A142991 a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)*(n+3)*a(n).

1, 9, 89, 936, 10560, 127800, 1657080, 22965120, 339252480, 5326819200, 88651670400, 1559600179200, 28929882240000, 564490975104000, 11560712397696000, 247991610230784000, 5561409662613504000

Offset: 1

Author

Peter Bala, Jul 17 2008

Keywords

• easy
• nonn

Comments

This is the case m = 3 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 A142990 (m=2).

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..445

Crossrefs

Cf. A142979, A142983, A142988, A142989, A142990.

Programs

Maple

p := n -> (2*n^3-3*n^2+7*n-3)/15: 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]==9,a[n+2]==9a[n+1]+(n+1)(n+3)a[n]},a,{n,20}] (* Harvey P. Dale, Jul 18 2020 *)

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^3-3*n^2+7*n-3)/15. Recurrence: a(1) = 1, a(2) = 9, a(n+2) = 9*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) = 9, b(2) = 84. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/9)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/(9+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 167/6 - 40*log(2).