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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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22
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0, 1, 5, 49, 820, 21076, 773136, 38402064, 2483133696, 202759531776, 20407635072000, 2482492033152000, 359072203696128000, 60912644957448192000, 11977654199703478272000, 2702572249389834608640000
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OFFSET
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0,3
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COMMENTS
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Coefficient of x^2 in Product_{k=0..n}(x + k^2). - Ralf Stephan, Aug 22 2004
p divides a(p-1) for prime p > 3. p divides a((p-1)/2) for prime p > 3. For prime p, p^2 divides a(n) for n > 2*p+1. - Alexander Adamchuk, Jul 11 2006; last comment corrected by Michel Marcus, May 20 2020
The ratio a(n)/A001044(n) is the partial sum of the reciprocals of squares. E.g., a(4)/A001044(4) = 820/576 = 1/1 + 1/4 + 1/9 + 1/16. - Pierre CAMI, Oct 30 2006
a(n) is the (n-1)-st elementary symmetric function of the squares of the first n numbers. - Anton Zakharov, Nov 06 2016
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REFERENCES
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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
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FORMULA
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a_n = (n!)^2 * Sum_{k=1..n} 1/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(0) = 0, a(n) = a(n-1)*n^2 + A001044(n-1). E.g., a(1) = 0*1 + 1 = 1 since A001044(0) = 1; a(2) = 1*2^2 + 1 = 5 since A001044(1) = 1; a(3) = 5*3^2 + 4 = 49 since A001044(2) = 4; and so on. - 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). The sequence 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) = A013661 as n -> infinity, rewriting the Keane formula. - 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. - Mircea Merca, Apr 03 2012
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MATHEMATICA
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Table[Sum[1/i^2, {i, 1, n}]/Product[1/i^2, {i, 1, n}], {n, 1, 40}] (* Alexander Adamchuk, Jul 11 2006 *)
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PROG
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CROSSREFS
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Second right-hand column of triangle A008955.
Equals row sums of A162990(n)/(n+1)^2 for n >= 1.
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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