New sequence : Primes for which SQRT(A000040(n)) < A001223(n)

[David Wilson]

Let gap(p) be the gap following prime p (that is, gap(p) = nextPrime(p)-p). 
If your sequence includes prime p with gap(p) = g, then it also includes the 
smallest prime p with gap(p) = g, that is, it includes p = A000230(g/2).

Checking the values of b000230.txt, we find that the largest listed element 
of A000230 belonging to your sequence is p = A000230(7) = 113 with gap(p) = 
14. This means that there are no elements p of your sequence with 14 < 
gap(p) <= 1190, the largest gap listed in b000230.txt.

By definition, if p is in your sequence, then p < gap(p)^2. Therefore, if p 
is in your sequence and gap(p) <= 14, we must have p < 196. You can easily 
verify that your sequence is complete up to 195.

Letting p be the smallest prime with gap(p) = g, that is, p = A000230(g/2), 
empirical analysis of b000230.txt shows that as g grows, log(p)/log(g) grows 
more slowly with a bit of bumpiness. By the time we reach g = 1190, 
log(p)/log(g) is solidly greater than 5, strongly indicating that for gaps g 
 > 1190, not only is p > g^2, but indeed p > g^5. This provides very strong 
evidence that your sequence is finite and complete as you have given it. 
Conway would call this is a surety, that is, a statement with overwhelming 
empirical or statistical support but no proof.

Indeed, the evidence suggests that for any k, the number of primes with p < 
gap(p)^k is finite (your sequence being the special case k = 2).


 ----- Original Message ----- 
From: <reismann at free.fr>
To: <seqfan at ext.jussieu.fr>
Sent: Sunday, December 10, 2006 1:16 PM
Subject: New sequence : Primes for which SQRT(A000040(n)) < A001223(n)


> Dear Seqfans,
>
> I will submit this sequence in January :
>
> %I A000001
> %S A000001 3,7,13,23,31,113
> %N A000001 Primes for which SQRT(A000040(n)) < A001223(n).
> %C A000001 Conjecture : this sequence is finite and complete.
> %e A000001 a(1) = 3 because SQRT(3) < 2.
> a(6) = 113 because SQRT(113) < 14.
> %Y A000001 A000040(n), A001223(n).
> %O A000001 1,1
> %K A000001 ,fini,nonn,
> %A A000001 Remi Eismann (reismann at free.fr), Dec 10 2006
>
> Any comments on the sequence or on the conjecture ?
> Have a nice day.
>
> Rémi EISMANN
>
>
>
>
> -- 
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.432 / Virus Database: 268.15.15/581 - Release Date: 12/9/2006 
> 3:41 PM
>
>