[seqfan] Re: Question on primes
Creighton Kenneth Dement
creighton.k.dement at mail.uni-oldenburg.de
Wed Jun 10 12:30:25 CEST 2009
> I think it's hard; I'm also pretty sure it's true.
> Up to 10^8, there are no prime gaps big enough that (p+1)*(q-1) <=
> ((p+q)/2 - 1)^2.
> To violate this, one would need to have q >= p + 4 + sqrt(8*(p+1)).
> Prime gaps on the order of sqrt(p) are generally believed not to exist.
> Franklin T. Adams-Watters
Thank you! I propose a sequence:
A(n) = number of primes p less than A000040(n) (the n-th prime) such that
(p + 1)*(A000040(n) - 1) is a square.
By hand, I believe the sequence would start off
0, 0, 1, 1, 0, 2, ...
> -----Original Message-----
> From: Creighton Kenneth Dement
> <creighton.k.dement at mail.uni-oldenburg.de>
> Dear Seqfans,
> I have a quick question that is probably either trivial or really hard.
> Observe that (2 + 1)(13 - 1) = 36 is a square. 2 and 13 are primes, but
> they are not consecutive (i.e. there are other primes inbetween).
> If p and q are consecutive primes, do we have
> (p + 1)*(q - 1) is a square if and only if q = p + 2, i.e. p and q are
> twin primes?
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