0,3

Coefficient of x^2 in Prod[k=0..n, x+k^2]. - R. Stephan, Aug 22 2004

p divides a(p-1) for prime p>3. p divides a((p-1)/2) for prime p>3. Prime p^2 divides all a(n) for n>2p-1. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 11 2006

The ratio a(n)/A001044(n) is the partial sum of the reverses of squares. E.g. a(4)/A001044(4)=820/576=1/1+1/4+1/9+1/16 - Pierre CAMI (pierrecami(AT)tele2.fr), Oct 30 2006

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.

T. D. Noe, Table of n, a(n) for n=0..50

Index entries for sequences related to factorial numbers

a_n = (n!)^2 * sum[ k=1..n ] k^(-2) - Joe Keane (jgk(AT)jgk.org)

a(n) ~ 1/3*pi^3*n*e^(-2*n)*n^(2*n) - Joe Keane (jgk(AT)jgk.org), Jun 06 2002

Sum_{n>=0} a(n)*x^n/n!^2 = polylog(2, x)/(1-x). - Vladeta Jovovic (vladeta(AT)eunet.rs), Jan 23 2003

a(n) = Sum[ 1/i^2, {i,1,n}] / Product[ 1/i^2, {i,1,n}]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 11 2006

a(0)=0 then a(n)=a(n-1)*n^2+A001044(n-1) E.g. a(1)=0*1+1=1 A001044(0)=1, a(2)=1*2^2+1=5 A001044(1)=1, a(3)=5*3^2+4=49 A001044(2)=4, ... - Pierre CAMI (pierrecami(AT)tele2.fr), Oct 30 2006

Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+1)*a(n) - n^4*a(n-1). b(n) = n!^2 satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 1. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(1-1^4/(5-2^4/(13-3^4/(25-...-(n-1)^4/((2*n^2-2*n+1)))))), leading to the infinite continued fraction expansion zeta(2) = 1/(1-1^4/(5-2^4/(13-3^4/(25-...-n^4/((2*n^2+2*n+1)-...))))). Compare with A142995. Compare also with A024167 and A066989. - Peter Bala (pbala(AT)toucansurf.com), Jul 18 2008

Table[Sum[1/i^2, {i, 1, n}]/Product[1/i^2, {i, 1, n}], {n, 1, 40}] - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 11 2006

Cf. A002455.

Cf. A000254, A066989.

Second right-hand column of triangle A008955.

Cf. A007406, A007407.

Cf. A001044.

Cf. A024167, A066989, A142995.

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Jul 21 2009: (Start)

Equals row sums of A162990(n)/(n+1)^2 for n=>1.

(End)

Sequence in context: A062995 A104600 A002111 this_sequence A064618 A075986 A084765

Adjacent sequences: A001816 A001817 A001818 this_sequence A001820 A001821 A001822

nonn,easy,nice

N. J. A. Sloane (njas(AT)research.att.com).