OFFSET
0,3
COMMENTS
This is the case m = 3 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+m^2+m+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^2 for the constant zeta(2). See A142995 for remarks on the general case.
REFERENCES
Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
FORMULA
a(n) = n!^2*p(n)*sum {k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = (10*n^3+15*n^2+11*n+3)/3 = A005902(n) is the polynomial that gives the crystal ball sequence for the A_3 lattice. Recurrence: a(0) = 1, a(1) = 1, a(n+1) = (2*n^2+2*n+13)*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) = 13. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(13-1^4/(17-2^4/(25-3^4/(37-...-(n-1)^4/(2*n^2-2*n+13))))), for n >= 2. Thus the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = 1/(13-1^4/(17-2^4/(25-3^4/(37-...-n^4/((2*n^2+2*n+13)-...))))) = sum {k = 1..inf} 1/(k^2*p(k-1)*p(k)) = 2*(1-1/4+1/9) - zeta(2). The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).
MAPLE
p := n -> (10*n^3+15*n^2+11*n+3)/3: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Peter Bala, Jul 18 2008, Oct 16 2008
STATUS
approved