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A142999
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a(0) = 0, a(1) = 1; for n>1, a(n+1) = (2*n+1)*a(n) + n^4*a(n-1).
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9
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0, 1, 3, 31, 460, 12076, 420336, 21114864, 1325949696, 109027627776, 10771080883200, 1316468976307200, 187978181665996800, 31997755234356019200, 6232784237890147123200, 1409976507981835100160000, 359243973790625586216960000, 104259271562188189469245440000
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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 = 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..inf} (-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)*(2x+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 -> infinity a(n)/b(n) = sum {k = 1..inf} (-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..inf} 1/(m+k)^2. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 30].
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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*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. Lim n -> infinity a(n)/b(n) = 1/(1+ 1^4/(3+ 2^4/(5+ 3^4/(7+...+ n^4/((2*n+1)+...))))) = sum {k = 1..inf} (-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
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MAPLE
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a := n -> n!^2*add ((-1)^(k+1)/k^2, k = 1..n): seq(a(n), n = 0..20);
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MATHEMATICA
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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 *)
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 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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