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UPD: To make it clearer, here is a statement:

For sequence $A = \{i\in\mathbb{N}_+\mid a_i=a_{i-1}\times 10^{\lfloor(\lg(10n))\rfloor} + n\}$, where $\lg n=\log_{10} n$, show whether there is a prime number in the sequence. If there isn't, please prove it.

I ran into the problem accidentally.

To be clear, you can check the OEIS sequence A007908.

Using computer, I found no prime number up to 500, that is, $\overline{1234\cdots 499500}$.

Any form of help would be appreciated. Thanks in advance.

David H
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    That OEIS link says no primes in the first $5000$ terms – J. W. Tanner Mar 31 '21 at 14:20
  • @J. W. Tanner This is helpful, but what I wondered is that if there is a prime number in the sequence and, if the answer is no, how to prove it? – David H Mar 31 '21 at 14:23
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    There are no primes in the first 200000 terms. - Serge Batalov, Oct 24 2015 – Anindya Prithvi Mar 31 '21 at 14:28
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    No prime of this form is known, and the smallest prime of this form must be huge, if there is one at all. There is no way to decide the question than finding a prime of this form. If there is none, we will almost surely never be able to prove that. – Peter Mar 31 '21 at 15:03
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    Additionally, a primality-proof for a prime of this form will probably be infeasible, we have very likely to be content with a probable prime. – Peter Mar 31 '21 at 15:21
  • @Peter why is it unlikely that we will be able to prove the negative case? – Enrico Borba Apr 01 '21 at 19:40
  • @EnricoBorba We cannot rule out small factors, neither can we establish algebraic or aurifeuillan factors since the numbers are not of the form $f(n)$ , where $f(x)$ is a polynomial with integer coefficients. I can search for cases with very large smallest prime factors, if this is more convincing. – Peter Apr 02 '21 at 07:45
  • I agree with Peter. The search on mersenneforum.org apparently checked first $10^6$ terms (first million terms) without finding a prime (according to OEIS comments). Also, the largest known semiprime is the 9706th term in the sequence. – Vepir Apr 02 '21 at 14:29
  • Mahler has proved that the number $0.123456789101112....$ whose decimal part follows the sequence of the natural integers is a transcendental number. – Piquito Apr 06 '21 at 20:51
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    As somebody who commented on that sequence, I ruled it down to just 4 remainders modulo 30 long ago. I still can see how to get it down to 8, because these numbers have a property of even numbers being in even places, 0 mod 3 numbers being in 0 mod 3 places, and 0 mod 5 numbers being in 0 mod 5 places. This leaves us with places coprime to 30 which have remainders 1,7,11,13,17,19,23, and 29 modulo 30. Small factors can be dealt with (even on massive numbers) as computed at ...

     – Roddy MacPhee Jul 25 '22 at 18:00
  • https://math.stackexchange.com/a/3152669/903195 div(n,x)=if(n<=10^znorder(Mod(10,x)),return(n),return(n%(10^znorder(Mod(10,x)))+div(floor(n/(10^znorder(Mod(10,x)))),x))) – Roddy MacPhee Jul 25 '22 at 18:01

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