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%I #8 Aug 23 2017 13:07:34
%S 0,1,129,30251,11129080,5985956824,4501140404544,4577449694922432,
%T 6130634638657941504,10576998168036159614976,
%U 23068077837238986190848000,62571876653848665596203008000,208106990164246904792046895104000
%N a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+41)*a(n) - n^6*a(n-1).
%C This is the case m = 4 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 A143005 (m=3).
%D Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%H Seiichi Manyama, <a href="/A143006/b143006.txt">Table of n, a(n) for n = 0..179</a>
%F a(n) = n!^3*p(n)*sum {k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = (35*n^8 +140*n^7 +630*n^6 +1400*n^5 +2595*n^4 +3020*n^3 +2500*n^2 +1200*n +288)/288. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+41)*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) = 41. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(41- 1^6/(129- 2^6/(235- 3^6/(371-...- (n-1)^6/((2*n-1)*(n^2-n+41)))))), 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/(41- 1^6/(129- 2^6/(235- 3^6/(371-...- n^6/((2*n+1)*(n^2+n+41)-...))))) = zeta(3) - (1 + 1/2^3 + 1/3^3 + 1/4^3), where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 4].
%p p := n -> (35*n^8 +140*n^7 +630*n^6 +1400*n^5 +2595*n^4 +3020*n^3 +2500*n^2 +1200*n +288)/288: 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 Cf. A066989, A143003, A143004, A143005, A143007.
%K easy,nonn
%O 0,3
%A _Peter Bala_, Jul 19 2008
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