Can $\epsilon$ be a generating function? - MathOverflow [closed]

Can $\epsilon$ be a generating function?

2012-07-19T09:49:51Z

I would like to know if I could do something like:

$\epsilon (0)\text{:=}0$
$\epsilon (n)\text{:=}\frac{4}{2 n-(-1)^n+1}$

and use it instead of a constant.

As $n\rightarrow \infty$, $\epsilon \rightarrow 0.$

Edit for clarification:

In Hardy and Wright, sixth edition, page 494, (22.19.2), it states, "...there is always a prime $p$ satisfying"

$x < p < (1 + \epsilon) x$

Using this function:

$x(0)\text{:=}0$
$x(n)\text{:=}\frac{1}{8} \left(2 n (n+2)-(-1)^n+1\right)$

I want to have:

$x(n) < p \leq (1+\epsilon(n)) x(n)$

Edit after the closure: The answer I was looking for is "YES." I have located several proofs that have $\epsilon$ dependent on $x,$ meaning it can vary under my control. Now, I can't use it that way because neither $x$ nor $\epsilon$ are dependent on each other, but both are dependent on $n.$ But, I hope to adapt.

I did not sign off on quid's answer because it was not exactly what I needed. However, I do like the answer very much.

Thanks.

Answer by Gerry Myerson for Can $\epsilon$ be a generating function?

2012-07-19T12:21:22Z

I don't have H & W handy, but I assume what they say is something like, for every positive $\epsilon$, there is an $N$ depending on $\epsilon$, such that if $x\gt N$, then there is a prime $p$ with $x\lt p\lt(1+\epsilon)x$. So, for the particular $\epsilon$ you have chosen, there is some function $x(n)$ you can use, but it can't be any old function, and you'd certainly have some work to do to show that the $x(n)$ you have chosen in valid.

Answer by quid for Can $\epsilon$ be a generating function?

2012-07-19T12:48:04Z

If this is true for all small $n$ (which I did not check) it is likely true for all $n$. However, to prove this seems presently infeasible.

Leaving the precise details of the constants asside the $x(n)$ is quadratic in $n$ and the $\varepsilon (n)$ being roughly $1/n$, this question amounts to (dropping constants) asking whether there is some prime between $an^2$ and $an^2 + b n$ for certain $a,b$ or to put it differently $x$ and $x + c x^{1/2}$.

Much stronger things are widely believed to be true asymptotically, but the problem for $x$ and $x + c x^{1/2}$ is open (even under GRH).

Various information can be found on the Wikipedia page on Prime gaps

So, asymptoically, this should be true and follows from standard conjectures; but a (unconditional) proof seems out of reach.