## Re: Primes in the interval (Problem)

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• ... The discussion after Theorem 3.1 [Mertens] in Riesel s book http://www.amazon.com/Numbers-Computer-Factorization-Progress-Mathematics/dp/0817637435
Message 1 of 3 , Dec 4, 2010
<chitatel2000@...> wrote:

> Determine the largest value of the error

> It is interesting to see how Sergey fooled himself into
> making his false conjecture. His mistake was to believe
> that we may rely on the sieve of Eratosthenes to give
> the "right" constant in the prime number theorem. In fact,
> we should not: there is a mismatch by the celebrated factor
> 2*exp(-Euler) > 1

The discussion after Theorem 3.1 [Mertens] in Riesel's book
http://www.amazon.com/Numbers-Computer-Factorization-Progress-Mathematics/dp/0817637435
explains this rather clearly.

David
• ... Off-list, Sergey asked me for an example of a large error. Let p[n] = 10^1000 + 453. Then p[n-1] = 10^1000 - 1769 is the previous prime. Sergey defines Q
Message 2 of 3 , Dec 6, 2010

Off-list, Sergey asked me for an example of a large error.

Let p[n] = 10^1000 + 453.
Then p[n-1] = 10^1000 - 1769 is the previous prime.
Sergey defines Q as the number of primes
between p[n-1]^2 and p[n]^2.
Using the prime number theorem, we may estimate
Q =~ (p[n]^2 - p[n-1]^2)/log(p[n]^2) =~ 9.65*10^999

Sergey defines his "error" E by
E = (p[n]^2 - p[n-1]^2)*prod(k=1,n,1-1/p[k]) - Q

Using Mertens' theorem
http://mathworld.wolfram.com/MertensTheorem.html
we may easily estimate
E =~ 2*exp(-Euler)*Q - Q =~ 1.19*10^999
whose integer part has 1000 decimal digits.

Clearly the error is unbounded, since it is
of the same order as the n-th prime, p[n].