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Consider the digits numbers formed by the first $k$ digits in the decimal representation of $\pi, k \ge 1$

$$ 3\\31\\ 314 \\3141 \\31415\\314159 \\ 3141592 \\31415926 \\314159265 \\ ... $$

Out of the first $10^4$ such numbers $4000$ (approximately $40\%$) end in $1,3,7$ or $9$. Since all primes $> 5$ end in one of these four digits I checked how many of these $4000$ numbers were primes and I could find only corresponding to $k = 2,6,38$ which is much lower than what I anticipated.

Question: In general, assuming $0 < \alpha < 1$ to be a normal in base $10$, what is the expected density of primes among the first $k$ numbers formed by the digits of $\alpha$ as explained above?

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    $\begingroup$ If we assume random digits, we can estimate the number of primes with the $1/ln(n)$-approach. $\endgroup$ – Peter Jan 16 '20 at 13:54
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    $\begingroup$ I think, a high search limit has already been established. Maybe, you look it up at OEIS. $\endgroup$ – Peter Jan 16 '20 at 13:56
  • $\begingroup$ digits in order ? there's a thread on mersenne forum by david eddy you might like. $\endgroup$ – user645636 Jan 16 '20 at 18:48
  • $\begingroup$ oh and the digits prior to a 1 or 7 will never form a number that is 2 mod 3. all digits prior to 3 or 9 will never form a multiple of 3. $\endgroup$ – user645636 Jan 16 '20 at 19:08
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    $\begingroup$ you forgot 31 but got 13... $\endgroup$ – user645636 Jan 17 '20 at 15:48
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The Prime Number Theorem states, essentially, that the number of primes less than $N$ is approximately $\frac{N}{\ln{N}}$. That means (roughly) that the probability of $N$ itself being prime is about $\frac{1}{\ln{N}}$. Remember, though, that $N$ is the number itself, not the number of digits it has. Since $\ln{N}$ is roughly equal to the number of digits of $N$ multiplied by $\ln{10}$, and since a number's index in your sequence is just the number of digits it has, that suggests that the probability that the $k$th number is prime should be about $\frac{1}{k\ln{10}}$. Assuming that whether each number is prime is independent of the primality of the ones before it, the expected number of primes should be

$$\frac{1}{\ln{10}}\left(1 + \frac12 + \frac13 + \frac14 + \cdots + \frac1k\right)$$

By a result of Euler and Mascheroni, this happens to be almost exactly $\frac{\ln{k}}{\ln{10}} = \log_{10}k$. $\log_{10}{10000} = 4$, which is exactly what your data indicates!

This is a very cool question - I've never had a density estimate work out this perfectly!

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