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%I #11 Aug 23 2017 13:07:21
%S 0,1,45,4211,704120,191875384,79755181632,48072816950976,
%T 40372248180436992,45735898093934800896,68049684624570789888000,
%U 130036437291331549384704000,313117351023401464093212672000
%N a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+13)*a(n) - n^6*a(n-1).
%C This is the case m = 2 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1 )*a(n) - n^6*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^3 for Apery's constant zeta(3). For remarks on the general theory see A143003 (m=1). For other cases see A066989 (m=0), A143005 (m=3) and A143006 (m=4).
%D Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%H Seiichi Manyama, <a href="/A143004/b143004.txt">Table of n, a(n) for n = 0..180</a>
%F a(n) = n!^3*p(n)*sum {k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = (3*n^4+6*n^3+9*n^2+6*n+2)/2. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+13)*a(n) - n^6*a(n-1). The sequence b(n):= n!^3*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 13. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(13- 1^6/(45- 2^6/(95- 3^6/(175-...- (n-1)^6/((2*n-1)*(n^2-n+13)))))), for n >=2. The behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} 1/(k^3*p(k-1)*p(k)) = 1/(13- 1^6/(45- 2^6/(95- 3^6/(175-...- n^6/((2*n+1)*(n^2+n+13)-...))))) = zeta(3) - (1 + 1/2^3), where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 2].
%p p := n -> (3*n^4+6*n^3+9*n^2+6*n+2)/2: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..14)
%t RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==(2n+1)(n^2+n+13)a[n]-n^6 a[n-1]}, a,{n,20}] (* _Harvey P. Dale_, Jan 23 2012 *)
%Y Cf. A066989, A143003, A143005, A143006, A143007.
%K easy,nonn
%O 0,3
%A _Peter Bala_, Jul 19 2008
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