A275047 -id:A275047 - cp's OEIS frontend

A184423 a(n) = (2n)!(3*n)!/n!^5.

1, 12, 540, 33600, 2425500, 190702512, 15849497664, 1369618398720, 121821136479900, 11079206239530000, 1025579963180813040, 96310511463483233280, 9152842704012278107200, 878622906816654279840000

Offset: 0

Paul D. Hanna, Jan 13 2011

Denoted by h_3[n] by T. Piezas III. He also gives formulas for 1/Pi such as 1/Pi = 2 * Sum_{n>=0} a(n) * (-1)^n * (51*n + 7) / (12^3)^(n + 1/2). - Michael Somos, May 31 2012
Diagonal of the rational function R(x,y,z,w) = 1/(1-(w*y+w*z+x+y+z)). - Gheorghe Coserea, Jul 15 2016
From Peter Bala, Jun 28 2023: (Start)
The supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for primes p >= 5 and positive integers n and r. This follows from Meštrović equation 39, since a(n) = binomial(3*n,n) * binomial(2*n,n)^2.
Inductively define a family of sequences {a(i,n) : n >= 0}, i >= 1, by setting a(1,n) = a(n) and, for i >= 2, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. We conjecture that the sequences {a(i,n) : n >= 0}, i >= 2, also satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for primes p >= 5 and positive integers n and r. Cf. A362730 and A362732. (End)

			G.f.: A(x) = 1 + 12*x + 540*x^2 + 33600*x^3 + 2425500*x^4 +...
G.f. of A184424 equals A(x)^(1/2):
A(x)^(1/2) = 1 + 6*x + 252*x^2 + 15288*x^3 + 1089270*x^4 + 84963060*x^5 +...+ [(3^n/n!^2)*Product_{k=1..n} (6*k-4)*(6*k-5)]*x^n +...

Gheorghe Coserea, Table of n, a(n) for n = 0..200
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.
Titus Piezas III, 0013: Article 3 (Pi Formulas and the Monster Group).

Cf. A184424, A275047.

Related to diagonal of rational functions: A268545-A268555.

Table[((2n)!(3n)!)/(n!)^5,{n,0,20}] (* Harvey P. Dale, Dec 18 2018 *)

PARI
```
{a(n)=(3*n)!*(2*n)!/n!^5}
```

{a(n)=polcoeff(sum(m=0,n,3^m*prod(k=1,m,(6*k-4)*(6*k-5))/m!^2*x^m+x*O(x^n))^2,n)}

Self-convolution of A184424:
a(n) = Sum_{k=0..n} A184424(k)*A184424(n-k) where A184424(n) = (3^n/n!^2)*Product_{k=1..n} (6*k-4)*(6*k-5).
a(n) = 6 * (2*n - 1) * (3*n - 1) * (3*n - 2) / n^3 * a(n-1) if n>0. - Michael Somos, May 31 2012
0 = (x^2-108*x^3)*y''' + (3*x-486*x^2)*y''+ (1-348*x)*y' - 12*y, where y is g.f. - Gheorghe Coserea, Jul 15 2016
a(n) ~ 3^(1/2)/(2*Pi^(3/2)) * n^(-3/2) * 108^n. - Ilya Gutkovskiy, Jul 15 2016
a(n) = C(2*n,n)^2 * C(3*n,n) = ( [x^n](1 + x)^(2*n) )^2 * ( [x^n](1 + x)^(3*n) ) = [x^n]( F(x)^(12*n) ), where [x^n] is the coefficient extraction operator and where F(x) = 1 + x + 11*x^2 + 350*x^3  + 15293*x^4  + 794433*x^5  + 45958617*x^6 + ... appears to have integral coefficients. Cf. A000897 and A001451. - Peter Bala, Dec 30 2019

A361677 Constant term in the expansion of (1 + x + y + z + 1/(xy) + 1/(yz) + 1/(z*x))^n.

Original entry on oeis.org

1, 1, 1, 19, 73, 181, 1711, 10081, 38809, 256033, 1696861, 8388271, 49449511, 326195299, 1847392093, 10789655059, 69202030969, 418647580489, 2498113460881, 15735859252147, 97919649290053, 598317173139313, 3748943081117323

Offset: 0

Seiichi Manyama, Mar 20 2023

nonn

Cf. A201805, A361678.

Cf. A275047, A344560.

Table[Sum[(3*k)!/k!^3 * Binomial[3*k,k] * Binomial[n,3*k], {k,0,n/3}], {n,0,25}] (* Vaclav Kotesovec, Mar 22 2023 *)

a(n) = sum(k=0, n\3, (3*k)!/k!^3*binomial(3*k, k)*binomial(n, 3*k));

a(n) = Sum_{k=0..floor(n/3)} (3*k)!/k!^3 * binomial(3*k,k) * binomial(n,3*k).
From Vaclav Kotesovec, Mar 22 2023: (Start)
Recurrence: 2*n^3*(2*n - 3)*a(n) = 2*(10*n^4 - 32*n^3 + 38*n^2 - 22*n + 5)*a(n-1) - 2*(n-1)*(2*n - 3)*(10*n^2 - 24*n + 17)*a(n-2) + (n-2)*(n-1)*(769*n^2 - 2331*n + 1594)*a(n-3) - 2*(n-3)*(n-2)*(n-1)*(739*n - 1481)*a(n-4) + 733*(n-4)*(n-3)*(n-2)*(n-1)*a(n-5).
a(n) ~ sqrt(733/108 + 1/2^(2/3) + 9/2^(4/3)) * (1 + 9/2^(2/3))^n / (2 * Pi^(3/2) * n^(3/2)). (End)

A036914 a(n) = binomial(2n,n)binomial(3n,2n)^4.

Original entry on oeis.org

1, 162, 303750, 995742720, 4202607543750, 20493770553668412, 109738295483524291584, 627433021349790289920000, 3765656995768668039930646470, 23460102529588600192836492187500, 150552597141762184641565143623272500, 989711604190467147276644388444241920000

Offset: 0

N. J. A. Sloane

From Peter Bala, Aug 07 2016: (Start)
Compare with the identities:
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(3*n,k)^2*binomial(3*n - k,n)^2 = binomial(3n,n)^2*binomial(2*n,n) = A275047(n), and
Sum_{k = 0..2*n} (-1)^k*binomial(3*n,k)*binomial(3*n - k,n)^3 = binomial(3*n,n)*binomial(2*n,n) = (3*n)!/n!^3 = A006480(n). (Sprugnoli, Section 2.9, Table 10, p. 123).
Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)^2 = A000984(n). (End)

The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1972; Eq 21.1, page 72 (see the Formula section).

G. C. Greubel, Table of n, a(n) for n = 0..250
R. Sprugnoli, Riordan array proofs of identities in Gould's book

Cf. A000984, A005809, A006480, A275047.

[(n+1)*Binomial(3*n,2*n)^4*Catalan(n): n in [0..30]]; // G. C. Greubel, Jun 22 2022

seq((3*n)!^4/(n!^6*(2*n)!^3), n = 0..20); # Peter Bala, Aug 07 2016

Table[Binomial[2n, n]Binomial[3n, 2n]^4, {n,0,11}] (* Michael De Vlieger, Aug 07 2016 *)

b=binomial; [b(2*n,n)*b(3*n,2*n)^4 for n in (0..30)] # G. C. Greubel, Jun 22 2022

Sum_{k=0..2*n} (-1)^k*C(3*n, k)^3*C(3*n-k, n)^3 = (-1)^n*C(2*n, n)*C(3*n, 2*n)^4.
From Peter Bala, Aug 07 2016: (Start)
a(n) = (3*n)!^4/(n!^6*(2*n)!^3).
a(n) = A005809(n)^4 * A000984(n) = A005809(n)^3 * A006480(n) = A005809(n)^2 * A275047(n).
a(n) = {[x^n] (1 + x)^(3*n)}^4 * [x^n] (1 + x)^(2*n) = [x^n] G(x)^(162*n), where G(x) = 1 + x + 776*x^2 + 1633370*x^3 + 5060509158*x^4 + 19379170742458*x^5 + 84908023350007787*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^162, where F(x) = 1 + x + 938*x^2 + 2049791*x^3 + 6487994244*x^4 + 25309359070330*x^5 + 112932966264239483*x^6 + ... appears to have integer coefficients. (End)
a(n) ~ (9/16)*9^(6*n)/((Pi*n)^(5/2)*64^n). - Ilya Gutkovskiy, Aug 07 2016

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A184423 a(n) = (2n)!(3*n)!/n!^5.

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A361677 Constant term in the expansion of (1 + x + y + z + 1/(xy) + 1/(yz) + 1/(z*x))^n.

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A036914 a(n) = binomial(2n,n)binomial(3n,2n)^4.

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cp's OEIS Frontend

A184423 a(n) = (2*n)!*(3*n)!/n!^5.

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A361677 Constant term in the expansion of (1 + x + y + z + 1/(x*y) + 1/(y*z) + 1/(z*x))^n.

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A036914 a(n) = binomial(2*n,n)*binomial(3*n,2*n)^4.

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A184423 a(n) = (2n)!(3*n)!/n!^5.

A361677 Constant term in the expansion of (1 + x + y + z + 1/(xy) + 1/(yz) + 1/(z*x))^n.

A036914 a(n) = binomial(2n,n)binomial(3n,2n)^4.