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 A001819 Central factorial numbers: second right-hand column of triangle A008955. (Formerly M4008 N1661) 13
 0, 1, 5, 49, 820, 21076, 773136, 38402064, 2483133696, 202759531776, 20407635072000, 2482492033152000, 359072203696128000, 60912644957448192000, 11977654199703478272000, 2702572249389834608640000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS 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, 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, Oct 30 2006 REFERENCES J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217. 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..50 Mircea Merca, A Special Case of the Generalized Girard-Waring Formula J. Integer Sequences, Vol. 15 (2012), Article 12.5.7. FORMULA 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, Jan 23 2003 a(n) = Sum[ 1/i^2, {i,1,n}] / Product[ 1/i^2, {i,1,n}]. - Alexander Adamchuk, 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, 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, Jul 18 2008 a(n)/(n!)^2 -> zeta(2) = (pi^2)/6, I believe. - Najam Haq (njmalhq(AT)yahoo.com), Jan 13 2010 a(n)=s(n+1,2)^2-2*s(n+1,1)*s(n+1,3), where s(n,k) are Stirling numbers of the first kind, A048994. [From Mircea Merca, Apr 03 2012] MATHEMATICA Table[Sum[1/i^2, {i, 1, n}]/Product[1/i^2, {i, 1, n}], {n, 1, 40}] - Alexander Adamchuk, Jul 11 2006 Table[ n!^2*HarmonicNumber[n, 2], {n, 0, 15}](* From Jean-François Alcover, May 09 2012, after Joe Keane *) CROSSREFS 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, Jul 21 2009: (Start) Equals row sums of A162990(n)/(n+1)^2 for n=>1. (End) Sequence in context: A104600 A221972 A002111 * A064618 A193199 A224680 Adjacent sequences:  A001816 A001817 A001818 * A001820 A001821 A001822 KEYWORD nonn,easy,nice,changed AUTHOR STATUS approved

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