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

Vladimir Shevelev shevelev at bgu.ac.il
Fri Jan 9 15:35:59 CET 2015


Dear Bob and Seqfans,

For a prime p, denote by V_p(x) the number of positive integers
not exceeding x for which p is the smallest prime divisor. Then for p>=3,
lim{n->infinity}V_p(n)/n=(1/p)*prod{2<=q<p, q primes}(1 - 1/q)  (*)
In particular,
sum{p>=3} (1/p)*prod{2<=q<p, q primes}(1 - 1/q)=1/2.
An approximation of (*) (at least for large p) is e^(-c)/ (p*ln(p)), where c=0.577...
is the Euler constant.

Best regards,
Vladimir

________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Bob Selcoe [rselcoe at entouchonline.net]
Sent: 09 January 2015 01:04
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Question about limit of 1/(k*Pi(k))

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?

Cheers,
Bob

--------------------------------------------------
From: "Jack Brennen" <jfb at brennen.net>
Sent: Thursday, January 08, 2015 11:29 AM
To: <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Question about limit of 1/(k*Pi(k))

> Here's your sum:
>
>    http://oeis.org/A124012
>
> Or 0.848969034043...
>
> - Jack
>
>
> On 1/8/2015 9:11 AM, Olivier Gerard wrote:
>> On Thu, Jan 8, 2015 at 5:10 PM, Bob Selcoe <rselcoe at entouchonline.net>
>> wrote:
>>
>>> Hello Seqfans,
>>>
>>> Is the following true, and if so, a known property?
>>>
>>> Where Pi(k) is the k-th prime and k = 1..inf:
>>>
>>> Sum_(1/(k*Pi(k))) = 1
>>>
>>> i.e., 1/2 + 1/6 + 1/15 + 1/28 + 1/55...
>>>
>>
>> No, this is not true, by a significant margin.
>>
>> This sum is strictly inferior to
>>
>> Sum_(1/(k*(k+1)))  which is equal to 1.
>>
>>
>> Olivier
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>>
>>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>

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