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?
> > [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? Then no, P cannot depend upon y. Please clarify what
you mean by the comment "... <= sqr 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?
> > I had heard thios is true. How can I persuade him? Incidentally, he
> > told me about some formulation for finding that value v. Was it
> > developed by Euler? Can anyone remind me of what I'm thinking of?
> >
How am I to know? You haven't even persuaded me you're saying anything.
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