%I #38 Apr 21 2024 09:59:15
%S 1,-1,3,-10,35,-126,462,-1716,6435,-24310,92378,-352716,1352078,
%T -5200300,20058300,-77558760,300540195,-1166803110,4537567650,
%U -17672631900,68923264410,-269128937220,1052049481860,-4116715363800,16123801841550
%N a(n) = binomial(2*n-1, n)*(-1)^n.
%H G. C. Greubel, <a href="/A110556/b110556.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = A110555(2*n,n), central terms in triangle A110555;
%F a(n) = A088218(n)*(-1)^n.
%F E.g.f.: E(x) = 1 - x/(G(0)+2*x) ; G(k) = (k+1)^2 - 2*x*(2*k+1) + 2*x*(2*k+3)*((k+1)^2)/G(k+1) ; (continued fraction). - _Sergei N. Gladkovskii_, Dec 21 2011
%F a(n) = coefficient of x^n of (1 / (1 + x))^n where 1 / (1 + x) is the g.f. of A033999. - _Michael Somos_, May 21 2013
%F HANKEL transform is A000027. HANKEL transform of a(n+1) is A033999(n+1). - _Michael Somos_, May 21 2013
%F E.g.f.: (1 + exp(-2*x) * BesselI(0,2*x)) / 2. - _Ilya Gutkovskiy_, Nov 03 2021
%F From _Peter Bala_, Feb 14 2024: (Start)
%F a(n) = binomial(-n, n).
%F O.g.f.: A(x) = (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) = 1/ (2 - c(-x)), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
%F The g.f. A(x) satisfies A(x/(1 - x)^2) = 1/(1 + x). (End)
%F a(n) ~ (-1)^n*2^(2*n-1)/sqrt(n*Pi). - _Stefano Spezia_, Apr 18 2024
%e 1 - x + 3*x^2 - 10*x^3 + 35*x^4 - 126*x^5 + 462*x^6 - 1716*x^7 + 6435*x^8 - ...
%t Table[Binomial[2n-1,n](-1)^n,{n,0,30}] (* _Harvey P. Dale_, Apr 01 2012 *)
%t a[ n_] := (-1)^n Binomial[ 2 n - 1, n] (* _Michael Somos_, May 21 2013 *)
%t a[ n_] := Binomial[ -n, -2 n] (* _Michael Somos_, May 21 2013 *)
%t a[ n_] := SeriesCoefficient[(1 + x)^-n, {x, 0, n}] (* _Michael Somos_, May 21 2013 *)
%o (PARI) {a(n) = if( n<0, 0, polcoeff( (1 + x)^-n + x * O(x^n), n))} /* _Michael Somos_, May 21 2013 */
%Y Another version of A001700, which is the main entry.
%Y Cf. A000027, A000108, A000984, A007318, A033999, A088218, A110555.
%K sign,easy
%O 0,3
%A _Reinhard Zumkeller_, Jul 27 2005