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

[Dean Hickerson]

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