A142999 a(0) = 0, a(1) = 1; for n > 1, a(n+1) = (2*n + 1)*a(n) + n^4*a(n-1).

0, 1, 3, 31, 460, 12076, 420336, 21114864, 1325949696, 109027627776, 10771080883200, 1316468976307200, 187978181665996800, 31997755234356019200, 6232784237890147123200, 1409976507981835100160000, 359243973790625586216960000, 104259271562188189469245440000

Offset: 0

Author

Peter Bala, Jul 18 2008

Keywords

• nonn
• easy

Comments

This is the case m = 0 of the general recurrence a(0) = 1, a(1) = 1, a(n+1) = (2*m + 1)*(2*n + 1)*a(n) + n^4*a(n-1) (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^2 for the constant 1/2*zeta(2). For other cases see A143000 (m = 1), A143001 (m = 2) and A143002 (m = 3).

The solution to the general recurrence may be expressed as a sum: a(n) = n!^2*p_m(n)*Sum_{k = 1..n} (-1)^(k+1)/(k^2*p_m(k-1)*p_m(k)), where p_m(x) := Sum_ {k = 0..m} C(m,k)*C(x,k)*C(x+k,k). Note that the polynomial q_m(x) := Sum_{k = 0..m} C(m,k)*C(m+k,k)*C(x,k), obtained by interchanging the roles of m and x, may be variously described as the Ehrhart polynomial of the polytope formed from the convex hull of a root system of type A_m, the polynomial that generates the crystal ball sequence for the A_m lattice [Bacher et al.], or the discrete Chebyshev polynomial D_m(N;x) at N = -1 [Gogin & Hirvensalo]. Compare with the comments in A142995.

The first few values are p_0(x) = 1, p_1(x) = x^2 + x + 1, p_2(x) = (x^4 + 2*x^3 + 7*x^2 + 6*x + 4)/4 and p_3(x) = (x^6 + 3*x^5 + 22*x^4 + 39*x^3 + 85*x^2 + 66*x + 36)/36.

The polynomial p_m(x) is the unique polynomial solution of the difference equation (x + 1)^2*f(x+1) - x^2*f(x-1) = (2*m + 1)*(2*x + 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 (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!^2)*p_m(n) with initial conditions b(0) = 1, b(1) = 2*m + 1. 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^2*p_m(k-1)*p_m(k)) = 1/((2*m + 1) + 1^4/(3*(2*m + 1) + 2^4/(5*(2*m + 1) + ... + n^4/(((2*n + 1)*(2*m + 1) + ...)))) = (1/2)*Sum_{k >= 1} 1/(m + k)^2. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 30].

For results of a similar nature for the constants e, log(2), zeta(2) and zeta(3) see A000522, A142979, A142995 and A143003 respectively.

References

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

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et polynômes d'Ehrhart associées aux réseaux de racines, Annales de l'Institut Fourier, Tome 49 (1999) no. 3, pp. 727-762.
• R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
• D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 1-19.
• N. Gogin and M. Hirvensalo, On the generating function of discrete Chebyshev polynomials, Turku Centre for Computer Science Technical Report No. 819, (2007), 1-8.

Crossrefs

Cf. A000522, A142979, A142995, A143000, A143001, A143002, A143003.

Programs

Maple

a := n -> n!^2*add ((-1)^(k+1)/k^2, k = 1..n): seq(a(n), n = 0..20);

Mathematica

f[k_] := (k^2) (-1)^(k + 1)
t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[n - 1, t[n]]
Table[a[n], {n, 1, 18}]    (* A142999 signed *)
(* Clark Kimberling, Dec 30 2011 *)
RecurrenceTable[{a[0]==0,a[1]==1,a[n]==(2(n-1)+1)a[n-1]+(n-1)^4 a[n-2]},a,{n,20}] (* Harvey P. Dale, Apr 26 2014 *)

Formula

a(n) = (n!^2) * Sum_{k = 1..n} (-1)^(k+1)/k^2.

Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n + 1)*a(n) + (n^4)*a(n-1).

The sequence b(n) := n!^2 satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 1. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(1 + 1^4/(3 + 2^4/(5 + 3^4/(7 + ... + (n - 1)^4/(2*n - 1))))), for n >= 2.

Limit_{n -> oo} a(n)/b(n) = 1/(1 + 1^4/(3 + 2^4/(5 + 3^4/(7 + ... + n^4/((2*n + 1) + ...))))) = Sum_{k >= 1} (-1)^(k+1)/k^2 = 1/2*zeta(2).

Sum_{n>=0} a(n) * x^n / (n!)^2 = -polylog(2,-x) / (1 - x). - Ilya Gutkovskiy, Jul 15 2020

Extensions

a(0)=0 added by Vincenzo Librandi, Apr 27 2014