A002736 Apéry numbers: a(n) = n^2*C(2n,n).

0, 2, 24, 180, 1120, 6300, 33264, 168168, 823680, 3938220, 18475600, 85357272, 389398464, 1757701400, 7862853600, 34901442000, 153876579840, 674412197580, 2940343837200, 12759640231800, 55138611528000, 237371722628040, 1018383898440480

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

Comments

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

References

• 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).

Links

• T. D. Noe, Table of n, a(n) for n = 0..200
• Kunle Adegoke, Robert Frontczak, and Taras Goy, Fibonacci-Catalan Series, Integers: Electronic Journal of Combinatorial Number Theory, Vol. 22 (2022), #A110.
• Paul S. Bruckman, Problem B-871, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 37, No. 1 (1999), p. 85.
• J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
• J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 (Annotated scans of some selected pages).
• Indulis Strazdins, Solution to problem B-871, Fibonacci Quartely, Vol. 38, No. 1 (2000), pp. 86-87.
• Hans J. H. Tuenter, Walking into an absolute sum, arXiv:math/0606080 [math.NT], 2006. Published version on Walking into an absolute sum, The Fibonacci Quarterly, Vol. 40, No. 2 (May 2002), pp. 175-180.
• Alfred J. van der Poorten, A proof that Euler missed ... Apery's proof of the irrationality of zeta(3), Math. Intelligencer, Vol. 1 (1978/1979), pp. 195-203.

Crossrefs

Cf. A000108, A002544, A005258, A005259, A005429, A005430, A086467.

A diagonal of A331431.

Programs

Magma

[n^2*Binomial(2*n, n): n in [0..30]]; // Vincenzo Librandi, Aug 08 2014

Maple

seq(n^2*binomial(2*n,n), n=0..50); # Robert Israel, Aug 07 2014

Mathematica

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 *)

MuPAD

combinat::catalan(n)*(n+1)*n^2 $ n = 0..36 // Zerinvary Lajos, Apr 17 2007

PARI

my(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

Formula

G.f.: x*(4*x+2)/((1-4*x)^(5/2)). - Marco A. Cisneros Guevara, Jul 25 2011

Sum_{n>=1} 1/a(n) = Pi^2/18 (Euler). - Benoit Cloitre, Apr 07 2002

From Ilya Gutkovskiy, Jan 17 2017: (Start)

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) = (2n)!/(Gamma(n))^2. - Diego Rattaggi, Mar 30 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

Sum_{n>=1} x^n/a(n) = 2*arcsin(sqrt(x)/2)^2, for abs(x) < 4 (Adegoke et al., 2022, section 5, p. 10). - Amiram Eldar, Dec 07 2024

From Peter Bala, Aug 02 2025: (Start)

For n >= 1,

a(n) = 2*n*(2*n-1)/(n-1)^2 * a(n-1) with a(1) = 2 and

1/a(n) = Sum_{k = 0..n} (-1)^(n+k+1) * binomial(n, k)*binomial(n+k, k)/(n+k)^2. (End)

a(n) = 2 * A002544(n-1) for n>=1. - Alois P. Heinz, Aug 03 2025