|
%I #19 Mar 08 2024 07:51:36
%S 1,8,70,656,6648,72864,862128,10977408,149892480,2187106560,
%T 33985025280,560578268160,9786290088960,180315565516800,
%U 3497645442816000,71256899266560000,1521414754578432000,33975929212194816000
%N a(1) = 1, a(2) = 8, a(n+2) = 8*a(n+1) + (n + 1)*(n + 2)*a(n).
%C 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.
%D Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%H Reinhard Zumkeller, <a href="/A142986/b142986.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^4 + 2*n^2)/3 = A014820(n).
%F Recurrence: a(1) = 1, a(2) = 8, a(n+2) = 8*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) = 8 and b(2) = 66.
%F 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.
%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)/(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)]).
%p 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);
%t 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 *)
%o (Haskell)
%o a142986 n = a142986_list !! (n-1)
%o a142986_list = 1 : 8 : zipWith (+)
%o (map (* 8) $ tail a142986_list)
%o (zipWith (*) (drop 2 a002378_list) a142986_list)
%o -- _Reinhard Zumkeller_, Jul 17 2015
%Y Cf. A014820, A142983, A142984, A142985, A142987.
%Y Cf. A002378.
%K easy,nonn
%O 1,2
%A _Peter Bala_, Jul 17 2008
| 