|
%I #24 Mar 15 2024 12:50:47
%S 1,3,13,66,406,2868,23220,210192,2116656,23375520,281792160,
%T 3673814400,51599514240,775673176320,12440524320000,211848037632000,
%U 3820318338816000,72685037892096000,1455838255452672000
%N a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n+1)^2*a(n).
%C This is the case m = 1 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) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of Mercator's series for the constant log(2). For other cases see A024167 (m = 0), A142980 (m = 2), A142981 (m = 3) and A142982 (m = 4).
%C The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p_m(k-1)*p_m(k)), where p_m(x) = Sum_{k = 0..m} 2^k*C(m,k)*C(x,k) = Sum_{k = 0..m} C(m,k)*C(x+k,m), is the Ehrhart polynomial of the m-dimensional cross polytope (the hyperoctahedron).
%C The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
%C The sequence {p_m(k)},k>=0 is the crystal ball sequence for the product lattice A_1 x... x A_1 (m copies). The table of values [p_m(k)]m,k>=0 is the array of Delannoy numbers A008288.
%C The polynomial p_m(x) is the unique polynomial solution of the difference equation (x + 1)*f(x+1) - x*f(x-1) = (2*m + 1)*f(x), normalized so that f(0) = 1. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials p_m(x-1), m = 1,2,3,..., satisfy a Riemann hypothesis [BUMP et al., Theorems 4 and 6]. The o.g.f. for the p_m(x) is (1 + t)^x/(1 - t)^(x+1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ....
%C The general recurrence in the first paragraph above also has a second solution b(n) = n!*p_m(n) with initial conditions b(1) = 2*m + 1, b(2) = (2*m + 1)^2 + 1.
%C Hence 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_m(k-1)*p_m(k)) = 1/((2*m + 1) + 1^2/((2*m + 1) + 2^2/((2*m + 1) + 3^2/((2*m + 1) + ... + n^2/((2*m + 1) + ...))))) = (-1)^m * (log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
%C For other sequences defined by similar recurrences and related to log(2) see A142983 and A142988. See also A142992 for the connection between log(2) and the C_n lattices. For corresponding results for the constants e, zeta(2) and zeta(3) see A000522, A142995 and A143003 respectively.
%D Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%H Seiichi Manyama, <a href="/A142979/b142979.txt">Table of n, a(n) for n = 1..448</a>
%H D. Bump, K. Choi, P. Kurlberg and J. Vaaler, <a href="https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.40.5901">A local Riemann hypothesis, I</a>, Math. Zeit. 233, (2000), 1-19.
%F a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n + 1.
%F Recurrence: a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n + 1)^2*a(n).
%F The sequence b(n):= n!*p(n) satisfies the same recurrence with b(1) = 3, b(2) = 10.
%F Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/3)))), for n >= 2.
%F Limit_{n -> oo} a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/(3 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4k^2 - 1)) = 1 - log(2).
%F Thus a(n) ~ c*n*n! as n -> oo, where c = 2*(1 - log(2)).
%p p := n -> 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 RecurrenceTable[{a[1]==1,a[2]==3,a[n+2]==3a[n+1]+(n+1)^2 a[n]},a,{n,20}] _Harvey P. Dale_, May 20 2012
%Y Cf. A008288, A024167, A142980, A142981, A142982, A142983, A142988, A142992.
%K easy,nonn
%O 1,2
%A _Peter Bala_, Jul 17 2008
|
