A143001 a(0) = 0, a(1) = 1, a(n+1) = 5*(2*n+1)*a(n) + n^4*a(n-1).

0, 1, 15, 391, 14900, 770596, 51695280, 4358885616, 451036788480, 56192122503936, 8297504007091200, 1433159145783936000, 286297057932974899200, 65505120288597559296000

Offset

0,3

Comments

This is the case m = 2 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), A143000(m=1) and A143002 (m=3).

References

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

Links

Table of n, a(n) for n=0..13.

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^4+2*n^3+7*n^2+6*n+4)/4. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = 5*(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) = 5. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5+1^4/(15+2^4/(25+3^4/(35+...+(n-1)^4/(5*(2*n-1)))))), for n >=2. Lim n -> infinity a(n)/b(n) = sum {k = 1..inf} (-1)^(k+1)/(k^2*p(k-1)p(k)) = 1/(5+ 1^4/(15+ 2^4/(25+ 3^4/(35+ ... + n^4/(5*(2*n+1)+...))))) = 1/2*(zeta(2)-(1+1/4)). The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 30].

Maple

p := n -> (n^4+2*n^3+7*n^2+6*n+4)/4: 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);

Crossrefs

Cf. A142999, A143000, A143002.

Sequence in context: A286139 A069990 A157581 * A373872 A250950 A366300

Adjacent sequences: A142998 A142999 A143000 * A143002 A143003 A143004

Keyword

easy,nonn

Author

Peter Bala, Jul 18 2008

Status

approved