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A066802
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a(n) = binomial(6*n,3*n).
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11
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1, 20, 924, 48620, 2704156, 155117520, 9075135300, 538257874440, 32247603683100, 1946939425648112, 118264581564861424, 7219428434016265740, 442512540276836779204, 27217014869199032015600
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OFFSET
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0,2
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COMMENTS
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For the trisection of a sequence (here A000984) given by its real o.g.f. see a comment and a reference under A187357.
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LINKS
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FORMULA
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a(n) = Sum_{i=0..n} Sum_{j=0..n} Sum_{k=0..n} binomial(n, i)*binomial(n, j) *binomial(n, k)*binomial(3n, i+j+k). - Benoit Cloitre, Mar 08 2005
O.g.f. (with a(0):=1): (cb(x^(1/3)) + sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))+1+2*x^(1/3)))/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). - Wolfdieter Lang, Mar 24 2011
O.g.f. also 1 + 20*x*4F3(1,7/6,3/2,11/6; 4/3,5/3,2; 64*x). - R. J. Mathar, Sep 17 2012
n*(3n-1)*(3n-2)*a(n) = 8*(6n-5)*(6n-1)*(2n-1)*a(n-1). - R. J. Mathar, Sep 17 2012
a(n) = hypergeom([-3*n, -3*n], [1], 1). - Peter Luschny, Mar 19 2018
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k.
a(n) = [(x*y)^(3*n)] (1 + x + y)^(6*n). Cf. A001448. (End)
Conjecture: a(n) = [x^n] G(x)^(2*n), where G(x) = (1 + x)*(1 - 6*x + x^2)/(2*x) + (x^2 - 1)*sqrt(1 - 14*x + x^2)/(2*x) = 1 + 10*x + 81*x^2 + 720*x^3 + .... The algebraic function G(x) satisfies the quadratic equation x*G(x)^2 - (1 - 5*x - 5*x^2 + x^3)*G(x) + (1 + x)^4 = 0. Cf. A001450. - Peter Bala, Oct 27 2022
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MAPLE
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a := n -> hypergeom([-3*n, -3*n], [1], 1):
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MATHEMATICA
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PROG
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(PARI) { for (n=0, 100, write("b066802.txt", n, " ", binomial(6*n, 3*n)) ) } \\ Harry J. Smith, Mar 28 2010
(Magma) [Binomial(6*n, 3*n): n in [0..15]]; // G. C. Greubel, Feb 17 2020
(Sage) [binomial(6*n, 3*n) for n in (0..15)] # G. C. Greubel, Feb 17 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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