%I #35 Jul 12 2024 05:32:39
%S 2,20,252,3432,48620,705432,10400600,155117520,2333606220,35345263800,
%T 538257874440,8233430727600,126410606437752,1946939425648112,
%U 30067266499541040,465428353255261088,7219428434016265740
%N Bisection of A000984.
%H Vincenzo Librandi, <a href="/A099976/b099976.txt">Table of n, a(n) for n = 0..100</a>
%F a(n) = binomial(4n+2, 2n+1). - _Emeric Deutsch_, Dec 20 2004
%F G.f.: 2*sqrt(2)/sqrt(1-16*x)/sqrt(1+sqrt(1-16*x)) = 2 + 60*x/(G(0)-30*x) where G(k)= 2*x*(4*k+3)*(4*k+5) + (2*k+3)*(k+1)- 2*x*(k+1)*(2*k+3)*(4*k+7)*(4*k+9)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - _Sergei N. Gladkovskii_, Jul 14 2012
%F G.f. A(x) satisfies A(x^2) = F'(x)/F(x), where F(x) = C(x)/C(-x) and C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. - _Peter Bala_, May 15 2023
%F From _R. J. Mathar_, Jul 11 2024: (Start)
%F D-finite with recurrence n*(2*n+1)*a(n) -2*(4*n-1)*(4*n+1)*a(n-1)=0.
%F a(n) = 2*A002458(n).
%F G.f.: 2* 2F1(3/4,5/4; 3/2 ; 16*x).
%F Conjecture: A000265(a(n)) = A063079(n+1), odd part of a(n). (End)
%F a(n) / (2*n+2) = A024492(n). - _R. J. Mathar_, Jul 12 2024
%p seq(binomial(4*n+2,2*n+1),n=0..20); # _Emeric Deutsch_, Dec 20 2004
%t Array[Binomial[4*# + 2, 2*# + 1] &, 20, 0] (* _Paolo Xausa_, Jul 11 2024 *)
%o (Magma) [Binomial(4*n+2, 2*n+1): n in [0..20]]; // _Vincenzo Librandi_, May 22 2011
%Y Cf. A000108, A000265, A000984, A001448, A002458, A063079.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_, Nov 19 2004
%E More terms from _Emeric Deutsch_, Dec 20 2004