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 A142981 #31 Dec 12 2024 09:27:21
%S A142981 1,7,53,434,3886,38052,406260,4708368,58959216,794092320,11454567840,
%T A142981 176267145600,2883327788160,49972442123520,914939341344000,
%U A142981 17648374867200000,357763095454464000,7604722004802048000,169148296960860672000,3929342722459564032000,95164717841561217024000
%N A142981 a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1) + (n+1)^2*a(n).
%C A142981 This is the case m = 3 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.
%D A142981 Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%H A142981 Seiichi Manyama, <a href="/A142981/b142981.txt">Table of n, a(n) for n = 1..447</a>
%F A142981 a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = (4*n^3 + 6*n^2 + 8*n + 3)/3 = A001845(n) is the Ehrhart polynomial for the 3-dimensional cross polytope (the octahedron).
%F A142981 Recurrence: a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1) + (n + 1)^2*a(n). The sequence b(n):= n!*p(n) satisfies the same recurrence with b(1) = 7, b(2) = 50.
%F A142981 Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7 + 1^2/(7 + 2^2/(7 + 3^2/(7 + ... + (n-1)^2/7)))), for n >= 2.
%F A142981 The behavior of a(n) for large n is given by limit_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p(k-1)*p(k)) = 1/(7 + 1^2/(7 + 2^2/(7 + 3^2/(7 + ... + n^2/(7 + ...))))) = (1 - 1/2 + 1/3) - log(2); the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]). Thus a(n) ~ c*n^3*n! as n -> oo, where c = (10 - 12*log(2))/9.
%F A142981 E.g.f.: A(x) = (2*x*(4*x^2 + 3*x + 3) - 3*(x + 1)^3*log(1 + x) )/(3*(1 - x)^4) satisfies the differential equation 1 + (x + 7)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0. - _Peter Bala_, Dec 09 2024
%p A142981 p := n -> (4*n^3+6*n^2+8*n+3)/3: 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 A142981 Module[{a, n}, RecurrenceTable[{a[n+2] == 7*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 7}, a, {n, 25}]] (* _Paolo Xausa_, Dec 12 2024 *)
%Y A142981 Cf. A001845, A024167, A142979, A142980, A142982.
%K A142981 easy,nonn
%O A142981 1,2
%A A142981 _Peter Bala_, Jul 17 2008