A142986 a(1) = 1, a(2) = 8, a(n+2) = 8*a(n+1) + (n + 1)*(n + 2)*a(n).

1, 8, 70, 656, 6648, 72864, 862128, 10977408, 149892480, 2187106560, 33985025280, 560578268160, 9786290088960, 180315565516800, 3497645442816000, 71256899266560000, 1521414754578432000, 33975929212194816000

Offset: 1

Author

Peter Bala, Jul 17 2008

Keywords

• easy
• nonn

Comments

This is the case m = 4 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.

References

• Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..250

Crossrefs

Cf. A014820, A142983, A142984, A142985, A142987.

Cf. A002378.

Programs

Haskell

a142986 n = a142986_list !! (n-1)
a142986_list = 1 : 8 : zipWith (+)
                       (map (* 8) $ tail a142986_list)
                       (zipWith (*) (drop 2 a002378_list) a142986_list)
-- Reinhard Zumkeller, Jul 17 2015

Maple

p := n -> (n^4+2*n^2)/3: a := n -> n!*p(n+1)*sum ((-1)^(k+1)/(p(k)*p(k+1)), k = 1..n): seq(a(n), n = 1..20);

Mathematica

RecurrenceTable[{a[1]==1,a[2]==8,a[n]==8a[n-1]+n(n-1)a[n-2]},a,{n,20}] (* Harvey P. Dale, Apr 08 2015 *)

Formula

a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (n^4 + 2*n^2)/3 = A014820(n).

Recurrence: a(1) = 1, a(2) = 8, a(n+2) = 8*a(n+1) + (n + 1)*(n + 2)*a(n).

The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 8 and b(2) = 66.

Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(8 + 1*2/(8 + 2*3/(8 + 3*4/(8 + ... + n*(n - 1)/8)))), for n >= 2.

The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(p(k)*p(k+1)) = 1/(8 + 1*2/(8 + 2*3/(8 + 3*4/(8 + ... + n*(n - 1)/(8 + ...))))) = 17/3 - 8*log(2), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).