1,2

This is the case m = 1 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1)+(n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142990 (m=2) and A142991 (m=3).

a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (2*n-1)/3. Recurrence: a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 5, b(2) = 28. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5+1*3/(5+2*4/(5+3*5/(5+...+(n-1)*(n+1)/5)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(5+1*3/(5+2*4/(5+3*5/(5+...+(n-1)*(n+1)/(5+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 17/2-12*log(2).

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

Cf. A142979, A142983, A142988, A142990, A142991.

Sequence in context: A001887 A118803 A093427 this_sequence A084131 A084771 A153398

Adjacent sequences: A142986 A142987 A142988 this_sequence A142990 A142991 A142992

easy,nonn

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