square-free ternary numbers

[hv at crypt.org]

Close, but you dropped the ball at the end.  One expects half of all
cases to have an increase in the limit.  Since we're looking at the 
ones
that do, we would expect the limit of a(n)/n to be 2.

I don't have a proof of this, either; just the same intuitive argument
that Joshua presents.

Franklin T. Adams-Watters

-----Original Message-----
From: Joshua Zucker <joshua.zucker at gmail.com>

On Sun, Mar 2, 2008 at 1:55 PM, Maximilian Hasler
<maximilian.hasler at gmail.com> wrote:
>  >  So I wonder, can we determine the limit, as n ->
>  >  infinity,  of a(n)/n?

>  yes : it decreases steadily.

We're asking for places where an even number has fewer factors than a
neighboring odd number.  Intiutively (that is, nonrigorously)
speaking, as numbers have more and more factors, things tend to
"average out" -- that is, we know the density of primes drops, but so
does the density of numbers that are products of 2 primes, or of the
form p^2 -- in general, there are fewer small values of d(n).  So,
sticking a 2 in -- which, all else being equal, doubles d(n), in the
sense that for odd k, d(2k) = 2d(k) -- is going to have an increase
which is proportionally more and more important as the numbers get
larger.

So it makes sense to me that this would be asymptotic to 0.  Though of
course I suspect that a(n) keeps increasing, just more and more
slowly.

I'll leave it to someone who knows what they're talking about to give
a more precise idea about the asymptotics of this thing.

--Joshua Zucker








Re A033286,A016688
Does an n exist :  

n*p(n) / (n+p(n)) is an integer ?

p(n) is  n-th prime

Ctibor O. Zizka