[seqfan] Re: Question about limit of 1/(k*Pi(k))

[Olivier Gerard]

On Fri, Jan 9, 2015 at 12:04 AM, Bob Selcoe <rselcoe at entouchonline.net>
wrote:

> Hi Seqfans,
>
> Thanks again, Jack (and Olivier).
>
> Ultimately, I'm trying to determine ratios of the natural numbers, by sets
> whose smallest factor is Pi(k).
>
> It is certainly the case that the sum of these ratios, whatever they are,
> equals 1.
>
> So for Pi(1) = 2, the set is the even numbers.  1/2 of the natural numbers
> are even.
>
> Pi(2) = 3, the set is odd multiples of 3 == 1/6  of the natural numbers.
> Pi(3) = 5, the set is A084967:  {5, 25, 35, 55, 65, 85, 95, 115, 125, 145,
> 155, 175...} == 1/15
> Pi(4) = 7, the set is  A084968:  {7, 49, 77, 91, 119...}.  This appears to
> be == 4/105.
> Pi(5) = 11, the set is A084969.  The ratio is ???
>
> Is there some equation for this?  A way to estimate?
>
>
Yes, there is a recursive definition and you are very close.

Let's take your set for p=3. The idea is that you have 1 representing all
integers,
you subtract from 1 the previous sets (here it is only 1/2 for the even
integers)
and then you reason that every 3rd integer will be divisible by 3.

So your fractions PiF(n) are defined by

PiF(n) = (1 - sum( PiF(i), i=1..n-1))/prime(n)

 ( 1/2, 1/6, 1/15, 4/105, 8/385, 16/1001, 192/17017, 3072/323323,  ... )

you can also express this as a product

PiF(n) = prod( 1-1/prime(i), i=1..n-1)/prime(n)


The numerators of these fractions are A038110 and A038111 in the
Encyclopedia
where you will find other expressions and further references on this
subject.


Olivier