A005810 -id:A005810 - cp's OEIS frontend

A169960 a(n) = binomial(11*n,n).

1, 11, 231, 5456, 135751, 3478761, 90858768, 2404808340, 64276915527, 1731030945644, 46897636623981, 1276749965026536, 34898565177533200, 957150015393611193, 26327386978706181060, 725971390105457325456, 20062118235172477959495, 555476984964439251664995

Offset: 0

N. J. A. Sloane, Aug 07 2010

nonn

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

binomial(k*n,n): A000984 (k = 2), A005809 (k = 3), A005810 (k = 4), A001449 (k = 5), A004355 (k = 6), A004368 (k = 7), A004381 (k = 8), A169958 - A169961 (k = 9 thru 12).

[Binomial(11*n, n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

Table[Binomial[11 n, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

a(n) = C(11*n-1,n-1)*C(121*n^2,2)/(3*n*C(11*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
From Peter Bala, Feb 21 2022: (Start)
The o.g.f. A(x) is algebraic: (1 - A(x))*(1 + 10*A(x))^10 + (11^11)*x*A(x)^11 = 0.
Sum_{n >= 1} a(n)*( x*(10*x + 11)^10/(11^11*(1 + x)^11) )^n = x. (End)

A192205 a(n) = sum of absolute values of coefficients in (1-x-x^2+x^3)^n.

Original entry on oeis.org

1, 4, 12, 36, 116, 344, 1104, 3280, 10456, 31152, 98804, 295988, 935876, 2811540, 8870324, 26695724, 84060148, 253376840, 796635360, 2404558304, 7549431884, 22820942416, 71541295984, 216562743948, 677938097756, 2054922521644

Offset: 0

Paul D. Hanna, Jun 25 2011

nonn

Conjecture: limit a(n)^(1/n) = 16*sqrt(3)/9 = 3.079201..., which is substantiated by the observation that the sums of the coefficients squared in (1-x-x^2+x^3)^n equals binomial(4n,n) (cf. A005810).

			The triangle A227964 of coefficients in (1+x-x^2-x^3)^n begins:
n=0: [1];
n=1: [1, -1, -1, 1];
n=2: [1, -2, -1, 4, -1, -2, 1];
n=3: [1, -3, 0, 8, -6, -6, 8, 0, -3, 1];
n=4: [1, -4, 2, 12, -17, -8, 28, -8, -17, 12, 2, -4, 1];
n=5: [1, -5, 5, 15, -35, -1, 65, -45, -45, 65, -1, -35, 15, 5, -5, 1];
n=6: [1, -6, 9, 16, -60, 24, 116, -144, -66, 220, -66, -144, 116, 24, -60, 16, 9, -6, 1]; ...
This sequence gives the sums of the absolute values of the coefficients for n>=0.

Cf. A192210, A084611, A084613, A084615, A084775, A084776, A084777, A084778, A084779, A084780.

Cf. A227964 (triangle).

Table[Total[Abs[CoefficientList[Expand[(1-x-x^2+x^3)^n],x]]],{n,0,30}] (* Harvey P. Dale, Mar 03 2013 *)

{a(n)=sum(k=0,3*n,abs(polcoeff((1-x-x^2+x^3)^n,k)))}
for(n=0,30,print1(a(n),", "))

A268315 Decimal expansion of 256/27.

Original entry on oeis.org

9, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8

Offset: 1

Gheorghe Coserea, Feb 01 2016

			9.481481481481481481481481481481...

Exponential growth rate for A000260, A002293, A005810, A006633, A006634, A052203, A066381, A069271, A078516, A078995, A153396, A196678, A268316.

[9] cat &cat[[4, 8, 1]^^45]; // Vincenzo Librandi, Feb 04 2016

Join[{9}, PadRight[{}, 120, {4, 8, 1}]] (* Vincenzo Librandi, Feb 04 2016 *)

PARI
```
1.0 * 256/27
```

More digits from Jon E. Schoenfield, Mar 15 2018

A273628 a(n) = (7n)!/((5n)!*n!^2).

Original entry on oeis.org

1, 42, 6006, 1085280, 217567350, 46262007792, 10217700004512, 2317454130543552, 536022010184210550, 125863265857621191900, 29909151834298018538256, 7176685161839833601969280, 1735941935586019529116213920, 422752608090008019258722317800

Offset: 0

Peter Bala, Jul 15 2016

This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(3*n + k,n)*binomial(4*n - k,n) = (-1)^m*a(m) for n = 2*m. For similar results see A001451, A006480 and A273629. Note the related sums:
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(3*n + k,n)*binomial(4*n + k,n) = (-1)^n*(2*n)!*(4*n)!/(n!^3*(3*n)!) = (-1)^n*binomial(2*n,n)*binomial(4*n,n) = (-1)^n*A000984(n)*A005810(n);
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = (3*n)!/n!^3 = A006480(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n + k,n)*binomial(4*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = binomial(2*n,n) = A000984(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n + k,n)*binomial(4*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)*binomial(4*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n).

Cf. A000984, A001451, A005810, A006480, A008979, A245086, A273629.

[Factorial(7*n) div (Factorial(5*n)*Factorial(n)^2): n in [0..15]]; // Vincenzo Librandi, Jul 16 2016

Maple
```
seq((7*n)!/((5*n)!*n!^2), n = 0..20);
```

Table[(7 n)!/((5 n)! n!^2), {n, 0, 13}] (* or *)
Table[Binomial[7 n, n] Binomial[6 n, n], {n, 0, 13}] (* Michael De Vlieger, Jul 15 2016 *)

a(n) = (7*n)!/((5*n)!*n!^2) = binomial(7*n,2*n)*binomial(2*n,n).
a(n) = binomial(7*n,n)*binomial(6*n,n) = [x^n](1 + x)^(7*n) * [x^n](1 + x)^(6*n).
It appears that a(n) = [x^n] F(x)^(42*n), where F(x) = 1 + x + 30*x^2 + 2280*x^3 + 232715*x^4 + 27800465*x^5 + 3661895341*x^6 + ... has all integer coefficients. Cf. A273629 and A008979.
Recurrence: 5*n^2*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)*a(n) = 7*(7*n - 1)*(7*n - 2)*(7*n - 3)*(7*n - 4)*(7*n - 5)*(7*n - 6)*a(n-1).
a(n) ~ 5^(-5*n-1/2)*7^(7*n+1/2)/(2*Pi*n). - Ilya Gutkovskiy, Jul 15 2016
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(6*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

A273629 a(n) = (9n)!/((7n)!*n!^2).

Original entry on oeis.org

1, 72, 18360, 5920200, 2118223800, 803927196072, 316938365223480, 128313095514575400, 52976845635264939960, 22204947580777261872000, 9418997650746914743158360, 4034374193416822645489549632, 1741969558937890710303111545400

Offset: 0

Peter Bala, Jul 15 2016

This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(4*n + k,n)*binomial(5*n - k,n) = (-1)^m*a(m) for n = 2*m. The sum vanishes for n odd. For similar results see A001451, A006480 and A273628.
Note the related sums:
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(4*n - k,n)*binomial(5*n - k,n) = binomial(2*n,n)*binomial(4*n,n) = A000984(n)*A005810(n);
Sum_{k = 0..2*n} (-1)^k*binomial(n,k)*binomial(4*n + k,n)*binomial(5*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(n,k)*binomial(4*n - k,n)*binomial(5*n - k,n) = binomial(2*n,n) = A000984(n).
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(4*n + k,n)*binomial(5*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(4*n - k,n)*binomial(5*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n).

Cf. A000984, A001451, A004381, A005810, A006480, A008979, A169958, A245086, A273628.

[Factorial(9*n)/(Factorial(7*n)*Factorial(n)^2): n in [0..40]]; // Vincenzo Librandi, Jul 17 2016

Maple
```
seq((9*n)!/((7*n)!*n!^2), n = 0..20);
```

Table[Factorial[9 n] / (Factorial[7 n] Factorial[n]^2), {n, 0, 20}] (* Vincenzo Librandi, Jul 17 2016 *)

a(n) = (9*n)!/((7*n)!*n!^2) = binomial(9*n,2*n)* binomial(2*n,n).
a(n) = binomial(8*n,n)*binomial(9*n,n) = A004381(n)*A169958(n).
a(n) = [x^n](1 + x)^(8*n) * [x^n] (1 + x)^(9*n).
It appears that a(n) = [x^n] F(x)^(72*n), where F(x) = 1 + x + 56*x^2 + 7700*x^3 + 1422008*x^4 + 307144278*x^5 + 73118586828*x^6 + ... has all integer coefficients. Cf. A273628 and A008979.
Recurrence: 7*n^2*(7*n - 1)*(7*n - 2)*(7*n - 3)*(7*n - 4)*(7*n - 5)*(7*n - 6)*a(n) = 9*(9*n - 1)*(9*n - 2)*(9*n - 3)*(9*n - 4)*(9*n - 5)*(9*n - 6)*(9*n - 7)*(9*n - 8)*a(n-1).
a(n) ~ 3^(18*n+1)*7^(-7*n-1/2)/(2*Pi*n). - Ilya Gutkovskiy, Jul 15 2016
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(8*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

A277170 Numerator of 3F2([3n, -n, n+1],[2n+1, n+1/2], 1).

Original entry on oeis.org

1, -1, 1, -1, 1, -3, 1, -1, 25, -1, 1, -49, 1, -1, 9, -3, 1, -363, 3025, -1, 169, -169, 1, -3, 1, -49, 289, -289, 7225, -361, 361, -361, 1, -1, 1, -529, 529, -529, 330625, -148225, 3025, -675, 9, -3, 7569, -2523, 142129, -409757907, 808201, -961, 8649, -2883, 1, -147

Offset: 0

Seiichi Manyama, Oct 19 2016

Neil Calkin found the closed forms of 3F2([3*n, -n, n+1],[2*n+1, n+1/2], 1) in 2007.

Jonathan Borwein, David Bailey, Mathematics by Experiment, 2nd Edition: Plausible Reasoning in the 21st Century.

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A005810, A066802, A052203, A187364, A277520.

a[n_] := HypergeometricPFQ[{3n, -n, n+1}, {2n+1, n+1/2}, 1] // Numerator; Table[a[n], {n, 0, 53}] (* Jean-François Alcover, Oct 22 2016 *)

(s(n) =) 3F2([3*n, -n, n+1],[2*n+1, n+1/2], 1) = a(n) / A277520(n).
s(2k) = (A005810(k) / A066802(k))^2 = (((4k)! * (3k)!) / ((6k)! * k!))^2.
s(2k+1) = -1/3 * (A052203(k) / A187364(k))^2 = -1/3 * (((4k+1)! * (3k)!) / ((6k+1)! * k!))^2.

A277520 Denominator of 3F2([3n, -n, n+1],[2n+1, n+1/2], 1).

Original entry on oeis.org

1, 3, 25, 147, 1089, 20449, 48841, 312987, 55190041, 14322675, 100100025, 32065374675, 4546130625, 29873533563, 1859904071089, 4089135109921, 9399479144449, 22568149425822049, 1293753708921104809, 2835106739783283, 3289668853728536041

Offset: 0

Seiichi Manyama, Oct 19 2016

Neil Calkin found the closed forms of 3F2([3*n, -n, n+1],[2*n+1, n+1/2], 1) in 2007.

Jonathan Borwein, David Bailey, Mathematics by Experiment, 2nd Edition: Plausible Reasoning in the 21st Century.

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A005810, A052203, A066802, A187364, A277170 (numerators).

a[n_] := HypergeometricPFQ[{3n, -n, n+1}, {2n+1, n+1/2}, 1] // Denominator;
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Oct 22 2016 *)

(s(n) =) 3F2([3*n, -n, n+1],[2*n+1, n+1/2], 1) = A277170(n) / a(n).
s(2k) = (A005810(k) / A066802(k))^2 = (((4k)! * (3k)!) / ((6k)! * k!))^2.
s(2k+1) = -1/3 * (A052203(k) / A187364(k))^2 = -1/3 * (((4k+1)! * (3k)!) / ((6k+1)! * k!))^2.

A306290 a(n) = 1/(Integral_{x=0..1} (x^3 - x^4)^n dx).

Original entry on oeis.org

1, 20, 252, 2860, 30940, 325584, 3364900, 34337160, 347103900, 3483301360, 34754081648, 345120260940, 3413758188932, 33655718658800, 330869721936600, 3244839440755920, 31754250910172700, 310165459118369712, 3024542552887591120, 29449493278116018800, 286360607519186119920

Offset: 0

G. C. Greubel, Feb 03 2019

Necdet Batir, On certain series involving reciprocals of binomial coefficients, Journal of Classical Analysis, Vol. 2, No. 1 (2013), pp. 1-8.

Cf. A002457, A005810, A016813, A090816, A090957, A090969.

List([0..30], n -> Factorial(4*n+1)/(Factorial(n)*Factorial(3*n)));

[Factorial(4*n+1)/(Factorial(n)*Factorial(3*n)): n in [0..20]];

Mathematica
```
Table[1/Beta[3*n+1, n+1], {n, 0, 20}]
```

vector(20, n, n--; (4*n+1)!/(n!*(3*n)!))

Sage
```
[1/beta(3*n+1,n+1) for n in range(20)]
```

a(n) = 1/Beta(3*n+1,n+1) = (4*n+1)!/(n! * (3*n)!).
a(n) = Sum_{k = 0..n} (-1)^(n+k) * (3*n + 2*k + 1)*binomial(3*n+k, k).  This is the particular case m = 1 of the identity Sum_{k = 0..m*n} (-1)^k * (3*n + 2*k + 1) * binomial(3*n+k, k) = (-1)^(m*n) * (m*n + 1) * binomial((m+3)*n+1, 3*n). - Peter Bala, Nov 02 2024
From Amiram Eldar, Dec 09 2024: (Start)
a(n) = (4*n + 1) * binomial(4*n, n) = A016813(n) * A005810(n).
Formulas from Batir (2013):
Sum_{n>=0} 1/a(n) = f(c) = 1.05435362585114283076..., where f(x) = (x*(x^2-1)/(2*(2*x^2+1))) * log(abs((x+1)/(x-1))) + ((x-1)*(x^3+1)/(4*x*(2*x^2+1))) * sqrt(x/(x-2)) * (arctan(sqrt(x/(x-2))) + arctan(((3-x)/(x+1))*sqrt(x/(x-2)))) + ((x+1)*(x^3-1)/(4*x*(2*x^2+1))) * sqrt(x/(x+2)) * (arctan(((x+3)/(x-1))*sqrt(x/(x+2))) - arctan(sqrt(x/(x+2)))), and c = sqrt(1 + (16/sqrt(3))*cos(arctan(sqrt(229/27))/3)).
Sum_{n>=0} (-1)^n/a(n) = f(d) = 0.953648123517883351708..., where f(x) is defined above, and d = sqrt(1 + 16*(2/(3*(9+sqrt(849))))^(1/3) - 2*(2/3)^(2/3)*(9+sqrt(849))^(1/3)). (End)

A352276 a(0) = 1 and a(n) = Sum_{k = 0..3n} n/(n + 2k)binomial(n + 2k,k) for n >= 1.

Original entry on oeis.org

1, 9, 625, 58785, 5986993, 633580634, 68611922731, 7545931449401, 839183314181297, 94112350842056469, 10623982584664109750, 1205644823097085684641, 137414820511792364274091, 15718880489100999321142976, 1803621273322664188151352631, 207499462144488863314062180035

Offset: 0

Peter Bala, Mar 10 2022

The following identity can be easily verified using Maple's SumTools:-Summation procedure: for n >= 1, A005810(n) = binomial(4*n,n) = Sum_{k = 0..3*n} n/(n + k)*binomial(n + k,k).
The binomial coefficients A005810(n) are known to satisfy the supercongruences A005810(n*p^r) == A005810(n*p^(r-1)) (mod p^(3*r)) for primes p >= 5 and positive integers n and r (see Meštrović, equation 39).
Calculation suggests that the present sequence satisfies the same congruences.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r.
More generally, for m a positive integer, define a sequence u_m by setting u_m(n) = Sum_{k = 0..m*n} n/(n + 2*k)*binomial(n + 2*k,k) for n >= 1.
Then we conjecture that each sequence u_m satisfies the above supercongruences. This is the case m = 3. See A333093 (case m = 1) and A352275 (m = 2).

			Examples of supercongruences:
a(11) - a(1) = 1205644823097085684641 - 9 = (2^3)*3*(11^3)*43*2887*5059* 60096637 == 0 (mod 11^3)
a(3*5) - a(3) = 207499462144488863314062180035 - 58785 = 2*(5^4)*1801* 4959701*18583938263214197 == 0 (mod 5^4)

Paolo Xausa, Table of n, a(n) for n = 0..475
R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Cf. A005810, A333093, A352275.

nterms=25;Join[{1},Table[Sum[n/(n+2k)Binomial[n+2k,k],{k,0,3n}],{n,nterms-1}]] (* Paolo Xausa, Apr 10 2022 *)

a(n) ~ 7^(7*n + 3/2) / (37 * sqrt(Pi*n) * 2^(8*n + 3/2) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 15 2022

A352657 The number of lozenge tilings of a semiregular hexagon of side lengths n, n, 3n, n, n and 3n; equivalently, the number of plane partitions whose solid Young diagram fits inside an n X n X 3*n box.

Original entry on oeis.org

1, 4, 336, 572572, 19571505408, 13365232267026024, 182001937855822420050000, 49372092168218024268166702560000, 266640931683989945767062736068603511111680, 28657545169614835585678719963104037818950931553412096, 61277278161726929232430881966673334396569563602790616552072890176

Offset: 0

Peter Bala, Apr 22 2022

A lozenge is a unit rhombus with internal angles of 60 and 120 degrees. A hexagon is semiregular if its internal angles are 120 degrees and opposite sides are of equal length. Let S(n) = Product_{k = 0..n-1} k! = A000178(n-1) for n >= 1. S(n) equals the superfactorial of n-1. Then for a, b and c nonnegative integers a semiregular hexagon with side-lengths a, b, c, a, b, c can be tiled by lozenges in exactly S(a+b+c)*S(a)*S(b)*S(c)/(S(a+b)*S(a+c)*S(b+c)) ways.

The superfactorial ratio (S(a)*S(b)*S(c)*S(a+b+c))/(S(a+b)*S(a+c)*S(b+c)) is an integer (see MacMahon, Chapter II, Section 429, p. 182, with x -> 1) and can be viewed as the superfactorial analog of the binomial coefficient (a + b)!/(a!*b!). Setting a = b = n, c = 3*n gives the entries for the present sequence, a superfactorial analog of A005810(n) = binomial(4*n,n).

			Examples of supercongruences:
p = 5, n = 1, r = 1:
a(5) - a(1)^5 = 13365232267026024 - 4^5 = (2^3)*(5^5)*534609290681 == 0 (mod 5^5).
p = 7, n = 1, r = 1:
a(7) - a(1)^7 = 49372092168218024268166702560000 - 4^7 = (2^8)*(7^4)*42153329 *1905537621534581059 == 0 (mod 7^4).
p = 3, n = 1, r = 2:
a(3^2) - a(3)^3 = 28657545169614835585678719963104037818950931553412096 - 572572^3 = (2^6)*(3^9)*7*13*36206433373771931*6904632711001213215426713099 == 0 (mod 3^9).
exp(Sum_{n >= 1} a(n)*x^n/n) = 1 + 4*x + 176*x^2 + 191540*x^3 + 4893655248*x^4 + 2673066058559752*x^5 + 30333667002369040991520*x^6 + 7053156145366242954671905412736*x^7 + 33330116488711372656254906993570075436704*x^8 + 3184171685646079976603214029980784880572652377971904*x^9 + 6127727816185429609991005336553574169498938182021433716145181760*x^10 + ....

C. Krattenthaler, Advanced Determinant Calculus: A Complement, Linear Algebra Appl. 411 (2005), 68-166; arXiv:math/0503507v2 [math.CO], 2005.
P. A. MacMahon, Combinatory Analysis, vol. 2, Cambridge University Press, 1916; reprinted by Chelsea, New York, 1960.
Eric Weisstein's World of Mathematics, Plane Partition
Wikipedia, Superfactorial

Cf. A000178, A005810, A008793, A074962, A352656.

S := proc(n) local i; mul(i!, i = 0..n-1) end proc:
a := n -> S(n)^2*S(3*n)*S(5*n)/(S(2*n)*S(4*n)^2):
seq(a(n), n = 0..10);

Table[BarnesG[n + 1]^2 * BarnesG[3*n + 1] * BarnesG[5*n + 1] / (BarnesG[2*n + 1] * BarnesG[4*n + 1]^2), {n, 0, 10}] (* Vaclav Kotesovec, May 16 2022 *)

a(n) = S(n)^2*S(3*n)*S(5*n)/(S(2*n)*S(4*n)^2), where S(n) = Product_{k = 0..n-1} k! with S(0) = 1.
a(n) = Product_{i = 1..3*n} (2*n+i-1)!*(i-1)!/(n+i-1)!^2.
a(n) = Product_{i = 1..n} (4*n+i-1)!*(i-1)!/((3*n+i-1)!*(n+i-1)!).
a(n) = Product_{i = 1..3*n} Product_{1 <= j, k <= n} (i + j + k - 1)/(i + j + k - 2).
a(n) = Product_{i = 1..n} Product_{j = 1..n} (3*n + i + j - 1)/(i + j - 1).
a(n) = Product_{i = 1..3*n} Product_{j = 1..n} (n + i + j - 1)/(i + j - 1).
For n >= 1, a(n) = det( (binomial(4*n,n+i-j)) ) for 1 <= i, j <= n. Apply Krattenhaller, Theorem 4 with a = n, b = 3*n and c = n.
a(n) ~ 1/A*(32/(15*n))^(1/12)*exp(B*n^2 + 1/12), where A = 1.2824271291... is the Glaisher-Kinkelin constant A074962 and B = (25/2)*log(5) + (9/2)*log(3) - 34*log(2).
Conjecture 1): the Gauss congruences a(n*p^r) == a(n*p^(r-1)) (mod p^r) hold for all primes p and positive integers n and r. If true, then the expansion of exp(Sum_{n >= 1} a(n)*x^n/n) has integer coefficients.
Conjecture 2): the supercongruences a(n*p^r) == a(n*p^(r-1))^p (mod p^(4*r)) hold for all primes p and positive integers n and r.
a(n) ~ exp(1/12) * 3^(9*n^2/2 - 1/12) * 5^(25*n^2/2 - 1/12) / (A * n^(1/12) * 2^(34*n^2 - 5/12)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, May 16 2022
From Peter Bala, Feb 15 2023: (Start)
a(n) = Product_{i = 1..n} Product_{j = 3*n..4*n-1} (i+j) / Product_{j = 0..n-1} (i+j).
a(n) = Product_{i = 1..3*n} Product_{j = n..2*n-1} (i+j) / Product_{j = 0..n-1} (i+j). (End)

cp's OEIS Frontend

A169960 a(n) = binomial(11*n,n).

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A192205 a(n) = sum of absolute values of coefficients in (1-x-x^2+x^3)^n.

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A268315 Decimal expansion of 256/27.

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A273628 a(n) = (7*n)!/((5*n)!*n!^2).

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A273629 a(n) = (9*n)!/((7*n)!*n!^2).

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A277170 Numerator of 3F2([3*n, -n, n+1],[2*n+1, n+1/2], 1).

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A277520 Denominator of 3F2([3*n, -n, n+1],[2*n+1, n+1/2], 1).

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A306290 a(n) = 1/(Integral_{x=0..1} (x^3 - x^4)^n dx).

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A273628 a(n) = (7n)!/((5n)!*n!^2).

A273629 a(n) = (9n)!/((7n)!*n!^2).

A277170 Numerator of 3F2([3n, -n, n+1],[2n+1, n+1/2], 1).

A277520 Denominator of 3F2([3n, -n, n+1],[2n+1, n+1/2], 1).

A352276 a(0) = 1 and a(n) = Sum_{k = 0..3n} n/(n + 2k)binomial(n + 2k,k) for n >= 1.

A352657 The number of lozenge tilings of a semiregular hexagon of side lengths n, n, 3n, n, n and 3n; equivalently, the number of plane partitions whose solid Young diagram fits inside an n X n X 3*n box.