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A347855
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a(n) = (4*n)!/((2*n)!*(n)!) * (n/3)!/(4*n/3)!.
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17
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1, 9, 189, 4620, 120285, 3241134, 89237148, 2493521172, 70429218525, 2005604901300, 57481750139814, 1656023714623980, 47913489552349980, 1391243084942932620, 40519970408738302020, 1183237138556438547120
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OFFSET
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0,2
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COMMENTS
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Fractional factorials are defined using the Gamma function; for example, (n/3)! := Gamma(1 + n/3). The sequence defined by u(n) = (12*n)!*n! / ((6*n)!*(4*n)!*(3*n)!) is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 1). See A295431. Here we are essentially considering the sequence (u(n/3))n>=0. The sequence is conjectured to be integral.
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LINKS
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FORMULA
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a(n) = binomial(4*n,2*n)*binomial(2*n,n)/binomial(4*n/3,n).
D-finite with recurrence -n*(n-1)*(n-2)*(2*n-3)*a(n) + 216*(4*n-11)*(4*n-1)*(4*n-5)*(4*n-7)*a(n-3).
Asymptotics: a(n) ~ 1/(2*sqrt(Pi*n))*2^(10*n/3)*3^n as n -> infinity.
O.g.f.: A(x) = hypergeom([11/12, 7/12, 5/12, 1/12], [2/3, 1/2, 1/3], 27648*x^3) + 9*x*hypergeom([11/12, 5/4, 5/12, 3/4], [5/6, 4/3, 2/3], 27648*x^3) + 189*x^2*hypergeom([19/12, 13/12, 5/4, 3/4], [7/6, 5/3, 4/3], 27648*x^3) is conjectured to be algebraic over Q(x).
Conjectural congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k.
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EXAMPLE
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Congruence: a(11) - a(1) = 1656023714623980 - 9 = (3^2)*7*(11^3)*17* 1161713471 == 0 (mod 11^3).
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MAPLE
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seq( (4*n)!/((2*n)!*(n)!) * GAMMA(1+n/3)/GAMMA(1+4*n/3), n = 0..15);
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MATHEMATICA
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Table[Binomial[4n, 2n] Binomial[2n, n]/Binomial[4 n/3, n], {n, 0, 20}] (* Harvey P. Dale, Apr 09 2022 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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