A006480 -id:A006480 - cp's OEIS frontend

A229452 G.f.: exp( Sum_{n>=1} (3n)!/(3!n!^3) * x^n/n ).

1, 1, 8, 101, 1569, 27445, 518407, 10333243, 214320244, 4583132411, 100411556533, 2243625355010, 50955869372055, 1173262656151429, 27332509319090516, 643208905017756216, 15270427859720369204, 365356267775348553277, 8801688936499808334602

Offset: 0

Paul D. Hanna, Sep 23 2013

nonn

Self-convolution 6th power yields A229451.
From Peter Bala, Feb 16 2020: (Start)
The sequence defined by b(n) = [x^n] A(x)^n for n >= 1 begins [1, 17, 352, 7969, 189876, 4676768, 117905565, 3024222753, 78607893934, 2064924478892, 54710782664836, ...]. We conjecture that b(n) satisfies the supercongruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and all positive integers n and k [added 20 Oct 2024: more generally, for r a positive integer and s an integer we conjecture that the sequence {b(r,s;n) : n >= 1} defined by b(r,s; n) = [x^(r*n)] A(x)^(s*n) satisfies the same supercongruences].
More generally, for a positive integer m, set A_m(x) = exp( Sum_{n >= 1} (m*n)!/(m!*n!^m) * x^n/n ) and define a sequence b_m(n) := [x^n] A_m(x)^n for n >=1. Then we conjecture that b_m(n) is an integer sequence satisfying the same congruences. (End)
In general, for m >= 1, if g.f. = exp(m * Sum_{n>=1} (3*n)!/(3!*n!^3) * x^n/n), then a(n) ~ m * 2^(2*m-2) * 3^((m-1)/2) * Pi^(m-1) * A370293^m * 3^(3*n) / n^2, cf. A370289 (m=2), A370288 (m=3), A229451 (m=6). - Vaclav Kotesovec, Feb 14 2024

			G.f.: A(x) = 1 + x + 8*x^2 + 101*x^3 + 1569*x^4 + 27445*x^5 + ...,
where
log(A(x)) = x + 15*x^2/2 + 280*x^3/3 + 5775*x^4/4 + 126126*x^5/5 + 2858856*x^6/6 + ... + A060542(n)*x^n/n + ....

Vaclav Kotesovec, Table of n, a(n) for n = 0..700

Cf. A229451, A060542, A006480 (de Bruijn's S(3,n)), A061401, A370288, A370289, A370293, A370294, A370295.

CoefficientList[Series[Exp[Sum[(3*k)!/(3!*k!^3)*x^k/k, {k, 1, 20}]], {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 05 2020 *)
CoefficientList[Series[Exp[x*HypergeometricPFQ[{1, 1, 4/3, 5/3}, {2, 2, 2}, 27*x]], {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 11 2024 *)

{a(n)=polcoeff(exp(sum(k=1,n,(3*k)!/(3!*k!^3)*x^k/k) +x*O(x^n)),n)}
for(n=0,25,print1(a(n),", "))

a(n) ~ c * 3^(3*n) / n^2, where c = A370293 = 0.0490152812... - Vaclav Kotesovec, Feb 14 2024

a(n) = 1/n * Sum_{k = 0..n-1} b(n-k)*a(k) with a(0) = 1, where b(n) = (3*n)!/(6*n!^3). - Peter Bala, Oct 22 2024

A245086 Central values of the n-th discrete Chebyshev polynomials of order 2n.

Original entry on oeis.org

1, 0, -6, 0, 90, 0, -1680, 0, 34650, 0, -756756, 0, 17153136, 0, -399072960, 0, 9465511770, 0, -227873431500, 0, 5550996791340, 0, -136526995463040, 0, 3384731762521200, 0, -84478098072866400, 0, 2120572665910728000, 0, -53494979785374631680, 0

Offset: 0

Nikita Gogin, Jul 11 2014

In the general case the n-th discrete Chebyshev polynomial of order N is D(N,n;x) = Sum_{i = 0..n} (-1)^i*C(n,i)*C(N-x,n-i)*C(x,i). For N = 2*n , x = n, one gets a(n) = D(2n,n;n) = Sum_{i = 0..n} (-1)^i*C(n,i)^3 that equals (due to Dixon's formula) 0 for odd n and (-1)^m*(3m)!/(m!)^3 for n = 2*m. (Riordan, 1968) So, a(2*m) = (-1)^m*A006480(m).

John Riordan, Combinatorial Identities, John Willey&Sons Inc., 1968.

Nikita Gogin and Mika Hirvensalo, On the Generating Function of Discrete Chebyshev Polynomials, TUCS Technical Reports 819, Turku Centre for Computer Science, 2007.
Nikita Gogin and Mika Hirvensalo, Recurrent Constructions of the MacWilliams and Chebyshev Matrices, Fundamenta Informaticae, vol.116, issue 1-4, January 2012, pages 93-110.
H. W. Gould, Tables of Combinatorial Identities, Edited by J. Quaintance.
Darij Grinberg, Introduction to Modern Algebra (UMN Spring 2019 Math 4281 Notes), University of Minnesota (2019).
Y. Sun and X. Wang, A new proof of a curious identity, Math Gazette, Vol. 91, No. 520 (Mar 2007) 105-107.
Wikipedia, Dixon's identity

Cf. A000172, A002894, A005260, A006480, A103882, A183204, A273630, A364303.

Table[Coefficient[Simplify[JacobiP[n,0,-(2*n+1),(1+t^2)/(1-t^2)]*(1-t^2)^n],t,n],{n,0,20}]

from math import factorial
def A245086(n): return 0 if n&1 else (-1 if (m:=n>>1)&1 else 1)*factorial(3*m)//factorial(m)**3 # Chai Wah Wu, Oct 04 2022

a(n) is a coefficient at t^n in (1-t^2)^n*P(0,-(2*n+1);n;(1+t^2)/(1-t^2)), where P(a,b;k;x) is the k-th Jacobi polynomial (Gogin and Hirvensalo, 2007).
G.f.: Hypergeometric2F1[1/3,2/3,1,-27*x^2].
a(2*m+1) = 0, a(2*m) = (-1)^m*A006480(m).
From Peter Bala, Aug 04 2016: (Start)
a(n) = Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(2*n - k,n)*binomial(n + k,n) (Sun and Wang).
a(n) = Sum_{k = 0..n} (-1)^(n + k)*binomial(n + k, n - k)*binomial(2*k, k)*binomial(2*n - k, n) (Gould, Vol.5, 9.23).
a(n) = -1/(n + 1)^3 * A273630(n+1). (End)
From Peter Bala, Mar 22 2022: (Start)
a(n) = - (3*(3*n-2)*(3*n-4)/n^2)*a(n-2).
a(n) = [x^n] (1 - x)^(2*n) * P(n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A002894(n) = binomial(2*n,n)^2 = [x^n] (1 - x)^(2*n) * P(2*n,(1 + x)/(1 - x)). Cf. A103882. (End)
From Peter Bala, Jul 23 2023: (Start)
a(n) = [x^n] G(x)^(3*n), where the power series G(x) = 1 - x^2 + 2*x^4 - 14*x^6 + 127*x^8 - 1364*x^10 + ... appears to have integer coefficients.
exp(Sum_{n >= 1} a(n)*x^n/n) = F(x)^3, where the power series F(x) = 1 - x^2  + 8*x^4 - 101*x^6 + 1569*x^8 - 27445*x^10 + ..., appears to have integer coefficients. See A229452.
Row 1 of A364303. (End)
a(n) = Sum_{k = 0..n} (-1)^(n-k) * binomial(n+k, k)^2 * binomial(3*n+1, n-k). Cf. A183204.- Peter Bala, Sep 20 2024

A362733 a(n) = [x^n] F(x)^n, where F(x) = exp( Sum_{k >= 1} A362732(k)*x^k/k ).

Original entry on oeis.org

1, 6, 234, 10428, 492522, 24033006, 1197423396, 60530725380, 3092592004074, 159295600885794, 8258018380659234, 430335300869496072, 22521831447746893092, 1182951246247357578348, 62325193477833011143260, 3292376206935392483917428, 174323297281680647978503146, 9248680725006429075147528150

Offset: 0

Peter Bala, May 06 2023

Cf. A006480, A229451, A362722 - A362732.

E(n,x) := series(exp(n*add((3*k)!/k!^3*x^k/k, k = 1..20)), x, 21):
A362732(n) := coeftayl(E(n,x), x = 0, n):
F(n,x) := series(exp(n*add(A362732(k)*x^k/k, k = 1..20)), x, 21):
seq(coeftayl(F(n,x), x = 0, n), n = 0..20);
# alternative program
G(n,x) := series(exp(n*add((3*k)!/k!^3*x^(2*k)/k, k = 1..40)), x, 41):
seq((1/2)*coeftayl(G(2*n,x), x = 0, 2*n), n = 1..20); # Peter Bala, Oct 27 2024

a(n*p^r) == a(n*p^(r-1)) (mod p^r) (Gauss congruence) holds for all primes p and positive integers n and r.
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 3 and positive integers n and r.
For n >= 1, a(n) = (1/2) * [x^n] G(x)^(2*n), where G(x) = exp(Sum_{k >= 1} A006480(k)*x^k/k) is the g.f. of A229451. - Peter Bala, Oct 27 2024

A181544 Triangle in which the g.f. for row n is [Sum_{k>=0} C(n+k-1,k)^3x^k](1-x)^(3n+1), read by rows of k=0..2n terms.

Original entry on oeis.org

1, 1, 4, 1, 1, 20, 48, 20, 1, 1, 54, 405, 760, 405, 54, 1, 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1, 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1, 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1, 1, 490, 35623, 818300, 7917371, 37215794, 91789005, 123519792, 91789005, 37215794, 7917371, 818300, 35623, 490, 1, 1, 704, 73200, 2430400, 34657700, 246781248, 955910032, 2116980800, 2751843600, 2116980800, 955910032, 246781248, 34657700, 2430400, 73200, 704, 1

Offset: 0

Paul D. Hanna, Oct 30 2010

			Triangle begins:
 1;
 1, 4, 1;
 1, 20, 48, 20, 1;
 1, 54, 405, 760, 405, 54, 1;
 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1;
 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1;
 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1; ...
Row g.f.s begin:
 (1) = (1-x)*(1 + x + x^2 + x^3 + x^4 +...);
 (1 + 4*x + x^2) = (1-x)^4*(1 + 2^3*x + 3^3*x^2 + 4^3*x^3 +...);
 (1 + 20*x + 48*x^2 + 20*x^3 + x^4) = (1-x)^7*(1 + 3^3*x + 6^3*x^2 +...);
 (1 + 54*x + 405*x^2 + 760*x^3 + 405*x^4 + 54*x^5 + x^6) = (1-x)^10*(1 + 4^3*x + 10^3*x^2 + 20^3*x^3 + 35^3*x^4 +...); ...

Paul D. Hanna, Table of n, a(n) for n = 0..1088, as a flattened triangle of rows 0..32
Ilia Gaiur, Vladimir Rubtsov, and Duco van Straten, Product formulas for the Higher Bessel functions, arXiv:2405.03015 [math.AG], 2024. See p. 18.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 14.

Cf. A181543, A181545, A006480, A002897, A000172.

Cf. A183204 (central terms), A183205.

t[n_, k_] := SeriesCoefficient[Sum[Binomial[n+j, j]^3*x^j, {j, 0, n+k}]*(1-x)^(3*n+1), {x,0, k}]; Table[t[n, k], {n, 0, 9}, {k, 0, 2*n}] // Flatten (* Jean-François Alcover, Feb 04 2014, after PARI *)

{T(n,k)=polcoeff(sum(j=0,n+k,binomial(n+j,j)^3*x^j)*(1-x)^(3*n+1),k)}
for(n=0,10,for(k=0,2*n,print1(T(n,k),", "));print(""))

Row sums equal A006480(n) = (3n)!/(n!)^3, which is de Bruijn's s(3,n).
From Yahia Kahloune, Jan 30 2014: (Start)
Using these coefficients we can obtain formulas for the sums
Sum_{i=1..n} C(e-1+i,e)^3. Let us define b(k,e,3) = sum_{i=0..k-e} (-1)^i*C(3*e+1,i)*C(k-i,e)^3, where k=e+i.
For example:
b(e,e,3) = 1;
b(e+1,e,3) = (e+1)^3-(3*e+1) = e^2*(e+3);
b(e+2,e,3) = C(e+2,2)^3 - (3*e+1)*(e+1)^3 + C(3*e+1,2);
b(e+3,e,3) = C(e+3,e)^3 - (3*e+1)*C(e+2,e)^3 + C(3*e+1,2)*C(e+1,e)^3 - C(3*e+1,3);
b(e+4,e,3) = C(e+4,e)^3 - (3*e+1)*C(e+3,e)^3  + C(3*e+1,2)*C(e+2,e) - C(3*e+1,3)*C(e+1,e)^3 + C(3*e+1,4).
Then we have the formula: Sum_{i=1..n} C(e-1+i,e)^3 = Sum_{i=0..2*e} b(e+i,e,3)*C(n+e+i,3*e+1).
Example: Sum_{i=1..7} C(2+i,3)^3 = C(10,10) + 54*C(11,10) + 405*C(12,10) + 760*C(13,10) + 405*C(14,10) + 54*C(15,10) + C(16,10) = 820260. (End)
Let E be the operator D*x*D*x*D, where D denotes the derivative operator d/dx. Then (1/(n)!^3)  * E^n(1/(1 - x)) = (row n generating polynomial)/(1 - x)^(3*n+1) = Sum_{k >= 0} binomial(n+k, k)^3*x^k. For example, when n = 2 we have (1/2!)^3*E^3(1/(1 - x)) = (1 + 20 x + 48 x^2 + 20 x^3 + x^4)/(1 - x)^7. - Sergii Voloshyn, Dec 03 2024

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

A050983 de Bruijn's S(4,n).

Original entry on oeis.org

1, 14, 786, 61340, 5562130, 549676764, 57440496036, 6242164112184, 698300344311570, 79881547652046140, 9301427008157320036, 1098786921802152516024, 131361675994216221116836, 15863471168011822803270200, 1932252897656224864335299400

Offset: 0

Eric W. Weisstein

a(n) is divisible by (n+1). Prime p divides a(p-1). Prime p>2 divides all a(n) from a((p+1)/2) to a(p-1). - Alexander Adamchuk, Jul 05 2006

N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.

Seiichi Manyama, Table of n, a(n) for n = 0..471
T. Amdeberhan, De Bruijn's sequence is odd iff n = 2m - 1: Part I, MathOverflow 383035.
Eric Weisstein's World of Mathematics, Binomial Sums

Cf. A000984, A006480, A050984, A099601, A177322, A227357, A249332.

Sum[ (-1)^(k+n)Binomial[ 2n, k ]^4, {k, 0, 2n} ]
RecurrenceTable[{a[0] == 1, a[1] == 14, 4 (n + 1) (2 n + 1)^3 (48 n^2 + 162 n + 137) a[n] + (n + 2)^3 (2 n + 3) (48 n^2 + 66 n + 23) a[n + 2] == 2 (4 (n + 1)^2 (2 n + 3)^2 (408 n^2 + 969 n + 431) - (n + 1) (2 n + 3) (69 n + 31) + 57 n + 92) a[n + 1]}, a[n], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 26 2016 *)

a(n)=sum(k=0,2*n,(-1)^(k+n)*binomial(2*n,k)^4) \\ Charles R Greathouse IV, Dec 28 2011

a(n) = Sum_{k=-n..+n} (-1)^k*C(2*n,n+k)^4. - Benoit Cloitre, Mar 02 2005
a(n) = (-1)^n * HypergeometricPFQ[ {-2n, -2n, -2n, -2n}, {1, 1, 1}, -1]. - Alexander Adamchuk, Jul 05 2006
E.g.f.: Sum(n>=0,I^n*x^n/n!^4) * Sum(n>=0,(-I)^n*x^n/n!^4) = Sum(n>=0,a(n)*x^(2*n)/n!^4) where I^2=-1. - Paul D. Hanna, Dec 21 2011
a(n) ~ 0.125 k^(8n+3)/(Pi*n)^(3/2) where k = 2 cos(Pi/8) = A179260. This formula is due to de Bruijn 1958. - Charles R Greathouse IV, Dec 28 2011
Recurrence: a(0) = 1, a(1) = 14, 4 * (n + 1) * (2*n + 1)^3 * (48*n^2 + 162*n + 137) * a(n) + (n + 2)^3 * (2*n + 3) * (48*n^2 + 66*n + 23) * a(n+2) = 2 * (4 * (n + 1)^2 * (2*n + 3)^2 * (408*n^2 + 969*n + 431) - (n + 1) * (2*n + 3) * (69*n + 31) + 57*n + 92) * a(n+1). - Vladimir Reshetnikov, Sep 26 2016
From Peter Bala, Nov 02 2024; (Start)
a(n) = 1/n * Sum_{k = 0..2*n} (-1)^(n+k) * k * binomial(2*n, k)^4 for n >= 1.
a(n) = binomial(2*n, n) * Sum_{k = 0..n} binomial(2*n, n+k)^2 * binomial(2*n+k,k) = binomial(2*n, n) * Sum_{k = 0..n} (-1)^(n+k) * binomial(2*n, n+k) * binomial(2*n+k, k)^2. (End)

A060542 a(n) = (1/6)multinomial(3n;n,n,n).

Original entry on oeis.org

1, 15, 280, 5775, 126126, 2858856, 66512160, 1577585295, 37978905250, 925166131890, 22754499243840, 564121960420200, 14079683012144400, 353428777651788000, 8915829964229105280, 225890910734335847055, 5744976449471863238250, 146603287914300510042750

Offset: 1

Henry Bottomley, Apr 02 2001

Number of ways of dividing 3n labeled items into 3 unlabeled boxes with n items in each box.
From Antonio Campello (campello(AT)ime.unicamp.br), Nov 11 2009: (Start)
A060542(t) is the number of optimal [n,2,d] binary codes that correct at most t errors, i.e., having Hamming distance 2*t + 1 (achieved on length n = 3*t + 2). These codes are all isometric.
It is also the number of optimal [n,2,d] binary codes that detect 2*t + 1 errors, i.e., having Hamming distance 2t+2 (obtained by adding an overall parity check to the n = 3*t + 2 optimal codes). These codes are also all isometric.
For t = 0, we have the famous MDS, cyclic, simplex code {(000), (101), (110), (011)}. (End)
Also the number of distinct adjacency matrices of the complete tripartite graph K_{n,n,n}. - Eric W. Weisstein, Apr 21 2017

Harry J. Smith, Table of n, a(n) for n = 1..100
Eric Weisstein's World of Mathematics, Adjacency Matrix
Eric Weisstein's World of Mathematics, Complete Tripartite Graph

Row 3 of A060540.

Cf. A006480, A025035, A361717.

a:= n-> combinat[multinomial](3*n,n$3)/3!:
seq(a(n), n=1..18);  # Alois P. Heinz, Jul 29 2023

Table[(3*n)!/(n!^3*6),{n,1,20}] (* Vaclav Kotesovec, Sep 23 2013 *)
Table[Multinomial[n, n, n], {n, 20}]/6 (* Eric W. Weisstein, Apr 21 2017 *)

{ a=1/6; for (n=1, 100, write("b060542.txt", n, " ", a=a*3*(3*n - 1)*(3*n - 2)/n^2); ) } \\ Harry J. Smith, Jul 06 2009

a(n) = (3*n)!/((n!)^3*6) = a(n-1)*3*(3*n - 1)*(3*n - 2)/n^2 = A060540(3,n) = A006480(n)/6. - corrected by Vaclav Kotesovec, Sep 23 2013
a(n) ~ 3^(3*n-1/2)/(4*Pi*n). - Vaclav Kotesovec, Sep 23 2013
a(n) = 1/(8*n^3) * Sum_{k = 0..2*n} (-1)^(n+k) * k*(2*n-k)^2 * binomial(2*n, k)^3. - Peter Bala, Oct 11 2024
a(n) = 1/(n^2) * Sum_{k = 0..n} (-1)^(n+k+1) * (n-k)^2 * binomial(2*n, k)^3. - Peter Bala, Nov 03 2024

Definition revised by N. J. A. Sloane, Feb 02 2009

A064350 a(n) = (3*n)!/n!.

Original entry on oeis.org

1, 6, 360, 60480, 19958400, 10897286400, 8892185702400, 10137091700736000, 15388105201717248000, 30006805143348633600000, 73096577329197271449600000, 217535414131691079834009600000, 776601428450137155007414272000000, 3275704825202678519821273399296000000

Offset: 0

Karol A. Penson, Sep 18 2001

Also a(n) = (((n)!)^2)*A006480(n). [corrected by Johannes W. Meijer, Mar 02 2009]

a(n) is the number of ways to partition the set {1,2,...,3n} into n blocks of size 3 and then linearly order the elements within each block. - Geoffrey Critzer, Dec 30 2012

Harry J. Smith, Table of n, a(n) for n = 0..70
Karol A. Penson and Allan I. Solomon, Coherent states from combinatorial sequences, in: E. Kapuscik and A. Horzela (eds.), Quantum theory and symmetries, World Scientific, 2002, pp. 527-530; arXiv preprint, arXiv:quant-ph/0111151, 2001.

Cf. A006480, A248759.
From Johannes W. Meijer, Mar 07 2009: (Start)
Equals A001525*3!
Equals row sums of A157704 and A157705. (End)

Table[(3n)!/n!,{n,0,20}]  (* Geoffrey Critzer, Dec 30 2012 *)

{ t=f=1; for (n=0, 70, if (n, t*=3*n*(3*n - 1)*(3*n - 2); f*=n); write("b064350.txt", n, " ", t/f) ) } \\ Harry J. Smith, Sep 12 2009

Integral representation as n-th moment of a positive function on a positive half-axis: a(n) = Integral_{x=0..oo} x^n*BesselK(1/3, 2*sqrt(x/27))/(3*Pi*sqrt(x)) dx, n >= 0.
A recursive formula: a(n) = (27 * (n - 1)^2 + 27 * (n - 1) + 6) * a(n - 1) with a(0) = 1. An explicit formula following from the recursion equation: a(n) = (3/2)*27^n*GAMMA(n+2/3)*GAMMA(n+1/3)/(Pi*3^(1/2)). - Thomas Wieder, Nov 15 2004
E.g.f.: (of aerated sequence) 2*cos(arcsin((3*sqrt(3)*x/2)/3))/sqrt(4-27*x^2). - Paul Barry, Jul 27 2010
E.g.f.: (with interpolated zeros) exp(x^3). - Geoffrey Critzer, Dec 30 2012
Sum_{n>=1} 1/a(n) = A248759. - Amiram Eldar, Nov 15 2020

a(11) from Harry J. Smith, Sep 12 2009

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

A001451 a(n) = (5n)!/((3n)!n!n!).

Original entry on oeis.org

1, 20, 1260, 100100, 8817900, 823727520, 79919739900, 7962100660800, 808906548235500, 83426304143982800, 8707404737345073760, 917663774856743842200, 97491279924241456098300, 10427604345391237790688000, 1121786259855036145008408000

Offset: 0

N. J. A. Sloane

From Peter Bala, Jul 15 2016: (Start)
This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = (-1)^m*a(m) for n = 2*m. The sum vanishes for n odd. Cf. A273628 and A273629.
Note the similar results:
Sum_{k = 0..n} (-1)^k*binomial(n,k)* binomial(2*n + k,n)*binomial(3*n + k,n) = (-1)^n*(3*n)!/n!^3 = (-1)^n*A006480(n);
Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n)^2 = A002894(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n) = A000984(n);
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n). (End)
Choose three noncollinear step vectors to satisfy the zero sum, 3*v_1 + v_2 + v_3 = 0. Then a(n) is the number of loop plane walks of length 5*n which depart from and return to the origin. Equivalently, a(n) counts distinct permutations of a (5*n)-digit integer with digits 1,2,3 of multiplicity 3*n,n,n respectively. - Bradley Klee, Aug 12 2018

			G.f. = 1 + 20*x + 1260*x^2 + 100100*x^3 + 8817900*x^4 + 823727520*x^5 + ... - _Michael Somos_, Aug 12 2018

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

Cf. A000897, A002894, A002897, A006480, A008977, A186420, A188662, A273628, A273629.

List([0..15],n->Factorial(5*n)/(Factorial(3*n)*Fact0rial(n)*Factorial(n))); # Muniru A Asiru, Aug 12 2018

[Factorial(5*n)/(Factorial(3*n)*Factorial(n)*Factorial(n)): n in [0..30]]; // Vincenzo Librandi, May 22 2011

Maple
```
f := n->(5*n)!/((3*n)!*n!*n!);
```

Table[(5*n)!/((3*n)!*n!*n!), {n, 0, 20}] (* Vincenzo Librandi, Sep 04 2012 *)

a(n) = binomial(4*n,n)*binomial(5*n,n) = ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) = [x^n](F(x)^(20*n)), where F(x) = 1 + x + 12*x^2 + 390*x^3 + 16984*x^4 + 867042*x^5 + 48848541*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A186420 and A188662. - Peter Bala, Jul 14 2016
a(n) ~ 3^(-3*n-1/2)*5^(5*n+1/2)/(2*Pi*n). - Ilya Gutkovskiy, Jul 13 2016
G.f.: G(x) = 4F3(1/5,2/5,3/5,4/5;1/3,2/3,1;(5^5/3^3)*x). Let G^(n)(x) = d^n/dx^n G(x), and c = {120, 15000*x-6, 45000*x^2-114*x, 25000*x^3-135*x^2, 3125*x^4-27*x^3}, then Sum_{n=0..4} c_n*G^(n)(x) = 0. - Bradley Klee, Aug 12 2018
From Peter Bala, Mar 20 2022: (Start)
Right-hand side of the following identities valid for n >= 1:
Sum_{k = 0..3*n} 2*n*(2*n+k-1)!/(k!*n!^2) = (5*n)!/((3*n)!*n!^2);
Sum_{k = 0..n} 4*n*(4*n+k-1)!/(k!*n!*(3*n)!) = (5*n)!/((3*n)!*n!^2). Cf. A000897 and A113424. (End)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(4*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 12 2024

cp's OEIS Frontend

A229452 G.f.: exp( Sum_{n>=1} (3*n)!/(3!*n!^3) * x^n/n ).

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A245086 Central values of the n-th discrete Chebyshev polynomials of order 2n.

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A362733 a(n) = [x^n] F(x)^n, where F(x) = exp( Sum_{k >= 1} A362732(k)*x^k/k ).

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A181544 Triangle in which the g.f. for row n is [Sum_{k>=0} C(n+k-1,k)^3*x^k]*(1-x)^(3n+1), read by rows of k=0..2n terms.

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

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A050983 de Bruijn's S(4,n).

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A060542 a(n) = (1/6)*multinomial(3*n;n,n,n).

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A229452 G.f.: exp( Sum_{n>=1} (3n)!/(3!n!^3) * x^n/n ).

A181544 Triangle in which the g.f. for row n is [Sum_{k>=0} C(n+k-1,k)^3x^k](1-x)^(3n+1), read by rows of k=0..2n terms.

A060542 a(n) = (1/6)multinomial(3n;n,n,n).

A001451 a(n) = (5n)!/((3n)!n!n!).