1,2

This is the case m = 5 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^5+10*n^3+3*n)/15 = A069038(n). Recurrence: a(1) = 1, a(2) = 10, a(n+2) = 10*a(n+1)+(n+1)*(n+2)*a(n). The sequence b(n):= n!*p(n+1) satisfies the same recurrence with b(1) = 10, b(2) = 102. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(10 +1*2/(10 +2*3/(10 +3*4/(10 +...+(n-1)*n/10)))), 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/(10 +1*2/(10 +2*3/(10 +3*4/(10 +...+n*(n+1)/(10 +...))))) = 10*log(2) - 41/6, where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).

p := n -> (2*n^5+10*n^3+3*n)/15: 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. A069038, A142983, A142984, A142985, A142986.

Sequence in context: A068883 A087599 A068097 this_sequence A078192 A015589 A163192

Adjacent sequences: A142984 A142985 A142986 this_sequence A142988 A142989 A142990

easy,nonn

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