A142985 a(1) = 1, a(2) = 6, a(n+2) = 6*a(n+1) + (n + 1)*(n + 2)*a(n).

1, 6, 42, 324, 2784, 26424, 275472, 3132576, 38629440, 513708480, 7331489280, 111798455040, 1814503057920, 31234337164800, 568451665152000, 10906950910464000, 220060558384128000, 4657890328906752000

Offset: 1

Author

Peter Bala, Jul 17 2008

Keywords

• nonn
• easy

Comments

This is the case m = 3 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. A005900, A142983, A142984, A142986, A142987.

Cf. A002378.

Programs

Haskell

a142985 n = a142985_list !! (n-1)
a142985_list = 1 : 6 : zipWith (+)
                       (map (* 6) $ tail a142985_list)
                       (zipWith (*) (drop 2 a002378_list) a142985_list)
-- Reinhard Zumkeller, Jul 17 2015

Maple

p := n -> (2*n^3+n)/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]==6,a[n]==6a[n-1]+(n-1)n*a[n-2]},a,{n,20}] (* Harvey P. Dale, Sep 20 2013 *)

Formula

a(n) = n!*p(n+1)*Sum {k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (2*n^3 + n)/3 = A005900(n).

Recurrence: a(1) = 1, a(2) = 6, a(n+2) = 6*a(n+1) + (n + 1)*(n + 2)*a(n).

The sequence b(n):= n!*p(n+1) satisfies the same recurrence with b(1) = 6, b(2) = 38. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(6 + 1*2/(6 + 2*3/(6 + 3*4/(6 + ... + n*(n - 1)/6)))), for n >= 2.

The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k = 1..oo} (-1)^(k+1)/(p(k)*p(k+1)) = 1/(6 + 1*2/(6 + 2*3/(6 + 3*4/(6 + ... + n*(n - 1)/(6 + ...))))) = 6*log(2) - 4, where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).