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%I #20 May 03 2018 17:04:29
%S 0,1,3,8,24,74,237,781,2630,9020,31375,110442,392685,1408249,5087870,
%T 18501347,67662072,248703832,918291072,3404396173,12667520643,
%U 47292077070,177093735411,665005047259,2503548413211,9447352502685,35728169464702,135390957971502,514026687891806
%N a(n) = floor(C(n + 1/2)), where C = A000108.
%C A000108 interleaved with this sequence gives floor(C(n/2)).
%H Paolo P. Lava, <a href="/A303607/b303607.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) ~ 2^(2*n + 1) / (sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Apr 27 2018
%e C(n + 1/2)*Pi gives: 2^3/3, 2^6/(3*5), 2^10/(3*5*7), 2^13/(5*7*9), 2^18/(5*7*9*11), 2^21/(7*9*11*13), 2^25/(5*7*9*11*13), ...
%p P:=proc(n) floor(evalf(binomial(2*n+1,n+1/2)/(n+3/2),1200)); end: seq(P(i),i=0..28); # _Paolo P. Lava_, May 03 2018
%t Table[Floor[CatalanNumber[n + 1/2]], {n, 0, 30}]
%Y Cf. A000108, A000796.
%K nonn
%O 0,3
%A _Bruno Berselli_, Apr 27 2018
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