A066802 -id:A066802 - cp's OEIS frontend

A381422 Expansion of g.f. = exp( Sum_{n>=1} A066802(n)*x^n/n ).

1, 20, 662, 26780, 1205961, 58050204, 2924165436, 152231599628, 8125577046740, 442293253888592, 24457749066666142, 1370114821790970340, 77591333270514869230, 4434803157977731784808, 255492958449660158603448, 14820943641891118200315756, 864962304943085638764540396

Offset: 0

Karol A. Penson, Apr 22 2025

nonn

Cf. A066802, A155200, A255881, A229451, A229452, A156216.

G.f. = 64/((1 + sqrt(1 - 4*x^(1/3)))^2*(1 + sqrt(1 + 4*(-1)^(1/3)*x^(1/3)))^2*(1 + sqrt(1 - 4*(-1)^(2/3)*x^(1/3)))^2).
The above g.f. denoted by h satisfies algebraic equation of order eight:
1 + (8*x - 1)*h + 4*x*(7*x + 3)*h^2 + 7*x^2*(8*x - 1)*h^3 + x^2*(70*x^2 - 40*x + 1)*h^4 + 7*x^4*(8*x - 1)*h^5 + 4*x^5*(7*x + 3)*h^6 + x^6*(8*x - 1)*h^7 + x^8*h^8 = 0.

A113424 a(n) = (6n)!/((3n)!(2n)!*n!).

Original entry on oeis.org

1, 60, 13860, 4084080, 1338557220, 465817912560, 168470811709200, 62588625639883200, 23717177328413240100, 9124964373613212524400, 3553261127084984957001360, 1397224499394244497967972800, 553883078634868423069470550800, 221068174083308549543680044926400

Offset: 0

Michael Somos, Oct 31 2005

Appears in Ramanujan's theory of elliptic functions of signature 6.
The family of elliptic curves "x=2*H=p^2+q^2-q^3, 0Bradley Klee, Feb 25 2018
The power series with coefficients a(n) * n! plays a central role in the Faber-Zagier relations on the moduli space of algebraic curves; see Pandharipande and Pixton, Section 0.2. - Harry Richman, Aug 19 2024

			G.f. = 1 + 60*x + 13860*x^2 + 4084080*x^3 + 1338557220*x^4 + ... - _Michael Somos_, Dec 02 2018

G. C. Greubel, Table of n, a(n) for n = 0..250
J. Cremona, Elliptic Curves over Q, LMFDB 2017.
Alin Bostan, Armin Straub, and Sergey Yurkevich, On the representability of sequences as constant terms, arXiv:2212.10116 [math.NT], 2022.
H. J. Brothers, Pascal's Prism: Supplementary Material.
S. Hassani, J.-M. Maillard, and N. Zenine, On the diagonals of rational functions: the minimal number of variables (unabridged version), arXiv:2502.05543 [math-ph], 2025. See p. 23.
Bradley Klee, Geometric G.F. for Ramanujan Periods, seqfans mailing list, 2017.
Bradley Klee, On LMFDB period data, LMFDB-support mailing list, 2018.
Bradley Klee, Weierstrass Solution of Cubic Anharmonic Oscillation, Wolfram Demonstrations Project, 2018.
Romeo Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
R. Pandharipande and A. Pixton, Relations in the tautological ring of the moduli space of curves, arXiv:1301.4561 [math.AG], 2020.
S. Ramanujan, Modular Equations and Approximations to Pi, Quarterly Journal of Mathematics, XLV (1914), 350-372.
L. C. Shen, A note on Ramanujan's identities involving the hypergeometric function 2F1(1/6,5/6;1;z), The Ramanujan Journal, 30.2 (2013), 211-222.

a(n) = A347304(6*n)
Elliptic Integrals: A002894, A006480, A000897. Factors: A005809, A066802.
Cf. A188662.

List([0..15],n->Factorial(6*n)/(Factorial(3*n)*Factorial(2*n)*Factorial(n))); # Muniru A Asiru, Apr 08 2018

a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/6, 5/6, 1, 432 x], {x, 0, n}];
Table[Multinomial[n, 2 n, 3 n], {n, 0, 15}] (* Vladimir Reshetnikov, Oct 12 2016 *)
a[ n_] := Multinomial[n, 2 n, 3 n]; (* Michael Somos, Dec 02 2018 *)

{a(n) = if( n<0, 0, (6*n)! / ((3*n)! * (2*n)! * n!))};

G.f.: hypergeometric2F1(1/6, 5/6; 1; 432 * x).
a(n) ~ 432^n/(2*Pi*n). - Ilya Gutkovskiy, Oct 13 2016
a(n) = A005809(n)*A066802(n). - Bradley Klee, Feb 25 2018
0 = a(n)*(-267483013447680*a(n+2) +25577192448000*a(n+3) -204669037440*a(n+4) +372142500*a(n+5)) +a(n+1)*(+408751349760*a(n+2) -57870650880*a(n+3) +546809652*a(n+4) -1088188*a(n+5)) +a(n+2)*(-17884800*a(n+2) +21466920*a(n+3) - 295844*a(n+4) +693*a(n+5)) for all n in Z. - Michael Somos, May 16 2018
From Peter Bala, Feb 28 2020: (Start)
a(n) = C(6*n,2*n)*C(4*n,n).
a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k (apply Mestrovic, equation 39).
(-1)^n*a(n) = [x^(2*n)*y^(2*n)] ( (1 + x + y)*(1 - x + y) )^(4*n).
a(n) = [x^n] ( F(x) )^(60*n), where F(x) = 1 + x + 56*x^2 + 7355*x^3 + 1290319*x^4 + 264117464*x^5 + 59508459679*x^6 + ... appears to have integer coefficients. We conjecture that for k >= 1 the sequence defined by b_k(n) := [x^n] F(x)^(k*n) satisfies the above supercongruences for primes p >= 7. (End)
From Peter Bala, Mar 20 2022: (Start)
Right-hand side of the following identities valid for n >= 1:
Sum_{k = 0..2*n} 4*n*(4*n+k-1)!/(k!*n!*(3*n)!) = (6*n)!/((3*n)!*(2*n)!*n!);
Sum_{k = 0..3*n} 3*n*(3*n+k-1)!/(k!*n!*(2*n)!) = (6*n)!/((3*n)!(2*n)!*n!).
Cf. A001451. (End)
From Peter Bala, Feb 26 2023: (Start)
a(n) = (4^n/n!^2) * Product_{k = n..3*n-1} 2*k + 1.
a(n) = (12^n/n!^2) * Product_{k = 0..n-1} (6*k + 1)*(6*k + 5). (End)
a(n) = 12*(6*n - 1)*(6*n - 5)*a(n-1)/n^2. - Neven Sajko, Jul 19 2023
From Karol A. Penson, Dec 26 2023: (Start)
a(n) = Integral_{x=0..432} x^n*W(x) dx, n>=0, where W(x) = sqrt(18)*MeijerG([[], [0, 0]], [[-1/6, -5/6], []], x/432)/(1296*Pi), where MeijerG is the Meijer G - function.
Apparently, W(x) cannot be represented by any other function. W(x) is positive on x = [0, 432], it diverges at x=0, and monotonically decreases for x>0. It appears that at x=432, W(x) tends to a constant value close to 0.000368414. This integral representation as the n-th power moment of the positive function W(x) on the interval [0, 432] is unique, as W(x) is the solution of the Hausdorff moment problem. (End)
W(x) can be represented in terms of two 2F1 hypergeometric functions, W(x) = hypergeom([1/6, 1/6], [1/3], x/432)/(6*sqrt(Pi)*Gamma(2/3)*Gamma(5/6)*x^(5/6)) - Gamma(2/3)*Gamma(5/6)*sqrt(3)*hypergeom([5/6, 5/6], [5/3], x/432)/(1152*Pi^(5/2)*x^(1/6)), x on (0, 432). - Karol A. Penson, May 16 2025

A001450 a(n) = binomial(5n,2n).

Original entry on oeis.org

1, 10, 210, 5005, 125970, 3268760, 86493225, 2319959400, 62852101650, 1715884494940, 47129212243960, 1300853625660225, 36052387482172425, 1002596421878664480, 27963143931814663880, 781879430625942976880, 21910242651571684460050, 615167304833936727234180

Offset: 0

N. J. A. Sloane

T. D. Noe, Table of n, a(n) for n=0..100
Peter Bala, A note on A001450
M. Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv preprint arXiv:1410.5747 [math.CO], 2014.
M. Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, Discrete Mathematics, Volume 339, Issue 3, 6 March 2016, Pages 1116-1139.

Cf. A001451, A001459, A001460, A000108, A060941, A066802, A259550.

[Binomial(5*n, 2*n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

Maple
```
f := n->(5*n)!/((3*n)!*(2*n)!);
```

Table[Hypergeometric2F1[-3n,-2n,1,1],{n,0,60}] (* John M. Campbell, Jul 15 2011 *)
Table[Binomial[5n,2n],{n,0,20}] (* Harvey P. Dale, Nov 09 2011 *)

a(n) = binomial(5*n,2*n) \\ Altug Alkan, Oct 06 2015

a(n) = (5*n)!/((3*n)!*(2*n)!).
a(n) = 2F1[-3n,-2n,1,1] (see Mathematica code below). - John M. Campbell, Jul 15 2011
G.f.: hypergeom([1/5, 2/5, 3/5, 4/5], [1/3, 1/2, 2/3], (3125/108)*x). - Robert Israel, Aug 07 2014
From Peter Bala, Oct 05 2015: (Start)
a(n) = [x^n] ( (1 + x)*C(x) )^(5*n), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
a(n) = 5*A259550(n) for n >= 1.
exp( (1/5) * Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*x + 23*x^2 + 377*x^3 + ... is the o.g.f. for the sequence of Duchon numbers A060941. (End)
a(n) = [x^(2*n)] 1/(1 - x)^(3*n+1). - Ilya Gutkovskiy, Oct 10 2017
D-finite with recurrence 6*n*(3*n-1)*(2*n-1)*(3*n-2)*a(n) -5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Feb 08 2021
a(n) = Sum_{k = 0..2*n} binomial(3*n+k-1, k). Cf. A066802. - Peter Bala, Jun 04 2024
Right-hand side of the identity Sum_{k = 0..2*n} (-1)^k*binomial(-n, k)* binomial(4*n-k, 2*n-k) = binomial(5*n, 2*n). Compare with the identity Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(4*n-k, 2*n-k) = binomial(3*n, n). - Peter Bala, Jun 05 2024
From Karol A. Penson, May 07 2025: (Start)
G.f. denoted by h(x) satisfies the following algebraic equation of order 10:
8 - 3125*x + 20*(-13 + 3125*x)*h(x) - 45*(-74 + 9375*x)*h(x)^2 + 5*(-4023 + 175000*x)*h(x)^2 + 5*(-4023 + 175000*x)*h(x)^3 + 25*(1809 + 53125*x)*h(x)^4 + (34375*x - 738)*(3125*x - 108)*h(x)^5 + 15*(3125*x + 297)*(3125*x - 108)*h(x)^6 + 5*(3125*x - 108)^2*h(x)^7 + 135*(3125*x - 108)^2*h(x)^8 + (3125*x - 108)^3*h(x)^10=0.
a(n) = Integral_{x=0..3125/108} x^n*W(x)*dx, n>=0, where W(x) = W1(x)+W2(x)+W3(x)+W4(x) can be expressed with four generalized hypergeometric functions of type 4F3:
W1(x) = sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([1/5, 8/15, 7/10, 13/15], [2/5, 3/5, 4/5], (108*x)/3125)/(10*Pi*x^(4/5)),
W2(x) = sqrt(5)*sec((3*Pi)/10)*sin(Pi/10)*hypergeom([2/5, 11/15, 9/10, 16/15], [3/5, 4/5, 6/5], (108*x)/3125)/(50*Pi*x^(3/5)),
W3(x) = sqrt(5)*sec((3*Pi)/10)*sin(Pi/10)*hypergeom([3/5, 14/15, 11/10, 19/15], [4/5, 6/5, 7/5], (108*x)/3125)/(125*Pi*x^(2/5)), and
W4(x) = (7*sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([4/5, 17/15, 13/10, 22/15], [6/5, 7/5, 8/5], (108*x)/3125))/(1250*Pi*x^(1/5)).
Using the formula for a(n) only, W(x) can be shown to be a positive function. It is singular at x=0 and at x=3125/108. This integral representation is unique since W(x) is the solution of the Hausdorff moment problem. (End)
From Peter Bala, Jun 21 2025: (Start)
a(n) = [x^(3*n)] 1/(1 - x)^(2*n+1).
a(n) = Sum_{k = 0..3*n} binomial(2*n+k-1, k). (End)

A187364 Trisection of A000984 (central binomial coefficients): binomial(2(3n+1),3n+1)/2, n>=0.

Original entry on oeis.org

1, 35, 1716, 92378, 5200300, 300540195, 17672631900, 1052049481860, 63205303218876, 3824345300380220, 232714176627630544, 14226520737620288370, 873065282167813104916, 53753604366668088230810, 3318776542511877736535400, 205397724721029574666088520

Offset: 0

Wolfdieter Lang, Mar 10 2011

See a comment under A187363 concerning trisection.

This appears also in the trisection of A001700 (central binomials in the odd numbered Pascal rows): binomial(2*(3*n)+1,3*n+1).

Seiichi Manyama, Table of n, a(n) for n = 0..554

Cf. A066802 (binomial(6n,3n)), A187365 (binomial(2(3n+2),3n+2)/3!).

Cf. A002458, A100033.

Table[c=3n+1;Binomial[2c,c]/2,{n,0,20}] (* Harvey P. Dale, May 10 2012 *)

a(n) = binomial(2*(3*n+1),3*n+1)/2, n>=0.
a(n) = binomial(2*(3*n)+1,3*n+1), n>=0.
O.g.f.: (cb(x^(1/3)) - sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))-(1+8*x^(1/3))/2))/(6*x^(1/3)), with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x)=1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z) the o.g.f. of the Legendre polynomials).
From Peter Bala, Mar 19 2023: (Start)
a(n) = (1/2)*Sum_{k = 0..3*n+1} binomial(3*n+1,k)^2.
a(n) = (1/2)*hypergeom([-1 - 3*n, -1 - 3*n], [1], 1).
a(n) = 8*(2*n - 1)*(6*n + 1)*(6*n - 1)/(n*(3*n + 1)*(3*n - 1)) * a(n-1). (End)
Right-hand side of the binomial sum identity (1/18) * Sum_{k = 0..6*n+3} (-1)^(n+k) * (k/(2*n + 1))^2 * binomial(6*n+3, k)^2 = a(n). - Peter Bala, Nov 05 2024

A036555 Hamming weight of 3n: number of 1's in binary expansion of 3n.

Original entry on oeis.org

0, 2, 2, 2, 2, 4, 2, 3, 2, 4, 4, 2, 2, 4, 3, 4, 2, 4, 4, 4, 4, 6, 2, 3, 2, 4, 4, 3, 3, 5, 4, 5, 2, 4, 4, 4, 4, 6, 4, 5, 4, 6, 6, 2, 2, 4, 3, 4, 2, 4, 4, 4, 4, 6, 3, 4, 3, 5, 5, 4, 4, 6, 5, 6, 2, 4, 4, 4, 4, 6, 4, 5, 4, 6, 6, 4, 4, 6, 5, 6, 4, 6, 6, 6, 6, 8, 2, 3, 2, 4, 4, 3, 3, 5, 4, 5, 2, 4, 4

Offset: 0

Dan Hoey

a(n) is also the largest integer such that 2^a(n) divides binomial(6n,3n)=A066802(n). - Benoit Cloitre, Mar 27 2002
a(n) = A000120(A008585(n)). - Reinhard Zumkeller, Nov 03 2010
a(A002450(n)) = 2*n.

T. D. Noe, Table of n, a(n) for n = 0..1000
S. R. Finch, P. Sebah and Z.-Q. Bai, Odd Entries in Pascal's Trinomial Triangle, arXiv:0802.2654 [math.NT], 2008.
Philippe Flajolet, Peter Grabner, Peter Kirschenhofer, Helmut Prodinger and Robert F. Tichy, Mellin transforms and asymptotics: digital sums, Theoret. Comput. Sci. 123 (1994), 291-314.
Michael Gilleland, Some Self-Similar Integer Sequences
D. J. Newman, On the number of binary digits in a multiple of three, Proc. Amer. Math. Soc. 21 (1969) 719-721.

Cf. A036556, A036557, A099033, A190944.

a036555 = a000120 . (* 3)  -- Reinhard Zumkeller, Sep 01 2013

t1:=[];
for n from 0 to 100 do t2:=convert(3*n,base,2); t3:=add(t2[i],i=1..nops(t2)); t1:=[op(t1),t3];od:
t1;

Total/@IntegerDigits[3Range[0,100],2] (* Harvey P. Dale, Oct 03 2011 *)

a(n) = hammingweight(3*n); \\ Michel Marcus, Mar 13 2014

Name edited by Michel Marcus, Mar 13 2014

A187365 Trisection of A000984 (central binomial coefficients): binomial(2(3n+2),3n+2)/3!, n>=0.

Original entry on oeis.org

1, 42, 2145, 117572, 6686100, 388934370, 22974421470, 1372238454600, 82653088824684, 5011211083256840, 305437356823765089, 18697712969443807572, 1148770108115543559100, 70797430141465286938140, 4374750896947475198160300, 270950190057528375091435920

Offset: 0

Wolfdieter Lang, Mar 10 2011

See a comment under A187357 concerning trisection.

This appears also in the trisection of A001700: binomial(2*(3*n+1)+1,(3*n+1)+1)/3.

Seiichi Manyama, Table of n, a(n) for n = 0..554

Cf. A066802 binomial(6n,3n), A187364 binomial(2*(3n+1),3n+1)/2, A002458, A100033.

a(n)=binomial(2*(3*n+2),3*n+2)/3!, n>=0.
a(n)=binomial(3*(2*n+1),3*n+2)/3, n>=0.
O.g.f.:(cb(x^(1/3)) - sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))-(1-4*x^(1/3))/2))/(18*x^(2/3)),
with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x)=1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z)the o.g.f. of the Legendre polynomials).
From Peter Bala, Mar 19 2023: (Start)
a(n) = (1/6)*Sum_{k = 0..3*n+2} binomial(3*n+2,k)^2.
a(n) = (1/6)*hypergeom([-2 - 3*n, -2 - 3*n], [1], 1).
a(n) = 8*(2*n + 1)*(6*n + 1)*(6*n - 1)/(n*(3*n + 1)*(3*n + 2)) * a(n-1). (End)

A023854 Sum of exponents in prime-power factorization of binomial(6n, 3n).

Original entry on oeis.org

0, 3, 5, 6, 7, 11, 11, 12, 13, 14, 15, 16, 15, 19, 20, 21, 22, 23, 23, 22, 25, 29, 25, 29, 28, 31, 32, 30, 31, 34, 34, 35, 35, 36, 36, 38, 38, 41, 41, 41, 40, 46, 44, 43, 44, 44, 46, 47, 46, 47, 50, 51, 49, 53, 49, 52, 53, 53, 56, 55, 56, 60, 60, 61, 57, 61, 61, 61, 65, 66, 63, 67, 66, 69, 69, 66, 69, 71, 70, 72, 72

Offset: 0

Clark Kimberling

nonn

Ivan Neretin, Table of n, a(n) for n = 0..10000

Cf. A001222, A022559, A023852, A023853, A066802.

Join[{0}, Total[Transpose[FactorInteger[Binomial[6 #, 3 #]]][[2]]]&/@Range[80]] (* Harvey P. Dale, May 14 2011 *)
a[n_] := PrimeOmega[Binomial[6*n, 3*n]]; Array[a, 100, 0] (* Amiram Eldar, Jun 11 2025 *)

a(n) = bigomega(binomial(6*n, 3*n)); \\ Amiram Eldar, Jun 11 2025

From Amiram Eldar, Jun 11 2025: (Start)
a(n) = A001222(A066802(n)).
a(n) = A022559(6*n) - 2*A022559(3*n). (End)

Corrected and extended by Harvey P. Dale, May 14 2011

a(0)=0 inserted by Amiram Eldar, Jun 11 2025

A275655 a(n) = binomial(6n,3n)binomial(2n,n).

Original entry on oeis.org

1, 40, 5544, 972400, 189290920, 39089615040, 8385425017200, 1847301025078080, 415026659401497000, 94660194875011205440, 21850091031597537252544, 5092815839064962373499680, 1196622940864849837505171824, 283073284848591452381449360000

Offset: 0

Peter Bala, Aug 04 2016

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(6*n + k,6*n - k)*binomial(2*k,k) *binomial(2*n - k,n) = binomial(6*n,3*n)*binomial(2*n,n).
We also note that Sum_{k = 0..6*n} (-1)^(n+k)*binomial(6*n + k,6*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(6*n,3*n)*binomial(2*n,n).
Compare with Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k, k)*binomial(2*n - k,n) = binomial(2*n,n)^2 = A002894(n). See also A275652, A275653 and A275654.

Cf. A000984, A002894, A066802, A275652, A275653, A275654.

seq((6*n)!*(2*n)!/((3*n)!*n!)^2, n = 0..20);

Table[Binomial[6 n, 3 n] Binomial[2 n, n], {n, 0, 13}] (* Michael De Vlieger, Aug 07 2016 *)

a(n) = (6*n)!*(2*n)!/((3*n)!*n!)^2.
a(n) = A066802(n) * A000984(n).
Recurrence: a(n) = 16*(2*n - 1)^2*(6*n - 1)*(6*n - 5)/(n^2*(3*n - 1)*(3*n - 2)) * a(n-1).
a(n) = [x^(3*n)] (1 + x)^(6*n) * [x^n] (1 + x)^(2*n) = [x^n] G(x)^(8*n) where G(x) = 1 + 5*x + 159*x^2 + 11690*x^3 + 1160817*x^4 + 135123516*x^5 + 17357714116*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^8, where F(x) = 1 + 5*x + 359*x^2 + 42270*x^3 + 6182313*x^4 + 1021669966*x^5 + 182605696304*x^6 + ... appears to have integer coefficients.
a(n) ~ 256^n/(sqrt(3)*Pi*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 23 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(5*n-k-1,n-k)*binomial(6*n,k)^2.
a(n) = [x^n] (1 - x)^(2*n) * P(6*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.
The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k. (End)

A066798 a(n) = Sum_{i=1..n} binomial(6i,3i).

Original entry on oeis.org

20, 944, 49564, 2753720, 157871240, 9233006540, 547490880980, 32795094564080, 1979734520212192, 120244316085073616, 7339672750101339356, 449852213026938118560, 27666867082225970134160

Offset: 1

Benoit Cloitre, Jan 18 2002

nonn

Harry J. Smith, Table of n, a(n) for n = 1..100

Cf. A006134, A066796, A066797, A066802.

s := RootOf((s+8)*s^3*x-s+1, s):
series( (1+8/s)^(3/2)*(s-4)*s^5/(3*(s^4+8*s^3-s+1)*(s^2+4*s-8)) - 1/(1-x), x=0, 30); # Mark van Hoeij, May 02 2013

Accumulate[Table[Binomial[6n,3n],{n,20}]] (* Harvey P. Dale, Apr 04 2020 *)

{ a=0; for (n=1, 100, write("b066798.txt", n, " ", a+=binomial(6*n, 3*n)) ) } \\ Harry J. Smith, Mar 28 2010

G.f.: (1+8/s)^(3/2)*(s-4)*s^5/(3*(s^4+8*s^3-s+1)*(s^2+4*s-8)) - 1/(1-x) where (s+8)*s^3*x-s+1 = 0. - Mark van Hoeij, May 02 2013
a(n) ~ sqrt(3) * 64^(n+1) / (189*sqrt(Pi*n)). - Vaclav Kotesovec, Jun 07 2019
D-finite with recurrence n*(3*n-1)*(3*n-2)*a(n) +(-585*n^3+873*n^2-370*n+40)*a(n-1) +8*(6*n-5)*(6*n-1)*(2*n-1)*a(n-2)=0. - R. J. Mathar, Jan 11 2025

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.

cp's OEIS Frontend

A381422 Expansion of g.f. = exp( Sum_{n>=1} A066802(n)*x^n/n ).

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A113424 a(n) = (6*n)!/((3*n)!*(2*n)!*n!).

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GAP

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A001450 a(n) = binomial(5*n,2*n).

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Magma

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A187364 Trisection of A000984 (central binomial coefficients): binomial(2(3n+1),3n+1)/2, n>=0.

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A036555 Hamming weight of 3n: number of 1's in binary expansion of 3n.

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Haskell

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A187365 Trisection of A000984 (central binomial coefficients): binomial(2(3n+2),3n+2)/3!, n>=0.

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A023854 Sum of exponents in prime-power factorization of binomial(6n, 3n).

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A275655 a(n) = binomial(6*n,3*n)*binomial(2*n,n).

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A113424 a(n) = (6n)!/((3n)!(2n)!*n!).

A001450 a(n) = binomial(5n,2n).

A275655 a(n) = binomial(6n,3n)binomial(2n,n).

A066798 a(n) = Sum_{i=1..n} binomial(6i,3i).

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