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A002736
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Apéry numbers: a(n) = n^2*C(2n,n).
(Formerly M2136 N0848)
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17
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0, 2, 24, 180, 1120, 6300, 33264, 168168, 823680, 3938220, 18475600, 85357272, 389398464, 1757701400, 7862853600, 34901442000, 153876579840, 674412197580, 2940343837200, 12759640231800, 55138611528000, 237371722628040, 1018383898440480
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OFFSET
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0,2
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COMMENTS
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Let H be the n X n Hilbert matrix H(i,j) = 1/(i+j-1) for 1 <= i,j <= n. Let B be the inverse matrix of H. The sum of the elements in row n-1 of B equals -a(n-1). - T. D. Noe, May 01 2011
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REFERENCES
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J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933, p. 93.
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).
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LINKS
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Paul S. Bruckman, Problem B-871, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 37, No. 1 (1999), p. 85.
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FORMULA
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a(n) ~ 4^n*n^(3/2)/sqrt(Pi).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(phi)^2 = A086467, where phi is the golden ratio. (End)
D-finite with recurrence: (-n+1)*a(n) +2*(n+4)*a(n-1) +4*(2*n-3)*a(n-2)=0. - R. J. Mathar, Jan 21 2020
a(n) = Sum_{k=0..2*n} binomial(2*n,k)*abs(n-k)^3 (Bruckman, 1999; Strazdins, 2000). - Amiram Eldar, Jan 12 2022
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MAPLE
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MATHEMATICA
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CoefficientList[ Series[x (4 x + 2)/(1 - 4 x)^(5/2), {x, 0, 20}], x] (* Robert G. Wilson v, Aug 08 2011 *)
Table[n^2 Binomial[2n, n], {n, 0, 30}] (* Harvey P. Dale, Jun 21 2017 *)
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PROG
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(MuPAD) combinat::catalan(n)*(n+1)*n^2 $ n = 0..36 // Zerinvary Lajos, Apr 17 2007
(PARI) x='x+O('x^100); concat(0, Vec(x*(4*x+2)/((1-4*x)^(5/2)))) \\ Altug Alkan, Mar 21 2016
(PARI) a(n) = n^2*binomial(2*n, n); \\ Michel Marcus, Mar 21 2016
(Sage) [n^2*(n+1)*catalan_number(n) for n in (0..30)] # G. C. Greubel, Mar 23 2022
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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