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%I #19 Mar 15 2024 07:26:13
%S 1,3,17,96,696,5448,49752,492480,5457600,65128320,850296960,
%T 11864240640,178442611200,2848854758400,48517709184000,
%U 872011090944000,16589133517824000,331426982928384000,6966369015201792000
%N a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n + 1)*(n + 3)*a(n).
%C This is the case m = 0 of the general recurrence a(1) = 1, a(2) = 2*m + 3, a(n+2) = (2*m + 3)*a(n+1) + (n + 1)*(n + 3)*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)*(k + 2)) = 2*(log(2) - 5/8). For other cases see A142989 (m = 1), A142990 (m = 2) and A142991 (m = 3).
%C The solution to the general recurrence may be expressed as a series: a(n) = (n+2)!*p_m(n+2)*Sum_{k = 1..n} (-1)^(k+1)/(k*(k + 1)*(k + 2)*p_m(k+1)*p_m(k+2)), where p_m(x) = 1/(2*x*(x - 1))*Sum_{k = 2..m+2} 2^k*C(m,k-2)*C(x,k).
%C The first few values are p_0(x) = 1, p_1(x) = (2*x - 1)/3, p_2(x) = (x^2 - x + 1)/3 and p_3(x) = (2*x^3 - 3*x^2 + 7*x - 3)/15.
%C The polynomial p_m(x) is the unique polynomial solution (up to multiplication by a constant) of the difference equation (x + 1)*f(x+1) - (x - 2)*f(x-1) = (2*m + 3)*f(x).
%C O.g.f. for the p_m(x): Sum_{k >= 0} p_m(x)*t^m = 1/(2*x*(x - 1)*t^2) * ((1 + t)^x/(1 - t)^(x-1) - 1 + t - 2*x*t) = 1 + (2*x - 1)/3*t + (x^2 - x + 1)/3*t^2 + ....
%C These polynomials satisfy a Riemann hypothesis: their zeros all lie on the vertical line Re x = 1/2 in the complex plane (adapt the proof of the lemma on p. 4 of [BUMP et al.]).
%C The general recurrence in the first paragraph above has a second solution b(n) = (1/2)*(n + 2)!*p_m(n+2), with b(1) = 2*m + 3, b(2) = 4*(m^2 + 3*m + 3).
%C Hence the behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = 2*Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)*(k + 2)*p_m(k+1)*p_m(k+2)) = 1/((2*m + 3) + 1*3/((2*m + 3) + 2*4/((2*m + 3) + 3*5/((2*m + 3) +...+ (n - 1)*(n + 1)/((2*m + 3) + ...))))).
%C The infinite continued fraction has the value (-1)^m*2*(m + 1)*(m + 2)*(log(2) - 5/8 - (1/2)*(1/(1*2*3) - 1/(2*3*4) + 1/(3*4*5) - ... + (-1)^(m+1)/(m*(m + 1)*(m + 2)))) for m > 0. This evaluation follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 34] (set l = n in Entry 34 and then let n tend to 1).
%C For related results see A142979 and A142983.
%D Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%H Seiichi Manyama, <a href="/A142988/b142988.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.
%H Hyun Kwang Kim, <a href="http://dx.doi.org/10.1090/S0002-9939-02-06710-2">On Regular Polytope Numbers</a>, Proc. Amer. Math. Soc., 131 (2003), 65-75.
%F a(n) = (n + 2)!*Sum_{k = 1..n} (-1)^(k+1)/(k*(k + 1)*(k + 2)).
%F Recurrence: a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n + 1)*(n + 3)*a(n).
%F The sequence b(n) := (1/2)*(n + 2)!*p(n+2) satisfies the same recurrence with b(1) = 3, b(2) = 12.
%F Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 + 1*3/(3 + 2*4/(3 + 3*5/(3 + ... + (n - 1)*(n + 1)/3)))), for n >= 2.
%F Limit_{n -> oo} a(n)/b(n) = 1/(3 + 1*3/(3 + 2*4/(3 + 3*5/(3 + ... + (n - 1)*(n + 1)/(3 + ...))))) = 2*Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)*(k + 2)) = 4*log(2) - 5/2.
%p a := n -> (n+2)!*sum ((-1)^(k+1)/(k*(k+1)*(k+1)), k = 1..n): seq(a(n), n = 1..20);
%Y Cf. A142979, A142983, A142989, A142990, A142991.
%K easy,nonn
%O 1,2
%A _Peter Bala_, Jul 17 2008
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