A010791 a(n) = n!*(n+2)!/2.

1, 3, 24, 360, 8640, 302400, 14515200, 914457600, 73156608000, 7242504192000, 869100503040000, 124281371934720000, 20879270485032960000, 4071457744581427200000, 912006534786239692800000, 232561666370491121664000000, 66977759914701443039232000000

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Also determinant of n X n matrix with m(i,j) = i^2 if i=j, otherwise 1. - Robert G. Wilson v, Jan 28 2002

Partial products of positive values of A005563. - Jonathan Vos Post, Oct 21 2008

This sequence has been shown to contain infinitely many squares. From the Hong and Liu abstract: Recently Cilleruelo proved that the product Product_{k=1..n} (k^2 + 1) is a square only for n = 3 which confirms a conjecture of Amdeberhan, Medina and Moll. In this paper, we show that the sequence Product_{k=2..n} (k^2 - 1) contains infinitely many squares. Furthermore, we determine all squares in this sequence. We also give a formula for the p-adic valuation of the terms in this sequence. - Jonathan Vos Post, Oct 21 2008

Equals (-1)^n * (1, 1, 3, 24, 360, ...) dot (1, -4, 9, -16, 25, ...). E.g., a(4) = (1, 1, 3, 24, 360) dot (1, -4, 9, -16, 25) = 1 - 4 + 27 - 384 + 9000 = 8640. - Gary W. Adamson, Apr 21 2009

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Javier Cilleruelo, Squares in (1^2+1)...(n^2+1), Journal of Number Theory 128:8 (2008), pp. 2488-2491.
• Shaofang Hong and Xingjiang Liu, Squares in (2^2-1)...(n^2-1) and p-adic valuation, arXiv:0810.3366 [math.NT], 2008-2009.
• Index entries for sequences related to factorial numbers.

Crossrefs

Cf. A005563, A010792, A010793, A010794, A010795, A175430, A229020.

Programs

Magma

[Factorial(n)* Factorial(n+2) / 2: n in [0..20]]; // Vincenzo Librandi, Jun 11 2013

Maple

f := n->n!*(n+2)!/2;

Mathematica

Table[n!(n+2)!/2,{n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Apr 03 2011 *)

PARI

a(n) = n!*(n+2)!/2; \\ Michel Marcus, Feb 03 2016

Formula

From Amiram Eldar, Sep 27 2022: (Start)

a(n) = A175430(n+1)/2.

Sum_{n>=0} 1/a(n) = 2*BesselI(2,2) = 2*A229020.

Sum_{n>=0} (-1)^n/a(n) = 2*BesselJ(2,2). (End)

a(n) = 1/([x^n] hypergeom([], [3], x)). - Peter Luschny, Sep 13 2024