1,2

This is the case m = 3 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1)+(n+1)*(n+2)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). See A142983 for remarks on the general case.

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

a(n) = n!*p(n+1)*sum {k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (2*n^3+n)/3 = A005900(n). Recurrence: a(1) = 1, a(2) = 6, a(n+2) = 6*a(n+1)+(n+1)*(n+2)*a(n). The sequence b(n):= n!*p(n+1) satisfies the same recurrence with b(1) = 6, b(2) = 38. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(6 +1*2/(6 +2*3/(6 +3*4/(6 +...+(n-1)*n/6)))), for n >=2. The behaviour of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} (-1)^(k+1)/(p(k)*p(k+1)) = 1/(6 +1*2/(6 +2*3/(6 +3*4/(6 +...+n*(n+1)/(6 +...))))) = 6*log(2) - 4, where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).

p := n -> (2*n^3+n)/3: a := n -> n!*p(n+1)*sum ((-1)^(k+1)/(p(k)*p(k+1)), k = 1..n): seq(a(n), n = 1..20);

Cf. A005900, A142983, A142984, A142986, A142987.

Sequence in context: A153293 A145301 A107266 this_sequence A118351 A033296 A082302

Adjacent sequences: A142982 A142983 A142984 this_sequence A142986 A142987 A142988

easy,nonn

Peter Bala (pbala(AT)toucansurf.com), Jul 17 2008