[seqfan] Re: Question on primes

[Creighton Kenneth Dement]

> 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, ...

Sincerely,
Creighton



> -----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?
>
> Sincerely,
> Creighton
>
>
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