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a(n) = Sum_{k=0..2n}(k!*(2n - k)!)/(floor(k/2)!*floor((2n - k)/2)!)^2.
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%I #18 Feb 08 2020 23:46:33

%S 1,5,28,160,896,4864,25600,131072,655360,3211264,15466496,73400320,

%T 343932928,1593835520,7314866176,33285996544,150323855360,

%U 674309865472,3006477107200,13331578486784,58823872086016,258385232527360,1130297953353728,4925812092436480

%N a(n) = Sum_{k=0..2n}(k!*(2n - k)!)/(floor(k/2)!*floor((2n - k)/2)!)^2.

%H Colin Barker, <a href="/A327999/b327999.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (12,-48,64).

%F a(n) = 4^n*(n^2 + n + 8)/8.

%F a(n) = [x^n] (-16*x^2 + 7*x - 1)/(4*x - 1)^3.

%F a(n) = n! [x^n] exp(4*x)*(2*x^2 + x + 1).

%F a(n) = a(n-1)*4*(8 + n + n^2)/(8 - n + n^2).

%F a(n) = A328000(2*n).

%F From _Colin Barker_, Feb 05 2020: (Start)

%F a(n) = 12*a(n-1) - 48*a(n-2) + 64*a(n-3) for n>2.

%F a(n) = 2^(2*n - 3)*(8 + n + n^2).

%F (End)

%t LinearRecurrence[{12, -48, 64}, {1, 5, 28}, 24] (* _Michael De Vlieger_, Feb 07 2020 *)

%o (PARI) Vec((1 - 7*x + 16*x^2) / (1 - 4*x)^3 + O(x^25)) \\ _Colin Barker_, Feb 05 2020

%o (PARI) apply( {A327999(n)=(n^2+n+8)<<(2*n-3)}, [0..25]) \\ _M. F. Hasler_, Feb 07 2020

%Y Even bisection of A328000.

%K nonn,easy

%O 0,2

%A _Peter Luschny_, Oct 01 2019