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A142988 a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1)+(n+1)*(n+3)*a(n). 5

%I

%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..inf} (-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). The solution to the general recurrence may be expressed as a sum: 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). 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..inf} 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). Hence the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = 2*sum {k = 1..inf} (-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)+...))))). The final 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 D. Bump, K. Choi, P. Kurlberg and J. Vaaler, <a href="http://citeseer.ist.psu.edu/175986.html">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. 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. Lim n -> infinity a(n)/b(n) = 1/(3+1*3/(3+2*4/(3+3*5/(3+...+(n-1)*(n+1)/(3+...))))) = 2*sum {k = 1..inf} (-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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