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Conjecture:

For any positive integer $n$, there exists a prime whose first digits are the digits of $n$.

Proof:

The infinite set of integers starting with the digits of integer $n$ is distributed as $1$ sequence of consecutive integers for each power of $10$. For example, the integers starting with the same digits as $2$ are those between $20$ and $30$, between $200$ and $300$, between $2000$ and $3000$ $…$ Now let's say $n=1245$, and let's look at much larger integers than $n$. For example, all integers $x$ in the following interval starts with digits $1245$: $$ 12450000000000000000000000000 ≤ x <12460000000000000000000000000 $$

A range of this magnitude is proven to contain at least $1$ prime number, so whether $n$ is a single digit or a million digits long, we only need to add the proper number of zeros to reach an interval where having no prime is impossible.

Questions:

  1. Is my proof valid ? Did I miss something ?
  2. Is there other existing proofs of that ?

Of course any suggestions/edits to make it more clear or professional are welcome, I'm still a newbie at proof writing!

François Huppé
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    "A range of this magnitude is proven to contain at least 1 prime number" when was that proven? – Rushabh Mehta Jan 04 '20 at 17:37
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    "A range of this magnitude is proven to contain at least 1 prime number": it's not quite that simple $-$ even though it's true, you certainly haven't proven it just by stating it as proven. You can use the Prime Number Theorem here, but you need to be more careful with your ranges. – TonyK Jan 04 '20 at 17:39
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    I don't understand your argument. For a proof of the desired claim, see, e.g., this – lulu Jan 04 '20 at 17:40
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    Possible duplicate of Proof that there are infinitely many prime numbers starting with a given digit string – Parcly Taxel Jan 04 '20 at 17:41
  • @TonyK Of course you are right, I think this one (proven) should be enough: There is always a prime between n-n^(23/42) and n. I'll look into that – François Huppé Jan 04 '20 at 17:45
  • You might want to use Yitang Zhang's proof that there are infinitely many prime gaps that do not exceed 70 million . Bounded gaps between primes. – The Demonix _ Hermit Jan 04 '20 at 17:49
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    @TheDemonix_Hermit: Have you thought that through? I don't see how it would help. Why should any of those prime gaps start with the desired digits? – TonyK Jan 04 '20 at 17:50
  • The duplicate is answering my second question, but my goal was more to gain proof writing skills by having it verified. – François Huppé Jan 04 '20 at 17:52
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    General note: I think the OP has made it quite clear that he is not so much interested in the proof as in the flaw with the argument sketched in the post. On that point, I think it's clear that the issue is that it is not so easy to link up results about prime gaps and actual intervals. This is why standard proofs invoke the Prime Number Theorem or the like. In any case, I don't think the proposed link is actually a duplicate (though it is clearly relevant). – lulu Jan 04 '20 at 18:12
  • An analogue statement is a bit easier and follows from Dirichlet's theorem : If we have a digit string (it can have zeros at the beginning) ending with $1,3,7$ or $9$, then there are infinite many primes that end with this digit string. – Peter Jan 04 '20 at 18:38
  • The limit of π(n) / (n ln n) is 1; from the definition of the limit we can show for every 0 < c < 1 there is a prime between n and n + cn, if we pick n large enough dependent on c. Then for a sequence of digits d pick n = d * 10^k, with k so large that there is a prime between d * 10^k and d * 10^k + (1/d) d*10^k. – gnasher729 Jan 04 '20 at 22:51

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