%I #4 Jun 16 2012 01:02:07
%S 1,1,3,1,9,3,1,3,3,3,1,3,3,3,15,3,3,9,3,9,1,1,15,27,9,3,15,1,5,5,3,9,
%T 9,9,3,3,1,3,45,9,15,1,15,15,15,45,9,81,9,5,5,1,25,5,3,3,5,5,3,15,27,
%U 9,21,3,81,9,3,15,1,3,75,81,9,9,135,27,25,15
%N A048784(n) / 2^A213594(n).
%C a(n) = tau(binomial(2*n,n)) / 2^k, where tau = number of divisors (A000005) and k is the greatest possible integer.
%e a(7) = 1 because A048784(7) / 2^5 = 32 / 32 = 1 is integer.
%p with(numtheory): for n from 1 to 100 do:ii:=0:for k from 500 by -1 to 1 while(ii=0) do: x:=evalf(tau(binomial(2*n,n))/2^k):if x=floor(x) then ii:=1: printf(`%d, `,floor(x)):else fi:od:od:
%Y Cf. A048784, A213594.
%K nonn
%O 1,3
%A _Michel Lagneau_, Jun 15 2012