Suppose $r:\Bbb N\to (0,1)$ is a function given by $r(n)$ is obtained as put a point at the beginning of $n$ instance $r(34880)=0.34880$ and $\Bbb P$ is the set prime numbers.

Theorem: For each subinterval $(a,b)$ of $[0.1,1),\,\exists m\in \Bbb N$ that $\forall k\in \Bbb N$ with $k\ge m$ then $\exists t\in (a,b)$ that $t\cdot 10^k\in \Bbb P$. ($\color{fuchsia}{\text{proof}}$ at my previous account and it is shown $\color{fuchsia}{\text{here}}$ too.)

Corollary: $r(\Bbb P)$ is dense in the $[0.1,1]$. $$\\$$ and let $\forall n\in\Bbb N,$ and for each subinterval $(a,b)$ of $[0.1,1),$ that $a\neq b,$

$\begin{cases} U_{(a,b)}:=\{n\in\Bbb N\mid a\le r(n)\le b\},\\ \\V_{(a,b)}:=\{p\in\Bbb P\mid a\le r(p)\le b\},\\ \\U_{(a,b),n}:=\{m\in U_{(a,b)}\mid m\le n\},\\ \\V_{(a,b),n}:=\{m\in V_{(a,b)}\mid m\le n\},\\ \\w_{(a,b),n}:=(\#U_{(a,b),n})^{-1}\cdot\#V_{(a,b),n}\cdot\log n,\\ \\w_{(a,b)}:=\lim _{n\to\infty} w_{(a,b),n}\end{cases}$ $$\\$$ Now regarding to Prime number theorem and above corollary I offer following conjecture so please guide me about possibility of this conjecture:

Conjecture: $\forall (a,b)\subset [0.1,1),\,w_{(a,b)}=0.9^{-1}\cdot (b-a)$.

Thanks in advance!

  • $\begingroup$ The mentioned density easily follows from Dirichlet's theorem implying that infinite many prime numbers begin with an arbitary given digit-string (of course not beginning with $0$). $\endgroup$ – Peter Mar 9 '18 at 8:18
  • $\begingroup$ I think there is no way except using prime number theorem to prove this density because there is no difference between a prime $p$ and its image $r(p)$ other than a sign or a mark as a point instance $911$ with $0.911$. $\endgroup$ – user538968 Mar 11 '18 at 20:31
  • $\begingroup$ Yes, I do not see any other way either. What I meant is that Dirichlet's theorem shows the density mentioned above. The conjecture is much stronger and more concrete because it deals with the equidistribution. $\endgroup$ – Peter Mar 11 '18 at 20:36
  • $\begingroup$ $@$Peter but another question, why Dirichlet and PNT haven't been combined yet? $\endgroup$ – user538968 Mar 11 '18 at 21:07
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    $\begingroup$ The prime number theorem wants to count the primes (or at least give very good approximations of the number of primes in a range), Dirichlet's theorem only guarantees that infinite many primes of linear forms can be found, if the coefficients are coprime. How many prime numbers we find in a range of a given form has surely be analyzed as well, but is not the intent of the initial theorem. I remember a paper asking how large the smallest prime will approximately be, but I neither remember the details nor the source. But you should find it , if you google "dirichlet's theorem" $\endgroup$ – Peter Mar 11 '18 at 21:14

Imagine a very large number $N$ and consider the range $[10^N,10^{N+1}]$. The natural logarithms of $10^N$ and $10^{N+1}$ only differ by $\ln(10)\approx 2.3$ Hence the reciprocals of the logarithms of all primes in this range virtually coincicde. Because of the approximation $$\int_a^b \frac{1}{\ln(x)}dx$$ for the number of primes in the range $[a,b]$ the number of primes is approximately the length of the interval divided by $\frac{1}{\ln(10^N)}$, so is approximately equally distributed. Hence your conjecture is true.

Benfords law seems to contradict this result , but this only applies to sequences producing primes as the Mersenne primes and not if the primes are chosen randomly in the range above.

  • $\begingroup$ $@$Peter welcome back! thank you so much! you are great! $\endgroup$ – user538968 Mar 11 '18 at 20:27

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