This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A142980 #31 Dec 12 2024 09:27:25
%S A142980 1,5,29,190,1414,11820,110004,1129200,12686256,154896480,2043108000,
%T A142980 28958014080,438997622400,7088892491520,121487996448000,
%U A142980 2202440792832000,42113131054848000,847071044402688000,17880009683784192000,395192695448291328000,9127967350755133440000
%N A142980 a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n + 1)^2*a(n).
%C A142980 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 A142980 Seiichi Manyama, <a href="/A142980/b142980.txt">Table of n, a(n) for n = 1..448</a>
%F A142980 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 A142980 Recurrence: a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n+1)^2*a(n).
%F A142980 The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 5, b(2) = 26.
%F A142980 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 A142980 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 A142980 Thus a(n) ~ c*n^2*n! as n -> oo, where c = 2*log(2) - 1.
%F A142980 From _Peter Bala_, Dec 09 2024: (Start)
%F A142980 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 A142980 Sum_{k = 1..n} Stirling2(n, k)*a(k) = A135148(n+1). (End)
%p A142980 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)
%t A142980 Module[{a, n}, RecurrenceTable[{a[n+2] == 5*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 5}, a, {n, 25}]] (* _Paolo Xausa_, Dec 12 2024 *)
%Y A142980 Cf. A024167, A135148, A142979, A142981, A142982.
%K A142980 easy,nonn
%O A142980 1,2
%A A142980 _Peter Bala_, Jul 17 2008