A090527 Smallest prime p such that floor((n^n)/p) is prime, or 0 if no such number exists.

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

Offset: 2

Author

Amarnath Murthy, Dec 07 2003

Keywords

• nonn

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

Crossrefs

Cf. A090525, A090526, A090528.

Programs

Mathematica

<Ryan Propper, Aug 02 2005 *)

Extensions

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

a(15)-a(51) from Ryan Propper, Aug 02 2005