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

%I A142999 #47 Jun 19 2024 01:51:46
%S A142999 0,1,3,31,460,12076,420336,21114864,1325949696,109027627776,
%T A142999 10771080883200,1316468976307200,187978181665996800,
%U A142999 31997755234356019200,6232784237890147123200,1409976507981835100160000,359243973790625586216960000,104259271562188189469245440000
%N A142999 a(0) = 0, a(1) = 1; for n > 1, a(n+1) = (2*n + 1)*a(n) + n^4*a(n-1).
%C A142999 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).
%C A142999 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.
%C A142999 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.
%C A142999 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.]).
%C A142999 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].
%C A142999 For results of a similar nature for the constants e, log(2), zeta(2) and zeta(3) see A000522, A142979, A142995 and A143003 respectively.
%D A142999 Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%H A142999 Vincenzo Librandi, <a href="/A142999/b142999.txt">Table of n, a(n) for n = 0..200</a>
%H A142999 R. Bacher, P. de la Harpe and B. Venkov, <a href="https://doi.org/10.5802/aif.1689">Séries de croissance et polynômes d'Ehrhart associées aux réseaux de racines</a>, Annales de l'Institut Fourier, Tome 49 (1999) no. 3, pp. 727-762.
%H A142999 R. Bacher, P. de la Harpe and B. Venkov, <a href="https://doi.org/10.1016/S0764-4442(97)83542-2">Séries de croissance et séries d'Ehrhart associées aux réseaux de racines</a>, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
%H A142999 D. Bump, K. Choi, P. Kurlberg and J. Vaaler, <a href="http://www.cecm.sfu.ca/~choi/paper/lrh.pdf">A local Riemann hypothesis, I</a>, Math. Zeit. 233, (2000), 1-19.
%H A142999 N. Gogin and M. Hirvensalo, <a href="http://tucs.fi/publications/attachment.php?fname=TR819.pdf">On the generating function of discrete Chebyshev polynomials</a>, Turku Centre for Computer Science Technical Report No. 819, (2007), 1-8.
%F A142999 a(n) = (n!^2) * Sum_{k = 1..n} (-1)^(k+1)/k^2.
%F A142999 Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n + 1)*a(n) + (n^4)*a(n-1).
%F A142999 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.
%F A142999 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).
%F A142999 Sum_{n>=0} a(n) * x^n / (n!)^2 = -polylog(2,-x) / (1 - x). - _Ilya Gutkovskiy_, Jul 15 2020
%p A142999 a := n -> n!^2*add ((-1)^(k+1)/k^2, k = 1..n): seq(a(n), n = 0..20);
%t A142999 f[k_] := (k^2) (-1)^(k + 1)
%t A142999 t[n_] := Table[f[k], {k, 1, n}]
%t A142999 a[n_] := SymmetricPolynomial[n - 1, t[n]]
%t A142999 Table[a[n], {n, 1, 18}]    (* A142999 signed *)
%t A142999 (* _Clark Kimberling_, Dec 30 2011 *)
%t A142999 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 *)
%Y A142999 Cf. A000522, A142979, A142995, A143000, A143001, A143002, A143003.
%K A142999 nonn,easy
%O A142999 0,3
%A A142999 _Peter Bala_, Jul 18 2008
%E A142999 a(0)=0 added by _Vincenzo Librandi_, Apr 27 2014