A006480 -id:A006480 - cp's OEIS frontend

A361030 a(n) = 20160(3n)!/(n!*(n+3)!^2).

560, 210, 504, 2352, 15840, 135135, 1361360, 15519504, 194699232, 2636552100, 38003792400, 577037174400, 9155656500480, 150853746558690, 2568167588473200, 44990491457326800, 808333317429976800, 14853124707775823700, 278470827854627007600, 5316261259042879236000

Offset: 0

Peter Bala, Mar 01 2023

Row 2 of square array A361027.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 60*A000984(n) is divisible by (n + 1)*(n + 2)*(n + 3) and the result 60*(2*n)!/(n!*(n+3)!) is the super ballot number A007272(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 20160*A006480(n) is divisible by ((n + 1)*(n + 2)*(n + 3))^2.
Equivalently, the central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that (1*3*5)*A000984(n+3) is divisible by (2*n + 1)*(2*n + 3)*(2*n + 5). The result is always an even integer. In fact, (1/2)*(1*3*5)/((2*n + 1)*(2n + 3)*(2n + 5))*A000984(n+3) = A007272(n).
Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that (1*2*4*5*7*8)*A006480(n+3) is divisible by (3*n + 1)*(3*n + 2)*(3*n + 4)*(3*n + 5)*(3*n + 7)*(3*n + 8). The result is always an integer divisible by 3.

Peter Bala, A361030 is an integer sequence

Row 2 of A361027. Cf. A006480, A007272, A361028, A361029, A361031, A361038.

a := proc(n) option remember; if n = 0 then 560 else 3*(3*n-1)*(3*n-2)/(n+3)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);

a(n) = 20160/((n+1)*(n+2)*(n+3))^2 * (3*n)!/n!^3.
a(n) = (1/3)*(1*2*4*5*7*8) * A006480(n+3)/((3*n + 1)*(3*n + 2)*(3*n + 4)*
(3*n + 5)*(3*n + 7)*(3*n + 8)), where A006480(n) = (3*n)!/n!^3.
a(n) = (1/5)*A007272(n)*A361038(n). Using this it can be shown that a(n) is always an integer.
a(n) = (1/3)*27^(n+3)*binomial(7/3, n+3)*binomial(8/3, n+3).
a(n) ~ sqrt(3)*10080*(27^n)/(Pi*n^7).
P-recursive: (n + 3)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 560.
The o.g.f. A(x) satisfies the differential equation x^2*(1 - 27*x)*A''(x) + x*(7 - 54*x)*A'(x) + (9 - 6*x)*A(x) - 5040 = 0, with A(0) = 560 and A'(0) = 210.

A361717 a(n) = Sum_{k = 0..n-1} binomial(n-1,k)^2*binomial(n+k,k).

Original entry on oeis.org

0, 1, 4, 27, 216, 1875, 17088, 160867, 1549936, 15195843, 151017780, 1517232189, 15379549056, 157058738343, 1614039427224, 16676755365555, 173118505001952, 1804500885273123, 18877476988765404, 198120856336103017, 2085303730716475960

Offset: 0

Peter Bala, Mar 26 2023

Compare with the Apery numbers A005258(n) = Sum_{k = 0..n} binomial(n,k)^2* binomial(n+k,k).
Conjecture 1: the supercongruence a(p) == 0 (mod p^4) holds for all primes p >= 5 (checked up to p = 199).
Conjecture 2: the supercongruence a(p-1) == 1 - 2*p - p^2 (mod p^3) holds for all primes except p = 3 (checked up to p = 199).

			a(5) = 3*(5^4); a(7) = (7^4)*67; a(11) = 3*(11^4)*34543; a(13) = (3^3)*(13^4)*203669.

Winston de Greef, Table of n, a(n) for n = 0..956

Cf. A005258, A006480, A060542, A361712, A361713, A361714, A361715.

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

A361717[n_]:=Sum[Binomial[n-1,k]^2Binomial[n+k,k],{k,0,n-1}];Array[A361717,30,0] (* Paolo Xausa, Oct 06 2023 *)

a(n) = sum(k=0, n-1, binomial(n-1,k)^2*binomial(n+k,k)) \\ Winston de Greef, Mar 27 2023

a(n) = hypergeom([1 + n, 1 - n, 1 - n], [1, 1], 1) for n >= 1.
P-recursive:
n*(n-1)*(5*n-7)*a(n) = (55*n^3-187*n^2+190*n-48)*a(n-1) + (n-1)*(n-3)*(5*n-2)* a(n-2) with a(0) = 0 and a(1) = 1.
a(n) ~ phi^(5*n - 3/2) / (2*5^(1/4)*Pi*n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Mar 27 2023
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1) * binomial(n-1, k) * binomial(n+k-1, k) * binomial(n+k, k+1) = (-1)^(n+1) * n * hypergeom([n, n + 1, 1 - n], [1, 2], 1). - Peter Bala, Sep 08 2023
a(n) = Sum_{k = 0..n-1} (-1)^k * binomial(n-2, k) * binomial(2*n-2-k, n-1-k)^2. - Peter Bala, Oct 09 2024
From Peter Bala, Jul 31 2025: (Start)
a(n) = n * Sum_{k = 0..n} 1/(k+1) * binomial(n-1, k)^2 * binomial(n+k-1,k).
a(n) = n * hypergeom([n, 1 - n, 1 - n], [1, 2], 1). (End)

A217800 Number of alternating permutations on 2n+1 letters that avoid a certain pattern of length 4 (see Lewis, 2012, Appendix, for precise definition).

Original entry on oeis.org

1, 2, 12, 110, 1274, 17136, 255816, 4124406, 70549050, 1264752060, 23555382240, 452806924752, 8939481277552, 180551099694400, 3719061442253520, 77933728043586630, 1658001861319441050, 35749633305661575300, 780123576993991461000, 17208112644166765652100

Offset: 0

N. J. A. Sloane, Oct 12 2012

nonn

1 together with A007724. - Omar E. Pol, Aug 22 2016

K. Gorska and K. A. Penson, Multidimensional Catalan and related numbers as Hausdorff moments, arXiv preprint arXiv:1304.6008 [math.CO], 2013.
J. B. Lewis, Pattern Avoidance for Alternating Permutations and Reading Words of Tableaux, Ph. D. Dissertation, Department of Mathematics, MIT, 2012.

Cf. A007724, A181197, A217799-A217830, A005789, A006480, A007724.

[Factorial(3*n+3)/((4*(n+1)^2-1)*Factorial((n+1))^2*Factorial(n+ 2)): n in [0..20]]; // Vincenzo Librandi, Aug 30 2014

a := n -> (-1)^n*hypergeom([-2-2*n, -2*n, -2*n-1], [2, 3], 1):
seq(round(evalf(a(n), 32)), n=0..20); # Peter Luschny, Aug 29 2014

Table[(3 n + 3)!/((4 (n + 1)^2 - 1) ((n + 1)!)^2 (n + 2)!), {n, 0, 20}] (* Vincenzo Librandi, Aug 30 2014 *)
Table[(-1)^n HypergeometricPFQ[{-2 - 2 n, -2 n, -2 n - 1}, {2, 3}, 1], {n, 0, 20}] (* Michael De Vlieger, Aug 22 2016 *)

a(n) = (3*n+3)!/((4*(n+1)^2-1)*((n+1)!)^2*(n+2)!); \\ Michel Marcus, Aug 10 2014

From Karol A. Penson, Aug 10 2014: (Start)
O.g.f.(in Maple notation): hypergeom([1/2, 1, 4/3, 5/3], [2, 5/2, 3], 27*z);a(n) ~ (1/93312)*sqrt(3)*27^n*(314928*n^4-1644624*n^3+5545260*n^2 -15387660*n+38310503)/(Pi*n^8), for n -> infinity.
Representation of a(n) as the n-th power moment of a positive function on the segment [0,27]:
a(n) = int(x^n*W(x),x=0..27),n=0,1,2..., where
W(x) = 1/(Pi*sqrt(x))+sqrt(x)/Pi-(9/20)*sqrt(3)*2^(1/3)* hypergeom([-2/3, -1/6, 1/3], [2/3, 11/6], (1/27)*x)*x^(1/3)/ (sqrt(Pi)*Gamma(5/6)*Gamma(2/3))-(27/56)*2^(2/3)*Gamma(5/6) *Gamma(2/3)*hypergeom([-1/3, 1/6, 2/3], [4/3, 13/6], (1/27)*x)* x^(2/3)/Pi^(5/2).
W(x) for x->0 has the singularity 1/sqrt(x), W(27)=0.
This is the solution of the Hausdorff moment problem and is unique.
a(n) = (1/2)*(n+3)!/((4*(n+1)^2-1)*(n+1)!)*A005789(n), where A005789(n) are the three-dimensional Catalan numbers (see the Gorska and Penson link).(End)
a(n) = A006480(n+1)/((2+n)*(1+2*n)*(3+2*n)). - Peter Luschny, Aug 15 2014
a(n) = (-1)^n*hypergeom([-2-2*n,-2*n,-2*n-1],[2,3],1). - Peter Luschny, Aug 29 2014
(2*n+3)*(n+2)*(n+1)*a(n) -3*(3*n+2)*(2*n-1)*(3*n+1)*a(n-1)=0. - R. J. Mathar, Jun 14 2016
a(n) ~ 3^(3*n + 7/2) / (8*Pi*n^4). - Vaclav Kotesovec, Jun 09 2019

More terms from Alois P. Heinz, Aug 22 2016

Merged with A241958 by R. J. Mathar, Jul 07 2023

A227357 de Bruijn's S(6,n).

Original entry on oeis.org

1, 62, 38466, 41312060, 56930297410, 90519385516812, 157933807781230404, 294111627143303836152, 574788682882785699423810, 1165869740380160987511514460, 2435635082278794046304453801716, 5211959633483650233198112526032152, 11377217758058088192513643732271022916

Offset: 0

Vaclav Kotesovec, Jul 09 2013

nonn

Generally, S(s,n) is asymptotic to (2*cos(Pi/(2*s)))^(2*n*s+s-1) *2^(2-s)*(Pi*n)^((1-s)/2)*s^(-1/2).

N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapter 4.7, p.72-75.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Eric Weisstein's World of Mathematics, Binomial Sums

Cf. A000984 (s=2), A006480 (s=3), A050983 (s=4), A050984 (s=5).

a:= n->add((-1)^(k+n)*binomial(2*n,k)^6, k=0..2*n):
seq(a(n), n=0..15);  # Alois P. Heinz, Jul 17 2013

Table[Sum[(-1)^(k+n)*Binomial[2*n,k]^6, {k, 0, 2*n}], {n, 0, 20}]

a(n) ~ (1+sqrt(3))^(12*n+5)/(sqrt(3)*2^(6*n+7)*(Pi*n)^(5/2)).

Recurrence: (n-1)*n^5*(2*n - 1)^3*(78037440*n^10 - 1398170800*n^9 + 11197027400*n^8 - 52776179300*n^7 + 162127296682*n^6 - 339174893304*n^5 + 489377694958*n^4 - 480894941069*n^3 + 308044053231*n^2 - 116166558141*n + 19587964597)*a(n) = (n-1)*(1726812472320*n^18 - 37845973351680*n^17 + 383495168176640*n^16 - 2385128962478080*n^15 + 10193794229981856*n^14 - 31763778392601840*n^13 + 74716717106494000*n^12 - 135540917163836728*n^11 + 192070195278504510*n^10 - 214041209444090466*n^9 + 187905640039584992*n^8 - 129585587008626217*n^7 + 69664459655905576*n^6 - 28800662692839270*n^5 + 8959012339689510*n^4 - 2025094914623067*n^3 + 313623932421492*n^2 - 29741972276520*n + 1302044058000)*a(n-1) - (2*n - 3)*(53979745847040*n^18 - 1291015588175040*n^17 + 14334324120939680*n^16 - 98074075137527840*n^15 + 462828677276119232*n^14 - 1597795252577443036*n^13 + 4175964673926667106*n^12 - 8435559969344133552*n^11 + 13328633341117570446*n^10 - 16565740193886252205*n^9 + 16202242092204003209*n^8 - 12416056458421188647*n^7 + 7385327565692140915*n^6 - 3358099721685530886*n^5 + 1140333781667693872*n^4 - 278997802954150098*n^3 + 46356206084424824*n^2 - 4676191704077040*n + 216042816276000)*a(n-2) + 8*(n-2)^3*(2*n - 5)^5*(2*n - 3)*(78037440*n^10 - 617796400*n^9 + 2125175000*n^8 - 4169616100*n^7 + 5152323982*n^6 - 4181430032*n^5 + 2256662768*n^4 - 801756137*n^3 + 180454862*n^2 - 23380182*n + 1331694)*a(n-3). - Vaclav Kotesovec, Sep 27 2016

A324467 Three times the binary weight of n: 3*A000120(n).

Original entry on oeis.org

0, 3, 3, 6, 3, 6, 6, 9, 3, 6, 6, 9, 6, 9, 9, 12, 3, 6, 6, 9, 6, 9, 9, 12, 6, 9, 9, 12, 9, 12, 12, 15, 3, 6, 6, 9, 6, 9, 9, 12, 6, 9, 9, 12, 9, 12, 12, 15, 6, 9, 9, 12, 9, 12, 12, 15, 9, 12, 12, 15, 12, 15, 15, 18, 3, 6, 6, 9, 6, 9, 9, 12, 6, 9, 9, 12, 9, 12, 12, 15, 6, 9, 9, 12

Offset: 0

N. J. A. Sloane, Mar 02 2019

nonn

Exponent of highest power of 2 that divides multinomial(4*n;n,n,n,n).

Vincenzo Librandi, Table of n, a(n) for n = 0..8191

Cf. A000120, A006480, A324465, A324466.

[3*Valuation(Binomial(2*n,n), 2): n in [0..120]]; // Vincenzo Librandi, Mar 11 2019

[seq(padic[ordp](multinomial(4*n,n,n,n,n),2),n=0..128)];

3 Log[2, Nest[Join[#, 2 #]&, {1}, 8]] (* Vincenzo Librandi, Mar 11 2019 *)

A352652 a(n) = ( binomial(7n,2n)binomial(7n/2,2n)binomial(2n,n)^2 ) / binomial(7n/2,n)^2.

Original entry on oeis.org

1, 30, 2860, 343200, 45643500, 6435891280, 942422020540, 141696569678400, 21724714133822700, 3381208130986900500, 532553441617598475360, 84695057996350934903680, 13578009523892192555221500, 2191530567314796197691108600, 355765014009052303028935320000

Offset: 0

Peter Bala, Apr 03 2022

We write x! as shorthand for Gamma(x+1) and binomial(x,y) as shorthand for x!/(y!*(x-y)!) = Gamma(x+1)/(Gamma(y+1)*Gamma(x-y+1)).
Given two sequences of numbers 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. Bober completed the classification of integral factorial ratio sequences of height 1. Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2. Usually, it is assumed that the c's and d's are integers but here we allow for some of the c's and d's to be rational numbers. See A276098 and the cross references for further examples of factorial ratios of this type.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k. The case n = k = 1 is easily proved.

			Examples of supercongruences:
a(11) - a(1) = 84695057996350934903680 - 30 = 2*(5^2)*(11^3)*23*593* 3671*5693*4464799 == 0 (mod 11^3)
a(2*7) - a(2) = 355765014009052303028935320000 - 2860 = (2^2)*5*(7^3)*11* 269*3307*375101*14129010228023 == 0 (mod 7^3)

Peter Bala, Some integer ratios of factorials
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, 2007, arXiv:0709.1977v1 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (2009), 422-444.
K. Soundararajan, Integral factorial ratios: irreducible examples with height larger than 1, Phil. Trans. R. Soc. A378: 2018044, 2019.

Cf. A275654, A276098, A276100, A276101, A276102, A352651.

a := n -> if n = 0 then 1 elif n = 1 then 30 else
7*(5*n-2)*(5*n-4)*(5*n-6)*(5*n-8)*(7*n-1)*(7*n-3)*(7*n-5)*(7*n-9)*(7*n-11)*(7*n-13)/(3*n^2*(n-1)^2*(3*n-2)*(3*n-4)*(5*n-1)*(5*n- 3)*(5*n -7)*(5*n-9)) *a(n-2) end if:
seq(a(n), n = 0..20);

from math import factorial
from sympy import factorial2
def A352652(n): return int(factorial(7*n)*factorial2(5*n)**2//factorial(5*n)//factorial2(7*n)//factorial2(3*n)//factorial(n)**2) # Chai Wah Wu, Aug 08 2023

a(n) = (5/3)*Sum_{k = 0..n} (-1)^(n+k)*binomial(7*n,n-k)*binomial(5*n+k-1,k)^2 for n >= 1 (this formula shows 3*a(n) is integral; how to show a(n) is integral?).
a(n) = (5/3)*Sum_{k = 0..n} binomial(4*n-k-2,n-k)*binomial(5*n-1,k)^2 for n >= 1.
a(n) = (7*n)!*(5*n/2)!^2/((5*n)!*(7*n/2)!*(3*n/2)!*n!^2!).
a(n) = (5/3) * [x^n] ( (1 - x)^(2*n) * P(5*n-1,(1 + x)/(1 - x)) ) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial.
a(n) = (5/3)*(-1)^n*binomial(7*n,n)*hypergeom([-n, 5*n, 5*n], [1, 6*n+1], 1) for n >= 1.
a(n) ~ sqrt(15)/Pi * 7^(7*n/2)/3^(3*n/2) * ( 1/(6*n) - 29/(945*n^2) + 841/(297675*n^3) + O(1/n^4) ).
a(n) = 7*(5*n-2)*(5*n-4)*(5*n-6)*(5*n-8)*(7*n-1)*(7*n-3)*(7*n-5)*(7*n-9)*(7*n-11)*(7*n-13)/(3*n^2*(n-1)^2*(3*n-2)*(3*n-4)*(5*n-1)*(5*n- 3)*(5*n -7)*(5*n-9)) * a(n-2) with a(0) = 1 and a(1) = 30.
a(n)*A275654(n) = (7*n)!/(n!^4*(3*n)!) = A071549(n)/A006480(n).
a(p) == 30 (mod p^3) for all primes p >= 5.

A361028 a(n) = 2(3n)!/(n!*(n+1)!^2).

Original entry on oeis.org

2, 3, 20, 210, 2772, 42042, 700128, 12471030, 233716340, 4557468630, 91752013080, 1896208270320, 40055997189600, 862021408906800, 18849534808095360, 417929529573239310, 9379553386892837940, 212776905535994934750, 4873239487455972633000, 112571832160232967822300

Offset: 0

Peter Bala, Feb 28 2023

Row 0 of square array A361027.

The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that A000984(n) is divisible by n + 1 and the result (2*n)!/(n!*(n+1)!) is the n-th Catalan number A000108(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 2*A006480(n) is divisible by (n+1)^2, leading to the present sequence. Do these numbers have a combinatorial interpretation?

Row 0 of A361027. Cf. A000108, A006480, A007226, A361029, A361030, A361031, A361033.

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

a(n) = (2/(n+1)^2) * (3*n)!/n!^3.
a(n) = (2/3) * A006480(n+1)/((3*n + 1)*(3*n + 2)), where A006480(n) = (3*n)!/n!^3.
a(n) = (1/3)*27^(n+1)*binomial(1/3, n+1)*binomial(2/3, n+1).
a(n) = 2*C(2*n,n)*C(3*n,n) - 3*C(2*n,n+1)*C(3*n,2*n) + 2*C(2*n,n)*C(3*n,2*n+2) + 2*C(2*n,n+1)*C(3*n,2*n+2). This formula shows that a(n) is an integer for all n.
a(n) = A007226(n) * A000108(n).
a(n) ~ sqrt(3)*27^n/(Pi*n^3).
P-recursive: (n + 1)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 2.
The o.g.f. A(x) satisfies the differential equation
x^2*(1 - 27*x)*A''(x) + x*(3 - 54*x)*A'(x) + (1 - 6*x)*A(x) - 2 = 0, with
A(0) = 2 and A'(0) = 3.

A361029 a(n) = 120(3n)!/(n!*(n+2)!^2).

Original entry on oeis.org

30, 20, 75, 504, 4620, 51480, 656370, 9237800, 140229804, 2259901800, 38230005450, 673210036800, 12262039956000, 229872375708480, 4417859720647350, 86767376381987400, 1736954330906081100, 35364582637561485000, 730985923118395894950, 15315895532004485418000

Offset: 0

Peter Bala, Mar 01 2023

Row 1 of the square array A361027.

The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 6*A000984(n) is divisible by (n + 1)*(n + 2) and the result 6*(2n)!/(n!*(n+2)!) is the super ballot number A007054(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 120*A006480(n) is divisible by ((n + 1)*(n + 2))^2, leading to the present sequence. Do these numbers have a combinatorial interpretation?

Row 1 of A361027. Cf. A006480, A007054, A361028, A361030, A361031, A361037.

a := proc(n) option remember; if n = 0 then 30 else 3*(3*n-1)*(3*n-2)/(n+2)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);

Table[120 (3n)!/(n!(n+2)!^2),{n,0,20}] (* Harvey P. Dale, Jul 02 2023 *)

a(n) = 120/((n+1)*(n+2))^2 * (3*n)!/n!^3.
a(n) = (1/3)*(1*2*4*5) * A006480(n+2)/((3*n + 1)*(3*n + 2)*(3*n + 4)*(3*n +
5)), where A006480(n) = (3*n)!/n!^3.
a(n) = (10*binomial(3*n,n) - 7*binomial(3*n,n+1) + binomial(3*n,n+2)) * (3*binomial(2*n,n) - 4*binomial(2*n,n+1) + binomial(2*n,n+2)), shows that a(n) is an integer for all n.
a(n) = (1/3)*27^(n+2)*binomial(4/3, n+2)*binomial(5/3, n+2).
a(n) ~ sqrt(3)*60*27^n/(Pi*n^5).
P-recursive: (n + 2)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 30.
The o.g.f. A(x) satisfies the  differential equation x^2*(1 - 27*x)*A''(x) + x*(5 - 54*x)*A'(x) + (4 - 6*x)*A(x) - 120 = 0, with A(0) = 30 and A'(0) = 20.

A001459 a(n) = (5n)!/((2n)!(2n)!*n!).

Original entry on oeis.org

1, 30, 3150, 420420, 62355150, 9816086280, 1605660228900, 269764879032000, 46225898052627150, 8042050347997165500, 1415997888807961859400, 251762943910387780962000, 45125969443194371927422500, 8143514687130622653091029120, 1478138194032735032800001630400

Offset: 0

N. J. A. Sloane

nonn

James Spahlinger, 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 (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Cf. A006480.

Maple
```
f := n->(5*n)!/((2*n)!*(2*n)!*n!);
```

Table[(5 n)!/((2 n)! (2 n)!*n!), {n, 0, 12}] (* Michael De Vlieger, Feb 07 2017 *)

From Ilya Gutkovskiy, Feb 07 2017: (Start)
O.g.f.: 4F3(1/5,2/5,3/5,4/5; 1/2,1/2,1; 3125*x/16).
E.g.f.: 4F4(1/5,2/5,3/5,4/5; 1/2,1/2,1,1; 3125*x/16).
a(n) ~ 5^(5*n+1/2)/(4*Pi*16^n*n). (End)
From Peter Bala, Sep 20 2021: (Start)
a(n) = 5*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)/(4*n^2*(2*n - 1)^2)*a(n-1).
a(n) = Sum_{k = n..3*n} binomial(3*n,k)^2*binomial(k,n). Cf. A006480.
Congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integers n and k (write a(n) as C(5*n,2*n)*C(3*n,n) and apply Mestrovic, equation 39, p. 12). (End)

A061401 From solution to a Picard-Fuchs equation.

Original entry on oeis.org

1, 6, 9, 56, -300, 3942, -48412, 639264, -8785098, 124733000, -1817441901, 27047510136, -409669978497, 6297861697848, -98055605511675, 1543575781461888, -24533312413314948, 393242952308487588, -6350814254230701986

Offset: 1

N. J. A. Sloane, Jun 10 2001

sign

M. Aganagic, A. Klemm and C. Vafa, Disk Instantons, Mirror Symmetry and the Duality Web, Equation 6.15, p. 44.
Sean A. Irvine, Java program (github)

Cf. A006480, A229451, A229452.

InverseSeries[z/Exp[6 z HypergeometricPFQ[{1, 1, 4/3, 5/3}, {2, 2, 2}, -27 z]] + O[z]^20, q] // CoefficientList[#, q]& // Rest (* Jean-François Alcover, Feb 18 2019 *)

From Peter Bala, Oct 20 2024: (Start)
The g.f. A(x) = x + 6*x^2 + 9*x^3 + 56*x^4 - 300*x^5 + ... = x*series_reversion(B(x)), where B(x) = exp( Sum_{n >= 1} (-1)^n*(3*n)!/n!^3*x^n/n ). See A229451.
[x*n] (x/A(x))^n  = (-1)^n * (3*n)!/n!^3 = (-1)^n * A006480(n).
The power series F(x) := (A(x)/x)^(1/6) = 1 + x - x^2 + 11*x^3 - 100*x^4 + 1101*x^5 - 13273*x^6 + 170860*x^7 - 2306884*x^8 + 32300950*x^9 - 465426461*x^10 + ... appears to have integer coefficients.
Conjecture: Let r be an integer and s a positive integer. The sequence defined by u(n) = [x^(s*n)] F(x)^(r*n) satisfies the supercongruence u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. (End)

More terms from Jean-François Alcover, Feb 18 2019

cp's OEIS Frontend

A361030 a(n) = 20160*(3*n)!/(n!*(n+3)!^2).

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A361717 a(n) = Sum_{k = 0..n-1} binomial(n-1,k)^2*binomial(n+k,k).

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A217800 Number of alternating permutations on 2n+1 letters that avoid a certain pattern of length 4 (see Lewis, 2012, Appendix, for precise definition).

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A227357 de Bruijn's S(6,n).

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A324467 Three times the binary weight of n: 3*A000120(n).

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A352652 a(n) = ( binomial(7*n,2*n)*binomial(7*n/2,2*n)*binomial(2*n,n)^2 ) / binomial(7*n/2,n)^2.

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A361028 a(n) = 2*(3*n)!/(n!*(n+1)!^2).

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A361030 a(n) = 20160(3n)!/(n!*(n+3)!^2).

A352652 a(n) = ( binomial(7n,2n)binomial(7n/2,2n)binomial(2n,n)^2 ) / binomial(7n/2,n)^2.

A361028 a(n) = 2(3n)!/(n!*(n+1)!^2).

A361029 a(n) = 120(3n)!/(n!*(n+2)!^2).

A001459 a(n) = (5n)!/((2n)!(2n)!*n!).