1,2

This is the case m = 2 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 A142989 (m=1) and A142991 (m=3).

Seiichi Manyama, Table of n, a(n) for n = 1..446

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) = (n^2-n+1)/3. Recurrence: a(1) = 1, a(2) = 7, a(n+2) = 7*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) = 7, b(2) = 52. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7+1*3/(7+2*4/(7+3*5/(7+...+(n-1)*(n+1)/7)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(7+1*3/(7+2*4/(7+3*5/(7+...+(n-1)*(n+1)/(7+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 24*log(2)-33/2.

p := n -> (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, A142989, A142991.

Sequence in context: A390335 A291537 A082413 * A202250 A147689 A248168

Adjacent sequences: A142987 A142988 A142989 * A142991 A142992 A142993

easy,nonn

Peter Bala, Jul 17 2008

approved