%I #25 Dec 10 2024 09:03:22

%S 1,5,29,190,1414,11820,110004,1129200,12686256,154896480,2043108000,

%T 28958014080,438997622400,7088892491520,121487996448000,

%U 2202440792832000,42113131054848000,847071044402688000

%N a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n + 1)^2*a(n).

%C This is the case m = 2 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.

%H Seiichi Manyama, <a href="/A142980/b142980.txt">Table of n, a(n) for n = 1..448</a>

%F a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n^2 + 2*n + 1 = A001844(n) is the Ehrhart polynomial for the 2-dimensional cross polytope (a square).

%F Recurrence: a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n+1)^2*a(n).

%F The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 5, b(2) = 26.

%F Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + (n-1)^2/5)))), for n >= 2.

%F The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + n^2/(5 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4*k^4 + 1)) = log(2) - (1 - 1/2); the final equality is a result of Glaisher.

%F Thus a(n) ~ c*n^2*n! as n -> oo, where c = 2*log(2) - 1.

%F From _Peter Bala_, Dec 09 2024: (Start)

%F E.g.f.: A(x) = ((1 + x)^2 *log(1 + x) - 2*x^2)/(1 - x)^3 satisfies the differential equation 1 + (x + 5)*A(x) + (x^2 - 1)*A(x)' with A(0) = 0.

%F Sum_{k = 1..n} Stirling2(n, k)*a(k) = A135148(n+1). (End)

%p p := n -> 2*n^2+ 2*n+1: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)

%Y Cf. A024167, A135148, A142979, A142981, A142982.

%K easy,nonn,changed

%O 1,2

%A _Peter Bala_, Jul 17 2008