A176285 -id:A176285 - cp's OEIS frontend

A002897 a(n) = binomial(2n,n)^3.

1, 8, 216, 8000, 343000, 16003008, 788889024, 40424237568, 2131746903000, 114933031928000, 6306605327953216, 351047164190381568, 19774031697705428416, 1125058699232216000000, 64561313052442296000000

Offset: 0

N. J. A. Sloane, Simon Plouffe

Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*z + x + y + z)). - Gheorghe Coserea, Jul 14 2016

Conjecture: The g.f. is also the diagonal of the rational function 1/(1 - (x + y)*(1 - 4*z*t) - z - t) = 1/det(I - M*diag(x, y, z, t)), I the 4 x 4 unit matrix and M the 4 x 4 matrix [1, 1, 1, 1; 1, 1, 1, 1; 1, 1, 1, -1; 1 , 1, -1, 1]. If true, then a(n) = [(x*y*z)^n] (1 + x + y + z)^(2*n)*(1 + x + y - z)^n*(1 + x - y + z)^n. - Peter Bala, Apr 10 2022

S. Ramanujan, Modular Equations and Approximations to pi, pp. 23-39 of Collected Papers of Srinivasa Ramanujan, Ed. G. H. Hardy et al., AMS Chelsea 2000. See page 36, equation (25).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..100
David H. Bailey, Jonathan M. Borwein, David Broadhurst and M. L. Glasser, Elliptic integral evaluations of Bessel moments, arXiv:0801.0891 [hep-th], 2008.
C. Domb, On the theory of cooperative phenomena in crystals, Advances in Phys., 9 (1960), 149-361.
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Yen Lee Loh, A general method for calculating lattice Green functions on the branch cut, arXiv:1706.03083 [math-ph], 2017.
Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.

Cf. A000897, A002894, A006480, A008977, A108625, A176285, A183204, A186420, A188662.

Related to diagonal of rational functions: A268545-A268555.

[Binomial(2*n, n)^3: n in [0..20]]; // Vincenzo Librandi, Nov 18 2011

a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/2, 1/2, 1/2}, {1, 1}, 64x], {x, 0, n}];
Table[Binomial[2n,n]^3,{n,0,20}] (* Harvey P. Dale, Dec 06 2017 *)

{a(n) = binomial(2*n, n)^3}; /* Michael Somos, Jan 31 2007 */

[binomial(2*n, n)**3 for n in range(21)] # Zerinvary Lajos, Apr 21 2009

Expansion of (K(k)/(Pi/2))^2 in powers of (kk'/4)^2, where K(k) is the complete elliptic integral of the first kind evaluated at modulus k. - Michael Somos, Jan 31 2007
G.f.: F(1/2, 1/2, 1/2; 1, 1; 64x) where F() is a hypergeometric function. - Michael Somos, Jan 31 2007
G.f.: hypergeom([1/4,1/4],[1],64*x)^2. - Mark van Hoeij, Nov 17 2011
D-finite with recurrence n^3*a(n) - 8*(2*n - 1)^3*a(n-1) = 0. - R. J. Mathar, Mar 08 2013
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)^3 = ( [x^n](1 + x)^(2*n) )^3 = [x^n](F(x)^(8*n)), where F(x) = 1 + x + 6*x^2 + 111*x^3 + 2806*x^4 + 84456*x^5 + 2832589*x^6 + 102290342*x^7 + ... appears to have integer coefficients. For similar results see A000897, A002894, A006480, A008977, A186420 and A188662. (End)
a(n) ~ 64^n/(Pi*n)^(3/2). - Ilya Gutkovskiy, Jul 13 2016
0 = (-x^2 + 64*x^3)*y''' + (-3*x + 288*x^2)*y'' + (-1 + 208*x)*y' + 8*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
a(n) = Sum_{k = 0..n} (2*n + k)!/(k!^3*(n - k)!^2). Cf. A001850(n) = Sum_{k = 0..n} (n + k)!/(k!^2*(n - k)!). - Peter Bala, Jul 27 2016
It appears that a(n) is the coefficient of (x*y*z)^(2*n) in the expansion of (1 + x*y + x*z - y*z)^(2*n) * (1 + x*y - x*z + y*z)^(2*n) * (1 - x*y + x*z + y*z)^(2*n).  Cf. A000172. - Peter Bala, Sep 21 2021
From Peter Bala, Sep 24 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)*binomial(2*n+k,n).
a(n) = the coefficient of (x*y*z*t^2)^n in the expansion of 1/(1 - x - y)*(1 - z - t) - x*y*z*t) (a(n) = A(n,n,n,2*n) in the notation of Straub, Theorem 1.2). (End)
a(n) = (8/5) * Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)*binomial(2*n+k-1,n) for n >= 1. - Peter Bala, Jul 09 2024
a(n) = Sum_{k = 0..n} binomial(n, k)^2 * A108625(2*n, k). Cf. A183204. - Peter Bala, Oct 12 2024
From Peter Bala, Oct 16 2024: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k)*binomial(2*n+k, k)*A108625(n, k) = 8 * Sum_{k = 0..n} (-1)^(n+k+1) * binomial(n-1, k)*binomial(2*n+k-1, k)*A108625(n, k) = (8/5) * Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k)*binomial(2*n+k-1, k)*A108625(n, k) for n >= 1. Cf. A176285. (End)

A176898 a(n) = binomial(6n, 3n)binomial(3n, n)/(2(2n+1)binomial(2n, n)).

Original entry on oeis.org

5, 231, 14586, 1062347, 84021990, 7012604550, 607892634420, 54200780036595, 4938927219474990, 457909109348466930, 43057935618181929900, 4096531994713828810686, 393617202432246696493436, 38142088615983865845923052, 3723160004902167033863327592

Offset: 1

Zhi-Wei Sun, Apr 28 2010

During April 26-28, 2010, Zhi-Wei Sun introduced this new sequence and proved that a(n) = binomial(6n,3n)*binomial(3n,n)/(2*(2n+1)*binomial(2n,n)) is a positive integer for every n=1,2,3,... He also observed that a(n) is odd if and only if n is a power of two, and that 3a(n)=0 (mod 2n+3). By Stirling's formula, we have lim_n (8n*sqrt(n*Pi)a(n)/108^n) = 1. It is interesting to find a combinatorial interpretation or recursion for the sequence.
From Tatiana Hessami Pilehrood, Dec 01 2015: (Start)
Zhi-Wei Sun formulated two conjectures concerning a(n) (see Conjectures 1.1 and 1.2 in Z.-W. Sun, "Products and sums divisible by central binomial coefficients" and Conjecture A89 in "Open conjectures on congruences"). The first conjecture states that Sum_{n=1..p-1} a(n)/(108^n) is congruent to 0 or -1 modulo a prime p > 3 depending on whether p is congruent to +-1 or +-5 modulo 12, respectively.
The second conjecture asks about an exact formula for a companion sequence of a(n). Both conjectures as well as many numerical congruences involving a(n) and (2n+1)a(n) were solved by Kh. Hessami Pilehrood and T. Hessami Pilehrood, see the link below. (End)
This is the even bisection of A078531 divided by 2. The odd bisection divided by 2 is A281733. - Akiva Weinberger, Dec 09 2024

			For n=2 we have a(2) = binomial(12,6)*binomial(6,2)/(2*(2*2+1)*binomial(4,2)) = 231.

Indranil Ghosh, Table of n, a(n) for n = 1..450
Khodabakhsh Hessami Pilehrood and Tatiana Hessami Pilehrood, Jacobi polynomials and congruences involving some higher-order Catalan numbers and binomial coefficients, preprint, arXiv:1504.07944 [math.NT], 2015.
Khodabakhsh Hessami Pilehrood and Tatiana Hessami Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, J. Int. Seq. 18 (2015) 15.11.7
Mark Roger Sepanski, On Divisibility of Convolutions of Central Binomial Coefficients, Electronic Journal of Combinatorics, 21 (1) 2014, #P1.32.
Chen Wang and Hui-Li Han, Supercongruences involving binomial coefficients and Euler polynomials, arXiv:2407.19882 [math.NT], 2024. See p. 2.
Zhi-Wei Sun, Products and sums divisible by central binomial coefficients, preprint, arXiv:1004.4623 [math.NT], 2010.
Zhi-Wei Sun, Open conjectures on congruences, preprint, arXiv:0911.5665 [math.NT], 2009-2011.
Brian Y. Sun and J. X. Meng, Proof of a Conjecture of Z.-W. Sun on Trigonometric Series, arXiv preprint arXiv:1606.08153 [math.CO], 2016.
Chen Wang and Hui-Li Han, Supercongruences involving binomial coefficients and Euler polynomials, arXiv:2407.19882 [math.NT], 2024. See p. 2.
Xiran Zhang and Guo-Shuai Mao, Congruences involving binomial coefficients and Legendre symbol, ResearchGate (2024). See p. 10.

Cf. A000984, A006013, A173774, A176285, A176477, A078531, A281733.

[Binomial(6*n, 3*n)*Binomial(3*n, n)/(2*(2*n+1)*Binomial(2*n, n)): n in [1..15]]; // Vincenzo Librandi, Dec 02 2015

ogf := eval((1-6*s)/((12*s-1)*(8*s-2)) - 1/2, s=RootOf(x+(3*s-1)*(12*s-1)^2*s*(4*s-1)^2,s));
series(ogf,x=0,30); # Mark van Hoeij, May 06 2013

S[n_]:=Binomial[6n,3n]Binomial[3n,n]/(2(2n+1)Binomial[2n,n]) Table[S[n],{n,1,50}]

a(n) = binomial(6*n, 3*n) * binomial(3*n, n) / (2*(2*n+1) * binomial(2*n, n)); \\ Indranil Ghosh, Mar 05 2017

import math
f=math.factorial
def C(n,r): return f(n)/f(r)/f(n-r)
def A176898(n): return C(6*n, 3*n) * C(3*n, n) / (2*(2*n+1) * C(2*n, n)) # Indranil Ghosh, Mar 05 2017

G.f.: (1-6*s)/((12*s-1)*(8*s-2)) - 1/2, where x+(3*s-1)*(12*s-1)^2*s*(4*s-1)^2 = 0. - Mark van Hoeij, May 06 2013
a(n) ~ 2^(2*n-3) * 3^(3*n) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jan 09 2023
From Peter Bala, Feb 21 2023: (Start)
a(n+1) = 6*(6*n + 1)*(6*n + 5)/((n + 1)*(2*n + 3))*a(n).
a(n) = (2^(2*n-1)) * Product_{1 <= i <= j <= 2*n-1} (2*i + j + 2)/(2*i + j - 1). Cf. A006013. (End)
D-finite with recurrence n*(2*n+1)*a(n) -6*(6*n-1)*(6*n-5)*a(n-1)=0. - R. J. Mathar, Nov 22 2024
a(n) = 2^(4n-1) * binomial(3n-1/2, 2n)/(2n+1). - Akiva Weinberger, Dec 09 2024
a(n) = A078531(2*n)/2. - Akiva Weinberger, Dec 09 2024

A176477 a(1)=2; for n >= 2, (2n+1)^3a(n) = 32n^3a(n-1) + (21n^3 + 22n^2 + 8n + 1)*binomial(2n-1,n)^4.

Original entry on oeis.org

2, 181, 23488, 3625081, 619898336, 113451041232, 21790823094272, 4339409873332321, 888730714063587232, 186141207745025911376, 39707252850926474171392, 8600444322930062324576656

Offset: 1

Zhi-Wei Sun, Apr 18 2010

nonn

On Apr 06 2010, Zhi-Wei Sun introduced this sequence and conjectured that each term a(n) is a positive integer. He also guessed that a(n) is odd if and only if n = 2, 2^2, 2^3, .... It is easy to see that 16*(2n+1)^3*binomial(2n,n)^3*a(n) equals Sum_{k=0..n} (21k^3 + 22k^2 + 8k + 1)*256^(n-k)*binomial(2k,k)^7. Sun also conjectured that for any prime p > 5 we have Sum_{k=0..p-1} (21k^3 + 22k^2 + 8k + 1)*binomial(2k,k)^7/256^k == p^3 (mod p^8). It is also remarkable that Sum_{n>=1} 256^n*(21n^3 - 22n^2 + 8n - 1)/(n^7*binomial(2n,n)^7) = Pi^4/8 as conjectured by J. Guillera.

			For n=2 we have a(2) = (32*2^3*a(1) + (21*2^3 + 22*2^2 + 8*2 + 1)*binomial(2*2-1,2)^4)/(2*2 + 1)^3 = 181.

Kasper Andersen, Re: A somewhat surprising conjecture
Jesús Guillera, About a new kind of Ramanujan-type series, Experiment. Math. 12(2003), 507-510.
Zhi-Wei Sun, Open Conjectures on Congruences, preprint, arXiv:0911.5665.
Zhi-Wei Sun, A somewhat surprising conjecture
Zhi-Wei Sun, Re: A somewhat surprising conjecture

Cf. A176285, A173774, A000984, A001700.

u[n_]:=u[n]=((21n^3+22n^2+8n+1)Binomial[2n-1,n]^4+32*n^3*u[n-1])/((2n+1)^3) u[1]=2 Table[u[n],{n,1,50}]

a(n) = (Sum_{k=0..n} (21k^3 + 22k^2 + 8k + 1)*256^(n-k)*binomial(2k,k)^7) / (16(2n+1)^3*binomial(2n,n)^3).

A374605 a(n) = Sum_{k = 0..n} binomial(n, k)^2binomial(n+k, k)binomial(3n+2k, n).

Original entry on oeis.org

1, 13, 621, 40864, 3116125, 258687513, 22695228864, 2069939892096, 194303918495709, 18648446389798225, 1821631879087498621, 180513102382789033728, 18101940249015916366528, 1833572727177462316881472, 187323995560940882748187200, 19279943156312884441303524864, 1997221716775275248175573251037

Offset: 0

Peter Bala, Jul 20 2024

Compare with the identity Sum_{k = 0..n} binomial(n, k)^2 * binomial(n+k, k) * binomial(2*n+k, n) = binomial(2*n, n)^3 = A002897(n).
It is easy to see that for odd prime p, binomial(2*n, n)^3 is divisible by p^3 for integer n in the interval [(p + 1)/2, p - 1]. A similar property appears to hold for the present sequence. We conjecture that for prime p >= 5, a(n) is divisible by p^3 for integer n in the interval [ceiling((2*p + 1)/3), p - 1] (checked up to p = 101).
More generally, for m >= 2, a similar divisibility property appears to hold for the sequence whose n-th term is equal to Sum_{k = 0..n} binomial(n, k)^2* binomial(n+k, k)*binomial((m + 1)*n + m*k, n).

			Factorization of a(8) thru a(10) showing divisibility by 11^3:
a(8) = (3^6)*11^3*10667*18773
a(9) = (5^2)*7*(11^3)*(13^3)*3607*10103
a(10) = (11^3)*(13^4)*31*22699*68099.

Cf. A176285, A374606.

seq(add(binomial(n, k)^2*binomial(n+k, k)*binomial(3*n+2*k, n), k = 0..n), n = 0..20);
# faster program for large n
seq(simplify(binomial(3*n, n)*hypergeom([-n, -n, (3*n+1)/2, (3*n+2)/2], [1, 1, n+1/2], 1)), n = 0..20);

a(n) = binomial(3*n, n)*hypergeom([-n, -n, (3*n+1)/2, (3*n+2)/2], [1, 1, n+1/2], 1).
P-recursive: 16*n^3*(5616*n^4 - 30888*n^3 + 63459*n^2 - 57709*n + 19600)*(4*n - 1)^2*(4*n - 3)^2*a(n) = 36*(72783360*n^11 - 655050240*n^10 + 2595613248*n^9 - 5966404272*n^8 + 8824615470*n^7 - 8803399545*n^6 + 6034085115*n^5 - 2836309905*n^4 + 893904075*n^3 - 179376410*n^2 + 20562360*n - 1019200)*a(n-1) +  27*n*(5616*n^4 - 8424*n^3 + 4491*n^2 - 991*n + 78)*(3*n - 4)^3*(3*n - 5)^3*a(n-2) with a(0) = 1, a(1) = 13.
a(n) ~ 3^(9*n/2) * (1 + sqrt(3))^(6*n + 3) / (Pi^(3/2) * n^(3/2) * 2^(9*n + 9/2)). - Vaclav Kotesovec, Jul 22 2024

A374606 a(n) = Sum_{k = 0..n} binomial(n, k)^2binomial(n+k, k)binomial(3n+2k+1, n).

Original entry on oeis.org

1, 16, 783, 52000, 3984625, 331782528, 29167556544, 2664193232448, 250366507021125, 24050203166498080, 2350951602405723031, 233101029618601148160, 23386851829632443099584, 2369867535548685346720000, 242199626637223381868264640, 24935593829720710016846519424, 2583792728153680286245574534061

Offset: 0

Peter Bala, Jul 13 2024

A companion sequence to A374605.
Conjecture: for prime p, a(n) is divisible by p^3 for integer n in the interval [(1/3)*(2*p + (p/3)), p - 1], where (p/3) denotes the Legendre symbol.
More generally, for m >= 2, a similar divisibility property appears to hold for the sequence whose n-th term is equal to Sum_{k = 0..n} binomial(n, k)^2* binomial(n+k, k)*binomial((m + 1)*n + m*k + 1, n).

			Factorization of a(9) thru a(12) showing divisibility by 13^3:
a(9) = (2^5)*5*(11^3)*(13^3)*1109*46351
a(10) = (11^3)*(13^3)*803962100633
a(11) = (2^8)*(3^3)*5*(13^3)*(17^3)*67*79*118057
a(12) = (2^6)*7*(13^3)*(17^3)*4836341163053.

Cf. A176285, A374605.

seq( add(binomial(n, k)^2*binomial(n+k, k)*binomial(3*n+2*k+1, n), k = 0..n), n = 0..20);
# faster program for large n
seq(simplify( binomial(3*n+1,n) * hypergeom([-n, -n, (3*n+2)/2, (3*n+3)/2], [1, 1, n+3/2], 1)), n = 0..20);

a(n) = binomial(3*n+1,n) * hypergeom([-n, -n, (3*n+2)/2, (3*n+3)/2], [1, 1, n+3/2], 1).
P-recursive: 16*(5616*n^4 - 22464*n^3 + 33183*n^2 - 21438*n + 5116)*(4*n - 1)^2*(4*n + 1)^2*n^3*a(n) = 36*(36391680*n^10 - 181958400*n^9 + 367112736*n^8 - 376700544*n^7 + 196483887*n^6 - 35225037*n^5 - 11205753*n^4 + 5552733*n^3 - 287502*n^2 - 163800*n + 20800)*(2*n - 1)*a(n-1) + 27*(n - 1)*(5616*n^4 - 513*n^2 + 13)*(3*n - 2)^3*(3*n - 4)^3*a(n-2) with a(0) = 1 and a(1) = 16.
a(n) ~ 3^((9*n + 3)/2) * (1 + sqrt(3))^(6*n+3) / (Pi^(3/2) * n^(3/2) * 2^(9*n + 13/2)). - Vaclav Kotesovec, Jul 16 2024

cp's OEIS Frontend

A002897 a(n) = binomial(2n,n)^3.

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A176898 a(n) = binomial(6*n, 3*n)*binomial(3*n, n)/(2*(2*n+1)*binomial(2*n, n)).

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A176477 a(1)=2; for n >= 2, (2n+1)^3*a(n) = 32n^3*a(n-1) + (21n^3 + 22n^2 + 8n + 1)*binomial(2n-1,n)^4.

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A374605 a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)*binomial(3*n+2*k, n).

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A374606 a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)*binomial(3*n+2*k+1, n).

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A176898 a(n) = binomial(6n, 3n)binomial(3n, n)/(2(2n+1)binomial(2n, n)).

A176477 a(1)=2; for n >= 2, (2n+1)^3a(n) = 32n^3a(n-1) + (21n^3 + 22n^2 + 8n + 1)*binomial(2n-1,n)^4.

A374605 a(n) = Sum_{k = 0..n} binomial(n, k)^2binomial(n+k, k)binomial(3n+2k, n).

A374606 a(n) = Sum_{k = 0..n} binomial(n, k)^2binomial(n+k, k)binomial(3n+2k+1, n).