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A142991 a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)*(n+3)*a(n).

%I A142991 #8 Sep 23 2021 01:26:10
%S A142991 1,9,89,936,10560,127800,1657080,22965120,339252480,5326819200,
%T A142991 88651670400,1559600179200,28929882240000,564490975104000,
%U A142991 11560712397696000,247991610230784000,5561409662613504000
%N A142991 a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)*(n+3)*a(n).
%C A142991 This is the case m = 3 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1) + (n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142989 (m=1) and A142990 (m=2).
%H A142991 Seiichi Manyama, <a href="/A142991/b142991.txt">Table of n, a(n) for n = 1..445</a>
%F A142991 a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (2*n^3-3*n^2+7*n-3)/15. Recurrence: a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 9, b(2) = 84. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/9)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/(9+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 167/6 - 40*log(2).
%p A142991 p := n -> (2*n^3-3*n^2+7*n-3)/15: a := n -> (n+2)!*p(n+2)*sum ((-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), k = 1..n): seq(a(n), n = 1..20);
%t A142991 RecurrenceTable[{a[1]==1,a[2]==9,a[n+2]==9a[n+1]+(n+1)(n+3)a[n]},a,{n,20}] (* _Harvey P. Dale_, Jul 18 2020 *)
%Y A142991 Cf. A142979, A142983, A142988, A142989, A142990.
%K A142991 easy,nonn
%O A142991 1,2
%A A142991 _Peter Bala_, Jul 17 2008