Automatic Sets, was: Re: Question on Sequence A046907 from Sloane's Database

[Antti Karttunen]

David W. Wilson wrote:
> 
> "James A. Sellers" wrote:...
> > I have a question on sequence A046907, which gives the values of n for which binomial(2n-1, n) is squarefree.   The 
sequence is 1,2,3,4,6,9,10,12,36.  I have
> >
> > James A. Sellers
> > Associate Professor, Mathematics
> > Cedarville College
> >
> > sellersj at cedarville.edu
> > http://www.cedarville.edu/dept/sm/jas_www.htm
> 
> This is one a broad set of problems which devolve to a unsolved general
> problem.  The general problem is:
> 
> Is the intersection of a given a-automatic and b-automatic set finite?
> 
> A k-automatic set is a set of integers whose base-k representations form
> a regular language, that is, a language accepted by a finite automaton
> or state machine.
> 
> If bases a and b are incompatible (do not have a common power), and if
> an a-automatic set Sa and b-automatic set Sb are both of density 0
> over the integers, it is believed that Sa intersect Sb is finite
> (or something along this line).
> 
> The current problem is whether f(n) = (2n-1 choose n) is squarefree for
> a finite number of n.  It turns out that f(n) is divisible by 4 unless
> n belongs to a 2-automatic set S2 (which happens to be the set of
> numbers
> whose binary representations contain at most two 1's).  Similarly, f(n)
> is divisibly by 9 unless n belongs to a 3-automatic set S3 (much harder
> to describe). 
I submitted the former set/sequence into Encyclopedia in 
July: A048645 1,2,3,4,5,6,8,9,10,12,16,17,18,20,24
for which a reasonably nice formula exists:
a(n) = 2^A003056[n]+2^A002262[n-1] (after n=1).
 
So how much harder is this latter, 3-automatic set to 
describe? Has it its own sequence in the Encyclopedia yet?
 
BTW. What is the exact definition of "density 0 over the 
integers" ?
lim count_of_terms_of_set_below(n)/n -> 0 as n -> infinity?

>  If f(n) is squarefree, therefore, n must belong to
> S = S2 intersect S3.  It is very probable that S is finite, but no
> proof is known. 
Is this just a necessary condition, or alone sufficient?
(I.e. I guess that infinitely many other k^2 could divide 
f(n) than just 2^2=4, 3^2=9, ((2*3)^2)=36 and so on?)
 
Is there any reference to the topic, on the Net or in the 
journals?
  
Yours,
 
Antti Karttunen