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Calculate the following number

$$\sup \{x\in[0, 1] \space | \space \#A_x = \infty \}$$

Where $A_x = \{n\in \mathbb{N} \space | \space \text{the fraction of $n$'s digit permutations that are prime } \geq x\}$.

The digit permutations of a number are all the permutations of its digits, for example for $n=2399$ the permutations are

$$\{ 2399, 2939, 2993, 3299, 3929, 3992, 9239, 9293, 9329, 9392, 9923, 9932 \}.$$

Seven out of these twelve numbers are prime: $\{2399, 2939, 3299, 3929, 9239, 9293, 9923\}$ so $n\in A_{\frac{7}{12}}$.

ploosu2
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  • I expect it is zero because the proportion for most $n$-digit numbers is below $10/n$ by the Prime Number Theorem, but I don't know how to rule out outliers. Zero is in the set because every number with all-even digits is in $A_0$. – Empy2 Nov 28 '18 at 10:44
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    Is there anything known about the prime numbers among $11\ldots1$? If there are infinitely many, then the supremum is $1$. Looking here https://math.stackexchange.com/questions/300805/when-is-the-number-11111-1-a-prime-number it seems this is suspected but not known. – Ingix Nov 28 '18 at 13:24

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