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

MATHEMATICA

<<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 *)

CROSSREFS

KEYWORD

nonn

AUTHOR

Amarnath Murthy, Dec 07 2003

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

STATUS

approved