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A142995
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a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 3)*a(n) - n^4*a(n-1), n >= 1.
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13
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0, 1, 7, 89, 1836, 56164, 2390832, 135213840, 9809203968, 888117094656, 98167241088000, 13010123816064000, 2036436482119680000, 371699564417796096000, 78251077775510986752000
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OFFSET
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0,3
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COMMENTS
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This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + m^2 + m + 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^2 for the constant zeta(2). For other cases see A001819 (m=0), A142996 (m=2), A142997 (m=3) and A142998 (m=4).
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^2*p_m(k-1)*p_m(k)), where p_m(x) := Sum_{k = 0..m} C(m,k)^2*C(x+k,m) = Sum_{k = 0..m} C(m,k)*C(m+k,k)*C(x,k) is the Ehrhart polynomial of the polytope formed from the convex hull of a root system of type A_m (equivalently, the polynomial that generates the crystal ball sequence for the A_m lattice [Bacher et al.]).
The first few are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 3*x^2 + 3*x + 1 and p_3(x) = (10*x^3 + 15*x^2 + 11*x + 3)/3. The o.g.f. for the p_m(x) is ((1-t^2)^x/(1-t)^(2x+1))*Legendre_P(x,(1+t^2)/(1-t^2)) = 1 + (2*x+1)*t + (3*x^2+3*x+1)*t^2 + ... [Gogin & Hirvensalo, Theorem 1 with N = -1].
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*x^2 + 2*x + m^2 + m + 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) = m^2 + m + 1. Hence the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = Sum_{k>=1} 1/(k^2*p_m(k-1)*p_m(k)) = 1/((m^2 + m + 1) - 1^4/((m^2 + m + 5) - 2^4/((m^2 + m + 13) - ... - n^4/((2*n^2 + 2*n + m^2 + m + 1) - ...)))) = 2*Sum_{k>=1} (-1)^(k+1)/(m+k)^2. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).
For related results see A142999. For corresponding results for the constants e, log(2) and zeta(3) see A000522, A142979 and A143003 respectively.
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REFERENCES
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Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
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LINKS
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FORMULA
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a(n) = n!^2*p(n)*Sum_{k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = 2*n+1. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 3)*a(n) - n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 3. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 - 1^4/(7 - 2^4/(15 - 3^4/(27 - ... - (n-1)^4/(2*n^2 - 2*n + 3))))), for n >= 2. Lim_{n -> infinity} a(n)/b(n) = 1/(3 - 1^4/(7 - 2^4/(15 - 3^4/(27 - ... - n^4/((2*n^2 + 2*n + 3) - ...))))) = Sum_{k>=1} 1/(k^2*(4*k^2 - 1)) = 2 - zeta(2).
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MAPLE
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p := n -> 2*n+1: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);
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MATHEMATICA
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a[n_] := -1/6*n!^2*(2*n*(Pi^2-12) + Pi^2 - 6*(2*n+1)*PolyGamma[1, n+1]) // Simplify; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Mar 06 2013 *)
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CROSSREFS
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Cf. A000522, A001819, A003215 (A_2 lattice), A005902 (A_3 lattice), A008384 (A_4 lattice), A008386 (A_5 lattice), A108625, A142979, A142996, A142997, A142998, A143003.
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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