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

[David Wilson]

----- Original Message ----- 
From: "Dean Hickerson" <dean at math.ucdavis.edu>
To: <seqfan at ext.jussieu.fr>
Sent: Sunday, December 10, 2006 4:07 PM
Subject: Re: New sequence : Primes for which SQRT(A000040(n)) < A001223(n)


> Mostly to Remi EISMANN:
>
>> 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 ?
>
> I suggest that you rewrite the definition so that people don't have to 
> look
> up both A000040 and A001223 to figure out what it means.  Something like
> "Primes p for which sqrt(p) < q-p, where q is the smallest prime larger
> than p." or "Primes p for which there are no primes between p and 
> p+sqrt(p).".
>
> You could add a comment that there are no other terms less than
> 218034721194214273.  (That's assuming that all of the terms in
> http://www.research.att.com/~njas/sequences/b002386.txt  are correct.)
>
> And you could state that finiteness of the sequence would follow from
> Cramer's conjecture that
>
>    lim sup (p(n+1)-p(n))/log(p(n))^2 = 1.
>
> Dean Hickerson
> dean at math.ucdavis.edu

Wouldn't the finiteness of this sequence follow from the prime number 
theorem?