login
a(n) is the nearest integer to binomial(n,n/2) = n!/((n/2)!)^2.
0

%I #11 Sep 06 2018 11:52:35

%S 1,1,2,3,6,11,20,37,70,132,252,482,924,1778,3432,6639,12870,24994,

%T 48620,94716,184756,360821,705432,1380533,2704156,5301248,10400600,

%U 20419624,40116600,78861995,155117520,305272239,601080390,1184086260,2333606220,4601020897,9075135300

%N a(n) is the nearest integer to binomial(n,n/2) = n!/((n/2)!)^2.

%C If we consider binomial(n,x) as a real-valued function of x, then a(n) is the maximum value of binomial(n,x) (always obtained by x = n/2), rounded. Here binomial(n,x) must be understood as Gamma(n+1)/(Gamma(x+1)*Gamma(n-x+1)).

%C A093581 actually has already mentioned a geometric interpretation of this sequence.

%F a(2n) = A000984(n), a(2n+1) = round((2^(4n+2)*(n!)^2)/(Pi*(2n+1)!)) = round(2^(4n+2)/(Pi*A002457(n))) = round(A093581(n)/(Pi*A001803(n))).

%F a(n) ~ 2^n/sqrt(n*Pi/2).

%e a(1) = round(1!/(0.5!)^2) = round(4/Pi) = round(1.2732395...) = 1.

%e a(3) = round(3!/(1.5!)^2) = round(32/(3*Pi)) = round(3.3953054...) = 3.

%e a(5) = round(5!/(2.5!)^2) = round(512/(15*Pi)) = round(10.8649774...) = 11.

%e a(7) = round(7!/(3.5!)^2) = round(4096/(35*Pi)) = round(37.2513512...) = 37.

%o (PARI) a(n) = round(gamma(n+1)/gamma(n/2+1)^2)

%Y Cf. A000984, A001803, A002457, A056040, A093581.

%K nonn

%O 0,3

%A _Jianing Song_, Sep 05 2018