%I
%S 1,0,1,8,129,3224,116065,5687184,363979777,29482361936,2948236193601,
%T 356736579425720,51370067437303681,8681541396904322088,
%U 1701582113793247129249,382855975603480604081024
%N a(n) = n^2*a(n1)+(1)^n.
%F a(n) = n!^2*Sum_{k=0..n} (1)^k/k!^2. BesselJ(0, 2*sqrt(x))/(1x) = Sum_{n>=0} a(n)*x^n/n!^2. a(n) = round(n!^2*BesselJ(0, 2)), n>0.
%F Recurrence: a(0) = 1, a(1) = 0, a(n) = (n^21)*a(n1) + (n1)^2*a(n1), n >= 2. The sequence b(n) := n!^2 satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 1. It follows that, for n >=3, a(n) = n!^2*(1/(4 + 4/(8 + 9/(15 +...+ (n1)^2/(n^21))))). Hence BesselJ(0,2) := sum {k = 0..inf} (1)^k/k!^2 = 1/(4 + 4/(8 + 9/(15 + ...+(n1)^2/(n^2+1 + ...)))) = 0.22388 90779 ... . Cf. A006040.  _Peter Bala_, Jul 09 2008
%t Join[{a = 1}, Table[a = a*n^2 + (1)^n, {n, 15}]] (* _Jayanta Basu_, Jul 08 2013 *)
%Y Cf. A000166, A006040.
%Y Cf. A006040.
%K nonn
%O 0,4
%A _Vladeta Jovovic_, Aug 30 2002
