 + 2*t*x - 1) - 6)*y.

A338075 Diagonal terms in the expansion of (1+xyz)/(1-x-y-z).

Original entry on oeis.org

1, 7, 96, 1770, 36330, 791406, 17909892, 416226096, 9864584730, 237338943270, 5778870222840, 142077992254380, 3521258757984240, 87862829835387600, 2205050763983594400, 55615552451285359680, 1408840444191389714010, 35825204161237194511830, 914089586182634239686000

Offset: 0

N. J. A. Sloane, Oct 22 2020

nonn

Expand the rational function (1+x*y*z)/(1-x-y-z) as Sum_i Sum_j Sum_k c(i,j,k)*x^i*y^j*z^k; a(n) = c(n,n,n).
If the numerator is changed to 1, we get A006480.
Suggested by Christol's Conjecture (see reference).

Abdelaziz, Youssef, C. Koutschan, and J. M. Maillard. "On Christol’s conjecture." Journal of Physics A: Mathematical and Theoretical 53.20 (2020): 205201; arXiv:1912.10259.

Robert Israel, Table of n, a(n) for n = 0..300
Y. Abdelaziz, C. Koutschan, and J-M. Maillard, On Christol's conjecture, arXiv:1912.10259 [math.NT], 2019-2020.

Other examples arising from diagonal terms of multivariate g.f.s: A000172, A006480, A338076.

N:= 25: # for a(0)..a(N)
F:=  (1+x*y*z)/(1-x-y-z):
S1:= series(F, x, N+1):
L1:= [seq(coeff(S1, x, i), i=0..N)]:
L2:= [seq(coeff(series(L1[i+1], y, i+1), y, i), i=0..N)]:
seq(coeff(series(L2[i+1], z, i+1), z, i), i=0..N); # Robert Israel, Oct 25 2020

nmax = 20; Flatten[{1, Table[Coefficient[Series[(1 + x*y*z)/(1 - x - y - z), {x, 0, n}, {y, 0, n}, {z, 0, n}], x^n*y^n*z^n], {n, 1, nmax}]}] (* Vaclav Kotesovec, Oct 23 2020 *)

{a(n) = if(n==0, 1, (3*(n-1))!/(n-1)!^3+(3*n)!/n!^3)} \\ Seiichi Manyama, Oct 31 2020

Conjectures from Robert Israel, Oct 25 2020: (Start)
G.f.: (x + 1)*LegendreP(-1/3, 1 - 54*x).
(-27*n^2 - 27*n - 6)*a(n + 1) + (-53*n^2 - 214*n - 173)*a(n + 2) + (-25*n^2 - 179*n - 319)*a(n + 3) + (n^2 + 8*n + 16)*a(n + 4) = 0. (End)
a(n) = (28*n^2 - 27*n + 6) * (3*n)! / (3 * (3*n - 1) * (3*n - 2) * n!^3). - Vaclav Kotesovec, Oct 28 2020
a(n) = A006480(n-1) + A006480(n) for n > 0. - Seiichi Manyama, Oct 31 2020

More terms from Vaclav Kotesovec, Oct 23 2020

A361031 a(n) = (3^3)(1245781011)(3n)!/(n!*(n+4)!^2).

Original entry on oeis.org

11550, 2772, 4620, 15840, 81675, 550550, 4492488, 42325920, 446185740, 5148297000, 63985977000, 846321189120, 11802213457650, 172255143129300, 2615726247519000, 41127042052404000, 666874986879730860, 11114583114662181000, 189866473537245687000, 3316382259894423720000

Offset: 0

Peter Bala, Mar 01 2023

Row 3 of A361027.

The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 840*A000984(n) is divisible by (n + 1)*(n + 2)*(n + 3)*(n + 4) and the result 840*(2*n)!/(n!*(n+4)!) is the super ballot number A348893(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 6652800 * A006480(n) is divisible by ((n + 1)*(n + 2)*(n + 3)*(n + 4))^2.

Row 3 of A361027. Cf. A006480, A348893, A361028, A361029, A361030.

a := proc(n) option remember; if n = 0 then 11550 else
3*(3*n-1)*(3*n-2)/(n+4)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);

a(n) = (3^3)*(1*2*4*5*7*8*10*11)/((n+1)*(n+2)*(n+3)*(n+4))^2 * (3*n)!/n!^3.
a(n) = (1/3)*(1*2*4*5*7*8*10*11) * A006480(n+4)/((3*n + 1)*(3*n + 2)*(3*n + 4)* (3*n + 5)*(3*n + 7)*(3*n + 8)*(3*n + 10)*(3*n + 11)), where A006480(n) = (3*n)!/n!^3.
a(n) = (1/3)*27^(n+4)*binomial(10/3, n+4)*binomial(11/3, n+4).
a(n) = (1/7)*A348893(n)*A361039(n). It can be shown from this that a(n) is always an integer.
a(n) ~ sqrt(3)*3326400*(27^n)/(Pi*n^9).
P-recursive: (n + 4)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 11550.
The o.g.f. A(x) satisfies the differential equation x^2*(1 - 27*x)*A''(x) + x*(9 - 54*x)*A'(x) + (16 - 6*x)*A(x) - 184800 = 0, with A(0) = 11550 and A'(0) = 2772.

A361716 a(n) = Sum_{k = 0..n-1} (-1)^kbinomial(n,k)^2binomial(n-1,k).

Original entry on oeis.org

0, 1, -3, -8, 45, 126, -840, -2400, 17325, 50050, -378378, -1100736, 8576568, 25069968, -199536480, -585307008, 4732755885, 13919870250, -113936715750, -335813478000, 2775498395670, 8194328596740, -68263497731520, -201822515032320

Offset: 0

Peter Bala, Mar 23 2023

Conjecture: the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and positive integers n and k.

Paolo Xausa, Table of n, a(n) for n = 0..1000
Wikipedia, Dixon's identity.

Cf. A006480, A245086, A361710, A361711.

seq(add((-1)^k*binomial(n,k)^2*binomial(n-1,k), k = 0..n-1), n = 0..20);

A361716[n_]:=Sum[(-1)^k*Binomial[n,k]^2Binomial[n-1,k],{k,0,n-1}];Array[A361716,30,0] (* Paolo Xausa, Oct 06 2023 *)

a(n) = sum(k = 0, n-1, (-1)^k*binomial(n,k)^2*binomial(n-1,k)); \\ Michel Marcus, Mar 26 2023

from math import comb
def A361716(n): return (sum(comb(n,k)**3*k if k&1 else -comb(n,k)**3*k for k in range(n+1)))//(n if n&1 else -n) if n else 0 # Chai Wah Wu, Mar 27 2023

a(n) = Sum_{k = 0..n} (-1)^(n+k) * (k/n) * binomial(n,k)^3.
a(2*n) = (-1)^n * (1/2) * (3*n)!/n!^3 for n >= 1; a(2*n+1) = (-1)^n * (3*n+1)/(2*n+1) * (3n)!/n!^3.
a(2*n) = (1/2)*A245086(2*n) = (1/2)*(-1)^n*A006480(n) for n >= 1.
a(2*n+1) = A361710(2*n+1) = A361711(2*n+1).
a(n) = hypergeom([1 - n, - n, - n], [1, 1], 1) for n >= 1.
P-recursive: n^2*(n-1)*(6*n^2-20*n+17)*a(n) = -( 6*(3*n^2-6*n+2)*(n-1)*a(n-1) + (3*n-6)*(3*n-5)*(3*n-4)*(6*n^2-8*n+3)*a(n-2) ).
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(3*n,n-k)*binomial(n+k,k)^2 for n >= 1.
a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)*binomial(n+k,n)*binomial(2*n-k-1,n). - Peter Bala, Sep 13 2023

A364507 a(n) = (5n)!(4n)! / ((3n)!^2 * (2n)! n!).

Original entry on oeis.org

1, 40, 5880, 1101100, 229265400, 50678855040, 11641642112100, 2746924727976000, 661097260785195000, 161538994454795003200, 39949572934939198410880, 9976687616280042928424700, 2511716999955421326631644900, 636662322699394050738883008000

Offset: 0

Peter Bala, Jul 27 2023

Row 2 of A364506.

			Examples of supercongruences:
a(7) - a(1) = 2746924727976000 - 40 = (2^3)*5*(7^4)*28601881799 == 0 (mod 7^4).
a(11) - a(1) = 9976687616280042928424700 - 40 = (2^2)*5*(11^3)*18397*3568463* 5708869513 == 0 (mod 11^3).

Paolo Xausa, Table of n, a(n) for n = 0..400
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. A001450, A005810, A006480, A364506, A364508.

seq( (5*n)!*(4*n)!*(2*n)! / ((3*n)!^2 * (2*n)!^2 * n!), n = 0..15);

A364507[n_]:=(5n)!(4n)!/((3n)!^2(2n)!n!);Array[A364507,15,0] (* Paolo Xausa, Oct 06 2023 *)

a(n) = Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k) * binomial(4*n, 2*n + k)^2 (showing a(n) to be integral). Compare with Dixon's identity, Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k)^3 = (3*n)!/n!^3 = A006480(n).
P-recursive: a(n) = (20/9)*(4*n-1)*(4*n-3)*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)/((3*n-1)^2*(3*n-2)^2*n^2) * a(n-1) with a(0) = 1.
a(n) ~ c^n * sqrt(10)/(6*Pi*n), where c = (2^6)*(5^5)/(3^6).
a(n) = [x^n] G(x)^(20*n), where the power series G(x) = 1 + 2*x + 69*x^2 + 5647*x^3 + 618860*x^4 + 79241349*x^5 + 11177111981*x^6 + 1684171189810*x^7 + 266238907746252*x^8 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^20, where the power series F(x) = 1 + 2*x + 149*x^2 + 18647*x^3 + 2913620*x^4 + 515276389*x^5 + 98628630997*x^6 + 19944410220744*x^7 + 4199273746072180*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 Aug 04 2023: the conjecture follows from Meštrović. equation 39].
a(n) = binomial(4*n,n)*binomial(5*n,2*n). - Christian Krause, Aug 03 2023

A364508 a(n) = (7n)!(6n)!(2n)! / ((4n)!^2 * (3n)!^2 n!).

Original entry on oeis.org

1, 350, 594594, 1299170600, 3164045050530, 8188909171581600, 22035578229399735000, 60924423899585957558848, 171839010049825493742617250, 492149504510899056782561257748, 1426695143534668869395862598229344, 4176678405144148418744441910948978000

Offset: 0

Peter Bala, Jul 27 2023

Row 3 of A364506.

			Examples of supercongruences:
a(7) - a(1) = 60924423899585957558848 - 350 = 2*(7^4)*12687301936606821649 == 0 (mod 7^4).
a(11) - a(1) = 4176678405144148418744441910948978000 - 350 = 2*(5^2)*7*(11^3)* 29*139*1181*1883311859633620981885519 == 0 (mod 11^3).

Paolo Xausa, Table of n, a(n) for n = 0..250
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2012), arXiv:1111.3057 [math.NT], (2011).

Cf. A006480, A364506, A364507.

seq( (7*n)!*(6*n)!*(2*n)! / ((4*n)!^2 * (3*n)!^2 * n!), n = 0..15);

A364508[n_]:=(7n)!(6n)!(2n)!/((4n)!^2(3n)!^2n!);Array[A364508,15,0] (* Paolo Xausa, Oct 06 2023 *)

a(n) = Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k) * binomial(6*n, 3*n + k)^2 (showing a(n) to be integral). Compare with Dixon's identity  Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k)^3 = (3*n)!/n!^3 = A006480(n).
P-recursive: a(n) = (7/4)*(6*n-1)*(6*n-5)*(7*n-1)*(7*n-2)*(7*n-3)*(7*n-4)*(7*n-5)*(7*n-6)/((3*n-1)*(3*n-2)*(4*n-1)^2*(4*n-3)^2*n^2) * a(n-1) with a(0) = 1.
a(n) ~ c^n * sqrt(21)/(12*Pi*n), where c = (7^7)/(2^8).
a(n) = [x^n] G(x)^(14*n), where the power series G(x) = 1 + 25*x + 12798*x^2 + 13543850*x^3 + 18933663145*x^4 + 30733263922830*x^5 + 54771428143877503*x^6 + 104061045049532102971*x^7 + 207134582792235253663131*x^8 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^14, where the power series F(x) = 1 + 25*x + 21548*x^2 + 31466125*x^3 + 57506245907*x^4 + 119069165444705*x^5 + 266966985031172547*x^6 + 632553825380957995891*x^7 + 1560815989686060202098169*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(7*n, 3*n)*binomial(6*n, 3*n) *binomial(2*n, n)/binomial(4*n, n)].

A364511 a(n) = (6n)!(5n)!(2n)! / ((4n)!^2 * (3n)! n!^2).

Original entry on oeis.org

1, 50, 8910, 2011100, 503909070, 133954543800, 36992598142500, 10491379251679040, 3034472729231379150, 891028813210575018980, 264787855164104281785160, 79455812929030151249454000, 24035311050907120054564683300, 7320107028326385998504601648000

Offset: 0

Peter Bala, Jul 28 2023

Row 4 of A364509.

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, A364509, A364510, A364512.

seq((6*n)!*(5*n)!*(2*n)! / ((4*n)!^2 * (3*n)! * n!^2), n = 0..15);

A364511[n_]:=(6n)!(5n)!(2n)!/((4n)!^2(3n)!n!^2);Array[A364511,15,0] (* Paolo Xausa, Oct 05 2023 *)

a(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^2 * binomial(6*n, 3*n+i).
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*binomial(6*n,2*n) * hypergeom([-2*n, -2*n, -4*n], [1, 2*n+1], 1).
P-recursive: a(n) = (15/4)*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)*(6*n-1)*(6*n-5)/((4*n-1)^2*(4*n-3)^2*n^2) * a(n-1) with a(0) = 1.
a(n) ~ c^n * sqrt(5)/(4*Pi*n), where c = (3^3)*(5^5)/(2^8).
a(n) = [x^n] G(x)^(10*n), where the power series G(x) = 1 + 5*x + 208*x^2 + 19960*x^3 + 2580710*x^4 + 390721786*x^5 + 65243160516*x^6 + 11646611942100*x^7 + 2182248792056787*x^8 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^10, where the power series F(x) =  1 + 5*x + 458*x^2 + 69285*x^3 + 13037740*x^4 + 2773287786*x^5 + 638122182196*x^6 + 155077758079485*x^7 + 39234250338228617*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(5*n, n)* binomial(2*n, n)/binomial(4*n, n)].
a(n) = (-1)^n * [x^(4*n)] (1 - x)^(8*n)*Legendre_P(2*n, (1 + x)/(1 - x)). - Peter Bala, Aug 14 2023

A367567 a(n) = Product_{k=0..n} (3*k)! / k!^3.

Original entry on oeis.org

1, 6, 540, 907200, 31434480000, 23788231346880000, 408042767492495815680000, 162838835029822082951032012800000, 1541352909587869227178909850805190656000000, 351233376660297011570511252132131832794456064000000000, 1949695346852822356399298814748829537555898997004605685760000000000

Offset: 0

Vaclav Kotesovec, Nov 23 2023

nonn

Cf. A000178, A006480, A268504, A061719, A268196.

Cf. A007685, A367568, A367569, A367570, A367571.

Table[Product[(3*k)!/k!^3, {k, 0, n}], {n, 0, 10}]
Table[Product[Binomial[3*k,k] * Binomial[2*k,k], {k, 0, n}], {n, 0, 10}]

a(n) = Product_{k=0..n} binomial(3*k,k) * binomial(2*k,k).
a(n) = A268504(n) / A000178(n)^3.
a(n) = A268504(n) / A061719(n).
a(n) = A007685(n) * A268196(n).
a(n) ~ A^(8/3) * Gamma(1/3)^(1/3) * 3^(3*n^2/2 + 2*n + 11/36) * exp(n - 2/9) / (n^(n + 13/18) * (2*Pi)^(n + 7/6)), where A is the Glaisher-Kinkelin constant A074962.

A071803 Number of paths in the lattice [0..n] X [0..n] X [0..n] which do not pass through the point (floor(n/2), floor(n/2), floor(n/2)). Number of paths through a lattice containing a "hole".

Original entry on oeis.org

54, 1140, 26550, 605556, 14330736, 340860960, 8264889270, 201651836100, 4978317147804, 123546256876224, 3090501687886704, 77632745316063840, 1961313438507566400, 49717549985405892480, 1265749551338006549430, 32312920048897640674500, 827693426702217868006500

Offset: 2

T. D. Noe, Jun 06 2002

Robert Israel, Table of n, a(n) for n = 2..700
Eric Weisstein's World of Mathematics, Lattice Path.

Cf. A006480, A071800, A071801.

g:= hypergeom([1/6,1/3,2/3,5/6],[ 1/2,1/2,1], 729*x^2) - hypergeom([1/3,1/3,2/3,2/3],[ 1,1,1], 729*x^2) + 6*x*( hypergeom([2/3,5/6,7/6,4/3],[ 1,3/2,3/2], 729*x^2) - hypergeom([1/3,2/3,4/3,5/3],[ 1,2,2], 729*x^2)):
S:= series(g,x,101):
seq(coeff(S,x,j),j=2..100); # Robert Israel, Oct 20 2016

Table[Factorial[3n]/Factorial[n]^3 - Factorial[3Floor[n/2]]Factorial[3 Ceiling[n/2]]/Factorial[Floor[n/2]]^3/Factorial[Ceiling[n/2]]^3, {n, 2, 20}]
Rest[Rest[CoefficientList[Series[HypergeometricPFQ[{1/6, 1/3, 2/3, 5/6}, {1/2, 1/2, 1}, 729 x^2] - HypergeometricPFQ[{1/3, 1/3, 2/3, 2/3}, {1, 1, 1}, 729 x^2] + 6 x (HypergeometricPFQ[{2/3, 5/6, 7/6, 4/3}, {1, 3/2, 3/2}, 729 x^2] - HypergeometricPFQ[{1/3, 2/3, 4/3, 5/3}, {1, 2, 2}, 729 x^2]), {x, 0, 20}], x]]] (* Benedict W. J. Irwin, Oct 20 2016 *)

A006480(n)=(3*n)!/n!^3
a(n) = A006480(n) - A006480(n\2)*A006480((n+1)\2) \\ Charles R Greathouse IV, Oct 20 2016

a(n) = s(3, n) - s(3, floor(n/2)) * s(3, ceiling(n/2)), where s(3,n) = A006480(n).

G.f.: 4F3(1/6,1/3,2/3,5/6; 1/2,1/2,1; 729*x^2) - 4F3(1/3,1/3,2/3,2/3; 1,1,1; 729*x^2) + 6*x*( 4F3(2/3,5/6,7/6,4/3; 1,3/2,3/2; 729*x^2) - 4F3(1/3,2/3,4/3,5/3; 1,2,2; 729*x^2)). - Benedict W. J. Irwin, Oct 20 2016

cp's OEIS Frontend

A248658 G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k)^3 * x^(2*k).

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A318107 Triangle read by rows: T(n,k) = (3*n - 2*k)!/((n-k)!^3*k!).

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A338075 Diagonal terms in the expansion of (1+x*y*z)/(1-x-y-z).

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A361031 a(n) = (3^3)*(1*2*4*5*7*8*10*11)*(3*n)!/(n!*(n+4)!^2).

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A361716 a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)^2*binomial(n-1,k).

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A364507 a(n) = (5*n)!*(4*n)! / ((3*n)!^2 * (2*n)! * n!).

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A364508 a(n) = (7*n)!*(6*n)!*(2*n)! / ((4*n)!^2 * (3*n)!^2 * n!).

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A318107 Triangle read by rows: T(n,k) = (3n - 2k)!/((n-k)!^3*k!).

A338075 Diagonal terms in the expansion of (1+xyz)/(1-x-y-z).

A361031 a(n) = (3^3)(1245781011)(3n)!/(n!*(n+4)!^2).

A361716 a(n) = Sum_{k = 0..n-1} (-1)^kbinomial(n,k)^2binomial(n-1,k).

A364507 a(n) = (5n)!(4n)! / ((3n)!^2 * (2n)! n!).

A364508 a(n) = (7n)!(6n)!(2n)! / ((4n)!^2 * (3n)!^2 n!).

A364511 a(n) = (6n)!(5n)!(2n)! / ((4n)!^2 * (3n)! n!^2).