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With the convention $0^0=1$, let

  • $Q(1) = 1^1 - 0^0=1-1=0$
  • $Q(2) = 2^2 - 1^1 - 0^0=4-1-1=2$
  • $Q(3) = 3^3 - 2^2 - 1^1 - 0^0=27-4-1-1=21$
  • $Q(4) = 4^4 - 3^3 - 2^2 - 1^1 - 0^0=256-27-4-1-1=223$
  • and so on...

I found that $Q(2)=2$, $Q(4)=223$ , and $Q(7)=773{,}473$ are prime , but after that I didn't find anymore primes up to $Q(1000)$. I found these some regular patterns:

  • $Q(4n+1)$ and $Q(4n+2)$ will always be even numbers for all integers $n \ge 0$
  • $3$ will always be the LEAST PRIME FACTOR of $Q(36n+3) , Q(36n+8) , Q(36n+16) , Q(36n+19) , Q(36n+24)$ and $Q(36n+35)$, for all integers $n \ge 0$

I found some semiprimes but I didn't find any primes besides the already known $Q(2), Q(4)$, and $Q(7)$. Could you find the next prime(s) of such form ?

Barry Cipra
  • 79,832
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    Is this problem directly related to the use of Wolfram Language or Mathematica software? – Αλέξανδρος Ζεγγ Jul 15 '20 at 07:40
  • I used this : Isprime n^n - 1 - ($sigma$ (n-1)^(n-1) , n = 2 to n) –  Jul 15 '20 at 08:03
  • Isprime is not a Mathematica function and that doesn't look like Mathematica syntax. – flinty Jul 15 '20 at 13:37
  • I passed $n= 9\ 100$ , no further primes. (+1 , since I am always interested in questions about prime numbers) – Peter Jul 17 '20 at 10:05
  • I arrived at $n=11\ 000$ without finding a further prime. – Peter Jul 23 '20 at 12:29
  • According to my calculations, in the range $11\ 000\le n\le 10^{11}$ , the expected number of primes of this form is slightly larger than $1$ , which could make it infeasible to find the next prime of this form. The naive $1/ln(n)$-approach predicts infinite many primes of this form since small factors are apparently not forced, and algebraic or similar factors seem not to exist either. – Peter Jul 23 '20 at 12:34

0 Answers0