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%I #20 Mar 08 2024 09:01:10
%S 1,4,22,136,984,8016,73392,742464,8254080,99838080,1307301120,
%T 18407831040,277570298880,4460506444800,76131788544000,
%U 1375048700928000,26208041287680000,525597067634688000,11065538390925312000
%N a(1) = 1, a(2) = 4, a(n+2) = 4*a(n+1) + (n + 1)*(n + 2)*a(n).
%C This is the case m = 2 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 Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%H Reinhard Zumkeller, <a href="/A142984/b142984.txt">Table of n, a(n) for n = 1..250</a>
%F a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = n^2.
%F Recurrence: a(1) = 1, a(2) = 4, a(n+2) = 4*a(n+1) + (n + 1)*(n + 2)*a(n).
%F The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 4 and b(2) = 18.
%F Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(4 + 1*2/(4 + 2*3/(4 + 3*4/(4 + ... + n*(n - 1)/(4))))), for n >= 2.
%F The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k^2*(k + 1)^2) = 1/(4 + 1*2/(4 + 2*3/(4 + 3*4/(4 + ... + n*(n - 1)/(4 + ...))))) = 3 - 4*log(2), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).
%F a(n) = n! * (((-1)^n * (2 * Psi(n/2 + 1) - 2 * Psi(n/2 + 3/2)) - 4 * log(2) + 3) * (n+1)^2 + (-1)^n * (2 * n + 1)). - _Robert Israel_, Mar 07 2024
%p a := n -> n!*(n+1)^2*sum ((-1)^(k+1)/(k^2*(k+1)^2), k = 1..n): seq(a(n), n = 1..20);
%o (Haskell)
%o a142984 n = a142984_list !! (n-1)
%o a142984_list = 1 : 4 : zipWith (+)
%o (map (* 4) $ tail a142984_list)
%o (zipWith (*) (drop 2 a002378_list) a142984_list)
%o -- _Reinhard Zumkeller_, Jul 17 2015
%Y Cf. A142983, A142985, A142986, A142987.
%Y Cf. A002378.
%K easy,nonn
%O 1,2
%A _Peter Bala_, Jul 17 2008
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