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

Bob Selcoe rselcoe at entouchonline.net
Fri Jan 9 06:37:12 CET 2015


Many thanks, Olivier.

I wouldn't have guessed the limit of the sum of such a function would be 1; 
and at first glance, it would appear that these ratios are inferior to 
1/(n*(n+1) (and hence sum < 1, as you pointed out).  But looking at some of 
the larger numbers (n=250, for example), this is clearly not the case.

Quite interesting!

I'll add a comment about the sum to one of the OEIS sequences you reference.

Best Wishes,
Bob


--------------------------------------------------
From: "Olivier Gerard" <olivier.gerard at gmail.com>
Sent: Thursday, January 08, 2015 10:57 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Question about limit of 1/(k*Pi(k))

> 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
>
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