2,1

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, 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 about the "Smarandache constant", e.g. described 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

<<NumberTheory`; Do[p = n^n; i = 1; While[ !ProvablePrimeQ[Floor[p/Prime[i]]], i++ ]; Print[Prime[i]], {n, 2, 100}] (Propper)

Cf. A090525, A090526, A090528.

Adjacent sequences: A090524 A090525 A090526 this_sequence A090528 A090529 A090530

Sequence in context: A090525 A126806 A121871 this_sequence A014220 A089544 A069648

nonn

Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Dec 07 2003

More terms from Weston Markham (WMarkham(AT)paradigmgenetics.com), Jun 15 2004

More terms from Ryan Propper (rpropper(AT)stanford.edu), Aug 02 2005