0,3

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1 )*a(n) - n^6*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} 1/k^3 for Apery's constant zeta(3). For other cases see A066989 (m=0), A143004 (m=2), A143005 (m=3) and A143006 (m=4).

The solution to the general recurrence may be expressed as a sum: a(n) = n!^3*p_m(n)*sum {k = 1..n} 1/(k^3*p_m(k-1)*p_m(k)), where p_m(x) = sum {k = 0..n} C(2*k,k)^2*C(n+k,2*k)*C(x+k,2*k) is a polynomial in x of degree 2*m.

The first few are p_0(x) = 1, p_1(x) = 2*x^2 + 2*x + 1, p_2(x) = (3*x^4 + 6*x^3 + 9*x^2 + 6*x + 2)/2 and p_3(x) = (10*x^6 + 30*x^5 + 85*x^4 + 120*x^3 + 121*x^2 + 66*x + 18)/18. For fixed n, the sequence [p_n(k)]k>=0 is the crystal ball sequence for the product lattice A_n x A_n. See A143007 for the table of values [p_n(k)] n,k >= 0. Observe that [p_n(n)] n >= 0 is the sequence of Apery numbers A005259.

The reciprocity law p_m(n) = p_n(m) holds for nonnegative integers m and n. In particular we have p_m(1) = 2*m^2+2*m+1 and p_m(2) = (3*m^4+6*m^3+9*m^2+6*m+2)/2.

The polynomial p_m(x) is the unique polynomial solution of the difference equation (x+1)^3*f(x+1) + x^3*f(x-1) = (2*x+1)*(x^2+x+2*m^2+2*m+1)*f(x), normalised so that f(0) = 1. The reciprocity law now yields the Apery-like recursion m^3*p_m(x) + (m-1)^3*p_(m-2)(x) = (2*m-1)*(m^2-m+1+2*x^2+2*x)*p_(m-1)(x).

The polynomial functions p_m(x) 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 (adapt the proof of the lemma on p.4 of [BUMP et al.]).

The general recurrence in the first paragraph above has a second solution b(n) = n!^3*p_m(n) with initial conditions b(0) = 1, b(1) = 2*m^2+2*m+1. Hence the behaviour of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} 1/(k^3*p_m(k-1)*p_m(k)) = 1/((2*m^2+2*m+1)- 1^6/(3*(2*m^2+2*m+3)- 2^6/(5*(2*m^2+2*m+7)- 3^6/(7*(2*m^2+2*m+13)-...)))) = sum {k = 1..inf} 1/(m+k)^3. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii)].

For the corresponding results for the constant zeta(2) see A142995. For corresponding results for the constant log(2) see A142979 and A142992 .

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 1-19.

a(n) = n!^3*p(n)*sum {k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = 2*n^2+2*n+1 = A001844(n). Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+5)*a(n) - n^6*a(n-1). The sequence b(n):= n!^3*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 5. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5- 1^6/(21- 2^6/(55- 3^6/(119-...- (n-1)^6/((2*n-1)*(n^2-n+5)))))), for n >=2. The behaviour of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} 1/(k^3*(4*k^4+1)) = 1/(5- 1^6/(21- 2^6/(55- 3^6/(119-...- n^6/((2*n+1)*(n^2+n+5)-...))))) = zeta(3) - 1, where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 1].

p := n -> 2*n^2+2*n+1: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..14)

Cf. A066989, A008288, A108625, A142979, A142992, A142995, A143004, A143005, A143006, A143007.

Sequence in context: A138473 A072678 A012153 this_sequence A012183 A012230 A086098

Adjacent sequences: A143000 A143001 A143002 this_sequence A143004 A143005 A143006

easy,nonn

Peter Bala (pbala(AT)toucansurf.com), Jul 19 2008