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a(n) = n*binomial(2n, n) - 4^n.
1

%I #12 Nov 14 2016 00:27:18

%S -1,-2,-4,-4,24,236,1448,7640,37424,175436,798984,3565448,15672656,

%T 68098936,293196944,1253020976,5322318944,22491436556,94632958664,

%U 396682105256,1657418948624,6905368852136,28697991157424,119000162557136

%N a(n) = n*binomial(2n, n) - 4^n.

%C Known to be >= 0 for n>3.

%D Hojoo Lee, Posting to Number Theory List, Feb 18 2002.

%H Harvey P. Dale, <a href="/A068554/b068554.txt">Table of n, a(n) for n = 0..1000</a>

%F From _Robert Israel_, Nov 13 2016: (Start)

%F a(n) = A005430(n) - A000302(n).

%F G.f.: (2*x-sqrt(1-4*x))/(1-4*x)^(3/2).

%F a(n) = ((16*(n-2))*(2*n-5)*a(n-3)-(4*(8*n^2-23*n+18))*a(n-2)+(2*(5*n-4))*(n-1)*a(n-1))/(n*(n-1)). (End)

%p seq(n*binomial(2*n,n)-4^n,n=0..40); # _Robert Israel_, Nov 13 2016

%t Table[n*Binomial[2n,n]-4^n,{n,0,30}] (* _Harvey P. Dale_, Nov 17 2012 *)

%Y Cf. A000302, A005430.

%K sign

%O 0,2

%A _N. J. A. Sloane_, Mar 23 2002