%I #22 Mar 15 2024 12:23:39

%S 1,9,85,846,8974,101916,1240308,16156656,224789616,3331795680,

%T 52465122720,875333381760,15432978107520,286828144485120,

%U 5606317009440000,114993185594112000,2470155824763648000,55464433059571200000,1299510384759562752000,31718253797341267968000

%N a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1) + (n + 1)^2*a(n).

%C This is the case m = 4 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 Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

%H Seiichi Manyama, <a href="/A142982/b142982.txt">Table of n, a(n) for n = 1..446</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^4 + 4*n^3 + 10*n^2 + 8*n + 3)/3 = A001846(n) is the Ehrhart polynomial for the 4-dimensional cross polytope (the 16-cell).

%F Recurrence: a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1) + (n + 1)^2*a(n).

%F The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 9, b(2) = 82.

%F Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(9 + 1^2/(9 + 2^2/(9 + 3^2/(9 + ... + (n-1)^2/9)))), 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)/(k*p(k-1)*p(k)) = 1/(9 + 1^2/(9 + 2^2/(9 + 3^2/(9 + ... + n^2/(9 + ...))))) = log(2) - (1 - 1/2 + 1/3 - 1/4); the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).

%F Thus a(n) ~ c*n^4*n! as n -> oo, where c = (12*log(2) - 7)/18.

%p p := n -> (2*n^4+4*n^3+10*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);

%Y Cf. A024167, A142979, A142980, A142981.

%K easy,nonn

%O 1,2

%A _Peter Bala_, Jul 17 2008