Search: id:A001819
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%I A001819 M4008 N1661
%S A001819 0,1,5,49,820,21076,773136,38402064,2483133696,202759531776,
%T A001819 20407635072000,2482492033152000,359072203696128000,
%U A001819 60912644957448192000,11977654199703478272000,2702572249389834608640000
%N A001819 Central factorial numbers.
%C A001819 Coefficient of x^2 in Prod[k=0..n, x+k^2]. - R. Stephan, Aug 22 2004
%C A001819 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
%C A001819 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
%D A001819 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 
               Academic Press, 1995 (includes this sequence).
%D A001819 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 
               (includes this sequence).
%D A001819 J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
%H A001819 T. D. Noe, <a href="b001819.txt">Table of n, a(n) for n=0..50</a>
%H A001819 <a href="Sindx_Fa.html#factorial">Index entries for sequences related 
               to factorial numbers</a>
%F A001819 a_n = (n!)^2 * sum[ k=1..n ] k^(-2) - Joe Keane (jgk(AT)jgk.org)
%F A001819 a(n) ~ 1/3*pi^3*n*e^(-2*n)*n^(2*n) - Joe Keane (jgk(AT)jgk.org), Jun 
               06 2002
%F A001819 Sum_{n>=0} a(n)*x^n/n!^2 = polylog(2, x)/(1-x). - Vladeta Jovovic (vladeta(AT)eunet.rs), 
               Jan 23 2003
%F A001819 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
%F A001819 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
%F A001819 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
%t A001819 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
%Y A001819 Cf. A002455.
%Y A001819 Cf. A000254, A066989.
%Y A001819 Second right-hand column of triangle A008955.
%Y A001819 Cf. A007406, A007407.
%Y A001819 Cf. A001044.
%Y A001819 Cf. A024167, A066989, A142995.
%Y A001819 Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Jul 21 
               2009: (Start)
%Y A001819 Equals row sums of A162990(n)/(n+1)^2 for n=>1.
%Y A001819 (End)
%Y A001819 Sequence in context: A062995 A104600 A002111 this_sequence A064618 A075986 
               A084765
%Y A001819 Adjacent sequences: A001816 A001817 A001818 this_sequence A001820 A001821 
               A001822
%K A001819 nonn,easy,nice
%O A001819 0,3
%A A001819 N. J. A. Sloane (njas(AT)research.att.com).

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