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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