%I A143005
%S A143005 0,1,81,12491,3176120,1235165464,697648230720,550023729068736,
%T A143005 586201214122536960,822460381655068717056,1485544574481829982208000,
%U A143005 3389058487000919282503680000,9606157364646714324010401792000
%N A143005 a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+25)*a(n) - n^6*a(n-1).
%C A143005 This is the case m = 3 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 the constant 
               zeta(3). For remarks on the general theory see A143003 (m=1). For 
               other cases see A066989 (m=0), A143004 (m=2) and A143006 (m=4).
%D A143005 Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%F A143005 a(n) = n!^3*p(n)*sum {k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = (10*n^6 
               +30*n^5 +85*n^4 +120*n^3 +121*n^2 +66*n +18)/18. Recurrence: a(0) 
               = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+25)*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) = 25. Hence we obtain the finite 
               continued fraction expansion a(n)/b(n) = 1/(25- 1^6/(81- 2^6/(155- 
               3^6/(259-...- (n-1)^6/((2*n-1)*(n^2-n+25)))))), for n >=2. The behaviour 
               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/(25- 1^6/(81- 2^6/(155- 3^6/
               (259-...- n^6/((2*n+1)*(n^2+n+25)-...))))) = zeta(3) - (1 + 1/2^3 
               + 1/3^3), where the final equality follows from a result of Ramanujan; 
               see [Berndt, Chapter 12, Entry 32(iii) at x = 3].
%p A143005 p := n -> (10*n^6+30*n^5+85*n^4+120*n^3+121*n^2+66*n+18)/18: a := n -> 
               n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..13);
%Y A143005 Cf. A066989, A143003, A143004, A143006, A143007.
%Y A143005 Sequence in context: A053913 A053895 A053905 this_sequence A017104 A143653 
               A116007
%Y A143005 Adjacent sequences: A143002 A143003 A143004 this_sequence A143006 A143007 
               A143008
%K A143005 easy,nonn
%O A143005 0,3
%A A143005 Peter Bala (pbala(AT)toucansurf.com), Jul 19 2008