0,3

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).

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

Seiichi Manyama, Table of n, a(n) for n = 0..180

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 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/(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 := 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);

RecurrenceTable[{a[0]==0, a[1]==1, a[n+1]==(2n+1)(n^2+n+25)a[n]-n^6 a[n-1]}, a, {n, 15}] (* Harvey P. Dale, Dec 04 2011 *)

Cf. A066989, A143003, A143004, A143006, A143007.

Sequence in context: A053895 A053905 A184690 * A187432 A017104 A243007

Adjacent sequences: A143002 A143003 A143004 * A143006 A143007 A143008

easy,nonn

Peter Bala, Jul 19 2008

approved