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A001819
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Central factorial numbers: second right-hand column of triangle A008955.
(Formerly M4008 N1661)
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13
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0, 1, 5, 49, 820, 21076, 773136, 38402064, 2483133696, 202759531776, 20407635072000, 2482492033152000, 359072203696128000, 60912644957448192000, 11977654199703478272000, 2702572249389834608640000
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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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 (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 (pierre-cami(AT)bbox.fr), Oct 30 2006
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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).
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LINKS
| T. D. Noe, Table of n, a(n) for n=0..50
Index entries for sequences related to factorial numbers
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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 (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 (pierre-cami(AT)bbox.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
a(n)/(n!)^2 -> zeta(2) = (pi^2)/6, I believe. - Najam Haq (njmalhq(AT)yahoo.com), Jan 13 2010
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MATHEMATICA
| 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
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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 (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 * A064618 A193199 A075986
Adjacent sequences: A001816 A001817 A001818 * A001820 A001821 A001822
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KEYWORD
| nonn,easy,nice
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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