cp's OEIS Frontend

Let c = 0.567148130202... (see A038458), the solution to 127^x - 113^x = 1. c is conjectured by Smarandache to be the smallest real number x such that A000040(n+1)^x - A000040(n)^x = 1 has a solution. This conjecture is equivalent to saying that the terms of the present sequence are always positive, but that if c were replaced by a larger real number, there would be zeros in the sequence. However, note that a(30) is not the last occurrence of 1: a(46) = a(61) = 1 as well.

Inspired by Andrica's conjecture.

Each of these terms, k, has been tested to at least 100*k. - Sean A. Irvine, Jul 29 2025

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

Inspired by Andrica's conjecture. If it is true, a(n)>1 for all n.

Cf. A038458, A074976, A078693