A090527 - OEIS

%I #14 Nov 20 2018 02:57:16

%S 2,2,11,29,11,137,79,149,13,17,181,7,71,41,53,541,197,61,149,149,19,

%T 541,1663,829,229,599,13,563,113,137,13,1129,421,1759,683,389,919,877,

%U 233,1933,2137,97,331,881,1753,193,137,521,1063,59

%N Smallest prime p such that floor((n^n)/p) is prime, or 0 if no such number exists.

%C Conjecture: No term is zero.

%C As long as p(j+1)/p(j) < 2 for all j, then for any integer n >= 4, there exists at least one p such that p and floor(n/p) are both prime. (I do not know a proof for the premise above; however, it seems quite weak compared to other conjectures and theorems about primes. It may be that it follows from the results in sequence A038458.) In fact, there exists a prime p such that either floor(n/p) = 2 or floor(n/p) = 3. Outline of proof: (1) If p is a prime number, then for all n with 2p <= n < 3p, floor(n/p) = 2, which is prime. (2) In addition, for all n with 3p <= n < 4p, floor(n/p) = 3, which is prime. So for any n >= 4, consider the largest prime, p, with 2p <= n. (3) floor(n/p) can't be less than 2, since 2 <= n/p. (4) If floor(n/p) = 2, then p and floor(n/p) are both prime, so we are done. (5) Similarly, if floor(n/p) = 3, we are done. The only remaining case is that 4p <= n. Let p_1 be the next prime after p. (6) p_n must not meet 2(p_1) <= n, since p is the largest that does. Therefore 2(p_1) > n. (7) 4p <= n < 2(p_1) (8) (p_1 / p) > 2 (9) As long as p(j+1)/p(j) < 2 for all j, the case of 4p <= n is not possible. - Weston Markham (WMarkham(AT)paradigmgenetics.com), Jun 15 2004

%t <<NumberTheory`; Do[p = n^n; i = 1; While[ !ProvablePrimeQ[Floor[p/Prime[i]]], i++ ]; Print[Prime[i]], {n, 2, 100}] (* _Ryan Propper_, Aug 02 2005 *)

%Y Cf. A090525, A090526, A090528.

%K nonn

%O 2,1

%A _Amarnath Murthy_, Dec 07 2003

%E a(7)-a(14) from Weston Markham (WMarkham(AT)paradigmgenetics.com), Jun 15 2004

%E a(15)-a(51) from _Ryan Propper_, Aug 02 2005