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%I #32 Jul 26 2022 12:45:57
%S 0,0,48,288,960,2400,5040,9408,16128,25920,39600,58080,82368,113568,
%T 152880,201600,261120,332928,418608,519840,638400,776160,935088,
%U 1117248,1324800,1560000,1825200,2122848,2455488,2825760,3236400,3690240
%N a(n) = sinh(2*arccosh(n))^2 = 4*n^2*(n^2 - 1).
%H Vincenzo Librandi, <a href="/A173121/b173121.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = 48*A002415(n) = 4*A047928(n).
%F G.f.: 48*x^2*(1+x)/(1-x)^5. - _Colin Barker_, Mar 22 2012
%F From _Amiram Eldar_, Jul 26 2022: (Start)
%F Sum_{n>=2} 1/a(n) = (21 - 2*Pi^2)/48.
%F Sum_{n>=2} (-1)^n/a(n) = (Pi^2 - 9)/48. (End)
%t Table[4 n^2*(n^2 - 1), {n, 0, 30}] (* or *) Table[Round[N[Sinh[2 ArcCosh[n]]^2, 100]], {n, 0, 50}]
%t LinearRecurrence[{5,-10,10,-5,1},{0,0,48,288,960},40] (* _Harvey P. Dale_, Jul 22 2015 *)
%o (Magma) [4*n^2*(n^2-1): n in [0..40]]; // _Vincenzo Librandi_, Jun 15 2011
%o (PARI) a(n)=4*n^2*(n^2-1) \\ _Charles R Greathouse IV_, Jul 01 2013
%Y Cf. A002415, A047928, A001079, A037270, A071253, A108741, A132592, A146311-A146313, A173115.
%K nonn,easy
%O 0,3
%A _Artur Jasinski_, Feb 10 2010
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