[seqfan] Re: The constants B_3 and b_3, as well as B = 1

Tomasz Ordowski tomaszordowski at gmail.com
Thu Mar 9 07:07:45 CET 2023

```At the end, an attempt to answer by analogy.

As with prime numbers, for natural numbers we have:
lim_{n->oo} (log(n) - 1/n Sum_{from k=1 to n} log(k)) = 1.
Because the Integral log(x) dx = x (log(x) - 1) + c, and the
definite Integral_{from 1 to n} log(x) dx = n (log(n) - 1) + 1.
Hence 1/n Sum_{from k=1 to n} log(k) = log(n) - 1 + O(1/n).

Note that p(n) = n (log p(n) - 1) + o(n).

T. Ordowski

śr., 8 mar 2023 o 07:37 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):

> P.S. However, my last question is hard,
> which I explain (in a new footnote),
>
> does lim_{x->oo} (log(x) - 1/pi(x) Sum_{p<=x} log(p)) = 1 ?
>
> In other words, lim_{n->oo} (log p(n) - 1/n Sum_{k=1..n} log p(k)) = 1,
> and equivalently, lim_{n->oo} (p(n)/n - 1/n Sum_{k=1..n} log p(k)) = 0 ?
>
> Maybe someone knows.
>
> Best regards,
>
> Thomas Ordowski
> _________________________
> The last two limits differ by one.
> This is a result of the proven fact that the present
> value of the historical Legendre's constant B = 1.
> If pi(x) = x / (log(x) - B(x)),
> then B = lim_{x->oo} B(x) = lim_{x->oo} (log(x) - x/pi(x)) = 1.
> See: https://en.wikipedia.org/wiki/Legendre%27s_constant
> So we have B = lim_{n->oo} (log p(n) - p(n)/n) = 1.
>
>
> pt., 3 mar 2023 o 20:05 Tomasz Ordowski <tomaszordowski at gmail.com>
> napisał(a):
>
>>
>> As is well known, the constant (related to Mertens' first theorem):
>> B_3 = lim_{x->oo} (log(x) - Sum_{p<=x} log(p)/p) = 1.332582...,
>> where p is a prime.
>> Cf. https://oeis.org/A083343 and the last formula in
>> https://mathworld.wolfram.com/MertensConstant.html
>> Let's define an analogous constant, namely (without proof):
>> b_3 = lim_{x->oo} (log(x) - Sum_{p<=x} (q - p)/p) = 0.0477967..,
>> where q is the next prime after p, so (q - p) is a prime gap.
>> The value of b_3 was calculated by Amiram Eldar.
>> Since (q - p)/p = q/p - 1, we have
>> b_3 = lim_{x->oo} (pi(x) + log(x) - Sum_{p<=x} q/p),
>> where q/p is a prime ratio.
>> Hence Sum_{p<=x} q/p = pi(x) + log(x) - b_3 + o(1).
>> Is this asymptotic known or provable?
>>
>> Finally, a somewhat similar but easier question:
>> does lim_{x->oo} (log(x) - 1/pi(x) Sum_{p<=x} log(p)) = 1 ?
>> In other words, lim_{n->oo} (log p(n) - 1/n Sum_{k=1..n} log p(k)) = 1,
>> and equivalently, lim_{n->oo} (p(n)/n - 1/n Sum_{k=1..n} log p(k)) = 0 ?
>> In forms more convenient for numerical calculations.
>>
>> Maybe someone will find the answers.
>> But first, please see the footnotes.
>>
>> Best regards,
>>
>> Thomas Ordowski
>> _________________
>> Mertens' first theorem (in modern notation):
>> Sum_{p<=x} log(p)/p = log(x) + O(1).
>> Cf. https://en.wikipedia.org/wiki/Mertens%27_theorems
>> and https://pl.wikipedia.org/wiki/Twierdzenia_Mertensa
>> "Ordowski's last conjecture" (in this notation):
>> Sum_{p<=x} (q/p - 1) = log(x) + O(1),
>> where q is the next prime after prime p.
>> _________________________
>> The answer to the last question seems to be YES.
>> This is a result of the proven fact that the present
>> value of the historical Lagendre's constant B = 1.
>> If pi(x) = x / (log(x) - B(x)),
>> then B = lim_{x->oo} B(x) = lim_{x->oo} (log(x) - x/pi(x)) = 1.
>> See: https://en.wikipedia.org/wiki/Legendre%27s_constant
>>
>>
```