A143000 a(0) = 0, a(1) = 1, a(n+1) = 3*(2*n+1)*a(n) + n^4*a(n-1).

0, 1, 9, 151, 3900, 143956, 7188048, 466900848, 38269041408, 3864146985216, 471339558835200, 68335862058777600, 11616056962961817600, 2288216707872948633600, 517111756256861311795200

Offset: 0

Author

Peter Bala, Jul 18 2008

Keywords

• easy
• nonn

Comments

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*m+1)*(2*n+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+1)/k^2 for the constant 1/2*zeta(2). For other cases see A142999 (m=0), A143001 (m=2) and A143002 (m=3).

References

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

Crossrefs

Cf. A142999, A143001, A143002.

Programs

Maple

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

Mathematica

RecurrenceTable[{a[0]==0,a[1]==1,a[n]==(6n-3)a[n-1]+(n-1)^4 a[n-2]},a,{n,15}] (* Harvey P. Dale, Jun 08 2014 *)

Formula

a(n) = n!^2*p(n)*sum {k = 1..n} (-1)^(k+1)/(k^2*p(k-1)*p(k)), where p(n) = n^2+n+1. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = 3*(2*n+1)*a(n) + n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 3. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3+ 1^4/(9+ 2^4/(15+ 3^4/(21+...+ (n-1)^4/(3*(2*n-1)))))), for n >=2. Lim n -> infinity a(n)/b(n) = sum {k = 1..inf} (-1)^(k+1)/(k^2*(1+k^2+k^4)) = 1/(3+ 1^4/(9+ 2^4/(15+ 3^4/(21+...+ n^4/(3*(2*n+1)+...))))) = 1/2*(zeta(2)-1). The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 30].