A142989 a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1)+(n+1)*(n+3)*a(n).

1, 5, 33, 240, 1992, 18360, 187416, 2093760, 25462080, 334592640, 4728412800, 71488811520, 1151817408000, 19699405286400, 356504125824000, 6805868977152000, 136702533123072000, 2881808345235456000

Offset: 1

Author

Peter Bala, Jul 17 2008

Keywords

• easy
• nonn

Comments

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).

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..447

Crossrefs

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

Programs

Maple

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);

Mathematica

RecurrenceTable[{a[1]==1,a[2]==5,a[n]==5a[n-1]+(n-1)(n+1)a[n-2]},a,{n,20}] (* Harvey P. Dale, Jun 17 2013 *)

Formula

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).