3 ms·

I'm surprised that "digitally delicate primes" are rare, why isn't this the majority of primes? That would be my intuition.

• dnautics 6y ago
  You must be able to change any digit into any other digit.
  • kzrdude 6y ago
    Yes, but if we turn it around, if digitally delicate is not the majority, then the majority of primes are one digits' edit distance away from another, i.e. all but one digit is shared with another prime.
    • dnautics 6y ago
      that shouldn't be surprising, because of pi(n).
• qsort 6y ago
  For a prime to be delicate as per the definition given, all numbers with "levhensthein distance" 1 must be composite. This puts a lot of conditions on what the prime can be, for p = a_0 + a_1 b + a_2 b^2 + ... + a_n b^n, if p is delicate then (p - a_0), (p - a_0 + 1), ..., (p - a_0 + b - 1) must all be composite, and so must be (p - b a_1), ..., (p - b a_1 + b (b-1)), and so forth. This is equivalent to proving that certain slices of some arithmetic successions contain no primes. Not a professional mathematician and I didn't read the paper, but I suspect this is related to prime gaps works by Tao, who is cited in the article.
  • breck 6y ago
    This is a great explanation, thanks!
• Someone 6y ago
  Napkin math (possibly with gross errors, and doing stats, so making the math rigid cannot lead to rigorous proof): A prime with N digits has 9N ‘neighbors’. There are about 10^N/ln(10^N) - 10^(N-1)/ln(10^(N-1)) primes with N digits, so the probability of a N-digit number being prime is about 1/ln(10^N), of it not being prime 1-1/ln(10^N). So, the question is about the behavior of (1-1/ln(10^N))^9N as N ⇒ ∞. Google gives me a plot that’s essentially zero, so yes, most primes should be digitally delicate.