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 A090527 Smallest prime p such that floor((n^n)/p) is prime, or 0 if no such number exists. 3
 2, 2, 11, 29, 11, 137, 79, 149, 13, 17, 181, 7, 71, 41, 53, 541, 197, 61, 149, 149, 19, 541, 1663, 829, 229, 599, 13, 563, 113, 137, 13, 1129, 421, 1759, 683, 389, 919, 877, 233, 1933, 2137, 97, 331, 881, 1753, 193, 137, 521, 1063, 59 (list; graph; refs; listen; history; text; internal format)
 OFFSET 2,1 COMMENTS Conjecture: No term is zero. 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 LINKS MATHEMATICA <

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Last modified April 17 02:16 EDT 2021. Contains 343059 sequences. (Running on oeis4.)