Expand Messages
• Jan 3, 2001
On Wed, 03 January 2001, Dick Boland wrote:
> Sorry Phil,
>
> It was a typo. k*(k+1)/2 primes <= (p(k^2)+1)/2 and k*(k-1)/2 primes between ((p(k^2)+1)/2+1) and p(k^2). You won't find a counter example there.

I know it was a typo, which is why in the data below I correct that typo. And with the typo corrected I do find counterexamples. Check again please.

Phil

> -Dick
>
>
>
>
> Phil Carmody <fatphil@...> wrote:
> On Wed, 03 January 2001, Dick Boland wrote:
> > Yes I can. The distribution function is simply stated as follows,
> >
> > For any integer k>4, the first k^2 primes will be exactly distributed as follows:
> >
> > k*(k+1) primes between 1 and (p(k^2)+1)/2, and the remaining k*(k-1) primes will be distributed between ((p(k^2)+1)/2+1) and p(k^2).
>
> k*(k+1) + k*(k-1) == 2k^2
>
> So you seem to be out by a factor of 2 somewhere.
>
> Factoring in that factor of two...
>
> Table[{k,
> k^2,
> Prime[k^2],
> (Prime[k^2]+1)/2,
> PrimePi[(Prime[k^2]+1)/2],
> k*(k+1)/2},
> {k, 4, 8}]
>
> {{4, 16, 53, 27, 9, 10},
> {5, 25, 97, 49, 15, 15},
> {6, 36, 151, 76, 21, 21},
> {7, 49, 227, 114, 30, 28},
> {8, 64, 311, 156, 36, 36}}
>
> You seem to be saying the last two columns are the same.
> I beg to differ.
>
> Let's look a bit further, at the data for k=100000-100005:
>
> {
> {100000, 10000000000, 252097800623, 126048900312, 5141644677,
> 5000050000},
> {100001, 10000200001, 252103045511, 126051522756, 5141747035,
> 5000150001},
> {100002, 10000400004, 252108316073, 126054158037, 5141850524, 5000250003},
> {100003, 10000600009, 252113577847, 126056788924, 5141953182, 5000350006},
> {100004, 10000800016, 252118846391, 126059423196, 5142056263, 5000450010},
> {100005, 10001000025, 252124112327, 126062056164, 5142159097, 5000550015}
> }
>
> The last 2 columns really aren't that similar.
>
>
> Phil
>
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Mathematics should not have to involve martyrdom;
Support Eric Weisstein, see http://mathworld.wolfram.com
Find the best deals on the web at AltaVista Shopping!
http://www.shopping.altavista.com
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