%I #7 Jun 27 2017 10:14:32

%S 0,1,11,193,5092,189916,9541872,622179216,51129292032,5172077028096,

%T 631719119232000,91679469784704000,15596136686979072000,

%U 3074102117690701824000,695050625746441101312000

%N a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+7)*a(n) - n^4*a(n-1), n >= 1.

%C This is the case m = 2 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+m^2+m+1 )*a(n) - n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} 1/k^2 for the constant zeta(2). See A142995 for remarks on the general case.

%D Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

%F a(n) = n!^2*p(n)*sum {k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = 3*n^2+3*n+1 = A003215(n) is the polynomial that gives the crystal ball sequence for the A_2 lattice. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+7)*a(n) - n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with b(0) = 1, b(1) = 7. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7-1^4/(11-2^4/(19-3^4/(31-...-(n-1)^4/(2*n^2-2*n+7))))), for n >=2. Thus the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = 1/(7-1^4/(11-2^4/(19-3^4/(31-...-n^4/((2n^2+2n+7)-...))))) = sum {k = 1..inf} 1/(k^2*(9*k^4-3*k^2+1)) = zeta(2)-2*(1-1/4). The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).

%p p := n -> 3*n^2+3*n+1: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);

%t RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==(2n^2+2n+7)a[n]-n^4 a[n-1]},a,{n,20}] (* _Harvey P. Dale_, Jun 27 2017 *)

%Y Cf. A001819, A003215, A142995, A142997, A142998.

%K easy,nonn

%O 0,3

%A _Peter Bala_, Jul 18 2008