Re: F(N)=The number of primes that do not divide N

Gerry Myerson wrote:
> In article <1114714563.897970.313970@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> davecornwell@xxxxxxxxxxx wrote:
>
> > You are correct, my notation is not very good.
> > I meant to state
> > S(k,p1,p2,p3,,pk)={N: p1 and p2 and .. and pk do not divide N,
> > and these are the only primes that do not divide N,
> > and these primes are less than sqrt(N)}
> >
> > For example consider
> > F(N)= 5,
> >
> > S(5,2,3,5,7,11)={N: such that there are 5 primes that do not divide
N
> > and they are 2,3,5,7 and 11}
> >
> > For general S:
> > Is S finite or infinite?
> > What is the smallest value in S?
> > What is the largest value?
>
> The product of the primes less than N is roughly exp(N).
> The product P of the primes less than sqrt N, not including
> the primes 2, 3, 5, 7, 11, is roughly (1/2310) exp(sqrt N).
> For N to be in your S(5; 2, 3, 5, 7, 11) it must be divisible
> by P, hence no less than P. But (1/2310) exp(sqrt N) grows
> much faster than N, so your S must be finite.
>
> I think you'll have a better idea of what's going on if
> you try to find all the elements of S(5; 2, 3, 5, 7, 11).
>
> --
> Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)

Thanks for your analysis. This question comes in to play in my earlier
entry concerning the Goldbach Conjecture a few days ago where we get
summations over primes that do not divide 2n. There I am interested in
keeping the set of primes that do not divide 2n fixed, in order to fix
the Dirichlet Characters involved in a certain sum involving the Zeta
function and linear combinations of Dirichlet L functions. I would like
to take a limit as s->1 (as in Zeta(s), L(s,Chi)) and also let 2n
->inf, over those 2n with fixed primes that do not divide it. If the
set S that I have been talking about is finite then I cannot take the
latter limit since there is an upper bound on 2n. Instead of trying to
show a certain sum over the inverses of Goldbach primes for 2n diverges
(hence showing there are an infinite number of them in the limit), I
would be left with the problem of showing it is non zero.

Dave Cornwell

.

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