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

%I A142985 #18 Mar 08 2024 07:51:41
%S A142985 1,6,42,324,2784,26424,275472,3132576,38629440,513708480,7331489280,
%T A142985 111798455040,1814503057920,31234337164800,568451665152000,
%U A142985 10906950910464000,220060558384128000,4657890328906752000
%N A142985 a(1) = 1, a(2) = 6, a(n+2) = 6*a(n+1) + (n + 1)*(n + 2)*a(n).
%C A142985 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.
%D A142985 Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%H A142985 Reinhard Zumkeller, <a href="/A142985/b142985.txt">Table of n, a(n) for n = 1..250</a>
%F A142985 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).
%F A142985 Recurrence: a(1) = 1, a(2) = 6, a(n+2) = 6*a(n+1) + (n + 1)*(n + 2)*a(n).
%F A142985 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.
%F A142985 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)]).
%p A142985 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);
%t A142985 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 *)
%o A142985 (Haskell)
%o A142985 a142985 n = a142985_list !! (n-1)
%o A142985 a142985_list = 1 : 6 : zipWith (+)
%o A142985                        (map (* 6) $ tail a142985_list)
%o A142985                        (zipWith (*) (drop 2 a002378_list) a142985_list)
%o A142985 -- _Reinhard Zumkeller_, Jul 17 2015
%Y A142985 Cf. A005900, A142983, A142984, A142986, A142987.
%Y A142985 Cf. A002378.
%K A142985 nonn,easy
%O A142985 1,2
%A A142985 _Peter Bala_, Jul 17 2008