## The number of primes in the interval (0, m)

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• David. You are skeptical about my result. But look: If the value (m) equals the sum of primes. We have: The error calculation of the number of primes in the
Message 1 of 2 , Jul 12, 2010
David. You are skeptical about my result. But look: If the value (m) equals the sum of primes. We have: The error calculation of the number of primes in the interval (0, m) (After n = 7) is negative and the steady growth of error. It is already possible to draw conclusions.

\sum\limits_{i = 1}^n {P_i } \cdot \prod\limits_{i = 1}^n {\frac{{P_i - 1}}{{P_i }}} + n - 1
• ... Dorogoi chitatel / Dear reader It s worse than that: you have no result for m 40291. For real m and positive integer n let Q(m,n) = m*prod(k=1,n,1 -
Message 2 of 2 , Jul 12, 2010
<chitatel2000@...> wrote:

> David. You are skeptical about my result.

It's worse than that: you have no "result" for m > 40291.

For real m and positive integer n let
Q(m,n) = m*prod(k=1,n,1 - 1/prime(k)) + n - 1,
pi(m) = number of primes not exceeding m.

In http://tinyurl.com/2w73pef you conjectured that
Q(m,pi(sqrt(m))) = pi(m) ................... [1]
has an infinite number of solutions:
> The Number Theory "Number of primes in intervals"
> "A million dollar problem"
> Friday, 2 April 2010
> Infinite number of figures (m_Q)

I claim that this is false, since
Q(m,pi(sqrt(m))) > pi(m), for m > 40291 .... [2]

You keep posting messages about "error in calculation".
But yours was an "error in understanding": the left
and right hand sides of [1] are clearly incommensurable,
at large m, for the very good reason that 2 > exp(Euler).

Little is served by comparing things that *always* differ,
for m > 40291.

Let's take an example, using http://primes.utm.edu/nthprime/

m = 29996224275833
pi(m) = 1000000000000
n = pi(sqrt(m)) = 379312
Q(m,n) = 1085398179034.020151103297662104429...
Q(m,n) - pi(m) > 85398179034

So here your "error in calculation" exceeds 85 billion.

Konets istorii? / End of story?

David
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