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%I #17 Nov 02 2016 12:41:48
%S 1,6,41,372,4077,53106,795645,13536360,257055705,5400196830,
%T 124170067665,3104906420700,83818724048325,2431059231544650,
%U 75354930324303525,2486926158748693200,87036225272850632625,3220532233879435917750,125594424461427237941625
%N The second column of the ED4 array A167584.
%H G. C. Greubel, <a href="/A167588/b167588.txt">Table of n, a(n) for n = 1..150</a>
%F a(n) = (1/2)*(-1)^(n)*(2*n-3)!!*(n+(4*n^2-1)*Sum_{k=0..n-1} ((-1)^(k+n)/(2*k+1))).
%F From _Peter Bala_, Nov 01 2016: (Start)
%F a(n) = (2*n + 1)!! * Sum_{k = 0..n-1} (-1)^(k-1)/((2*k - 1)*(2*k + 1)*(2*k + 3)).
%F a(n) ~ Pi * 2^(n-3/2) * ((n+1)/e)^(n+1).
%F E.g.f.: (4*x*sqrt(1 - 4*x^2) + 2*arcsin(2*x))/(8*(1 - 2*x)^(3/2)).
%F a(n) = 6*a(n-1) + (2*n - 5)*(2*n - 1)*a(n-2) with a(0) = 0, a(1) = 1.
%F The sequence b(n) := (2*n + 1)!! = (2*n + 2)!/((n + 1)!*2^(n+1)) satisfies the same recurrence with b(0) = 1 and b(1) = 3. This leads to the continued fraction representation a(n) = b(n)*[ 1/(3 - 3/(6 + 5/(6 + 21/(6 + ... + (2*n - 5)*(2*n - 1)/(6))))) ] for n >= 2.
%F As n -> infinity, a(n)/(A001147(n+1)) -> 1/2!*Pi/4 = 1/(3 - 3/(6 + 5/(6 + 21/(6 + ... + (2*n - 5)*(2*n - 1)/(6 + ...))))). Compare with the generalized continued fraction representation Pi = 3 + 1^2/(6 + 3^2/(6 + 5^2/(6 + ...))). See A142970. (End)
%t Table[(1/2)*(-1)^(n)*(2*n - 3)!!*((n) + (4*n^2 - 1)*Sum[(-1)^(k + n)/(2*k + 1), {k, 0, n - 1}]), {n, 1, 50}] (* _G. C. Greubel_, Jun 17 2016 *)
%Y Equals the second column of the ED4 array A167584.
%Y Other columns are A024199 and A167589.
%Y Cf. A007509 and A025547 (the sum((-1)^(k+n)/(2*k+1), k=0..n-1) factor), A001147, A142970.
%K easy,nonn
%O 1,2
%A _Johannes W. Meijer_, Nov 10 2009
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