"William Elliot" <marsh@[EMAIL PROTECTED]
> wrote in message
news:Pine.BSI.4.58.0807192335480.2040@[EMAIL PROTECTED]
> On Sat, 19 Jul 2008, Jack wrote:
>
>> > if you take the product of any set of primes p in a set P, and
>> > enumerate
>> > the number of integers, in the interval of that length, for which no
p
>
> The number of integers in [a,b] is max{ 0, b - a + 1 }
>
>> > divides n, then you get a value v, that, with increasing y for
>> > intervals
>
> "For which no p"? What's p? Do you actually mean for which no p in P?
> Then say so! What's n? A given integer?
You guess well.
>> > [1,y] and a set P whose members are all the primes <=sqrt(y),
>
> Just a nanosec now. P depends upon y? Is this the set P you first
> introduced?
Of course.
Then no, P cannot depend upon y. Please clarify what
> you mean by the comment "... <= sqr y"
I didn't say <= sqr y I, said <=sqrt(y), which is pretty obviously 'less
than or equal to the square root of y'.
>> > converges on being negligibly different from the number of primes
>> > found in that interval.
>
> That doesn't make sense.
>
>> Sorry, I should have said the *pro****tion* of integers, in the interval
>> of
>> product(P), for which no p divides n, converges on being negligibly
>> different from the *pro****tion* of primes found in that interval.
>>
> Never mind. I'll guess what you're trying to say.
>
> Let P be a set of primes.
> Let n(y,P) be the number of integers n, in [1,y]
> with for all p in P and p <= sqr y, p not divide n.
>
> Then n(y,P) converges on being negligibly different from the number of
> primes found in that interval. Hm. Mostly gibberish. Let's see.
>
> Let p(y) be the number of primes in [1,y]
> Then lim(y->oo) n(y,P)/p(y) = 1.
> Hm. n(y,P) = o(p(y)) or O(p(y)) ?
>
> Does that have any semblance to your utterances?
Yes, with a question mark over the 'Hm'...
Cheers.
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