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I'm looking at a page at PrimeCurios for the number 9999999900000001, at https://t5k.org/curios/page.php. (Search for that number where it says "Search Curios!") It makes several claims, including this one:

9999999900000001 is the 8th known prime, p, such that $(p)^2$ divides $(p)_3$. Can you find other cases like this?

I have no idea what $(p)_3$ means here. Can somebody define or explain this?

MiguelMunoz
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  • What do you tthink $(p)^2$ means? Seems they wouldn't use parrenthese if they just wanted the square, $p^2.$ – Thomas Andrews Sep 25 '23 at 18:51
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    Well, just a guess mind you, but if we say that $(p)^2$ means $p^2$ and that $(n)_m$ means "$n$ concactenated with itself $m$ times" then the claim is true. I mean $p^2$ does divide $p$ concactenated with itself $3$ times. See here – lulu Sep 25 '23 at 18:51
  • But, honestly, if they are going to use non-standard notation, they ought to have a convenient glossary. – lulu Sep 25 '23 at 18:52
  • But it looks like tthere is no real way to tell what the submitter Green meant by this. Any answer would be a guess. – Thomas Andrews Sep 25 '23 at 18:54
  • @lulu That was a good guess. You're probably right. I've certainly never seen that notation before, but that kind of thing comes up with prime curios a lot. Thanks. – MiguelMunoz Sep 25 '23 at 18:54
  • I have no idea what the first $7$ primes might be. $3^2,|,333$ would be the first one. And I think $13^2,|,131313$ is the second. Haven't searched for others. Hard to believe your $p$ is only the $8^{th}$ such. – lulu Sep 25 '23 at 18:56
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    The OEIS has a list – PrincessEev Sep 25 '23 at 18:57
  • @PrincessEev Ha! Thanks for that link. I am really surprised that this $p$ is only the $8^{th}$ but here we are. – lulu Sep 25 '23 at 18:58
  • @lulu You might want to answer this question instead of just commenting. That way I can close the question. – MiguelMunoz Sep 25 '23 at 19:00
  • @PrincessEev Thanks for that link. It has four members of the family of numbers I'm studying! The first from that list is 9901. – MiguelMunoz Sep 25 '23 at 19:01
  • I'm most interested by the statement at the bottom of the curios list: $$9999999900000001 = 99999999^2 + 99999999^1 + 99999999^0$$ – Eric Snyder Sep 25 '23 at 21:56

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To summarize the discussion in the comments:

The (somewhat unusual) notation means that $p^2$ divides the concatenation of $p$ with itself three times. For instance, $13$ passes the test since $13^2$ divides $131313$.

As confirmation, we have this OEIS list A147554 supplied by user @PrincessEev

lulu
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