4 deleted 133 characters in body

If you avoid trivial things**, then your

The slowest growing zipper depends will depend on the size of $p_{n+1}-p_n$ where $p_n$ is the $n^{th}$ prime number. There are many results regarding the size of the largest prime gap.

Unconditional: The work of Baker, Harman and Pintz shows that $$p_{n+1}-p_n \ll p_n^{0.525}$$ for some computable constant. This means that your zipper function may be taken to be $f(n)=Cn^{40/19}$ for some constant $C$. The $\frac{40}{19}$ appears in the exponent because $\frac{40}{19}=\frac{1}{1-0.525}$.

Conditional: If we assume the Riemann Hypothesis, then we have $$p_{n+1}-p_n \ll \sqrt {p_n}\log p_n,$$ and we may take $f(n)=n^2 \log n$. Assuming Cramer's conjecture, which says that $$p_{n+1}-p_n =O\left((\log p_n)^2\right),$$ would allows us to take $f(n)=Cn(\log n)^2$ for some constant $C$.

Also see this Wikipedia article on prime gaps.

Remark: Note that finding a prime zipper which grows slower than $f(n)=Cn^{40/19}$ would imply better bounds on the largest prime gap, so your question is equivalent to asking what is the largest prime gap.

**By your definition, if $p_n$ is the $n^{th}$ prime number, then * Avoid pointless functions such as $f(n)=p_n+1$ will be the slowest growing prime zipper, but that is not a meaningful statement.f(n)=p_n+1$. 3 added 27 characters in body; added 39 characters in body; added 212 characters in body If you avoid trivial things**, then your zipper depends on how large the size of$p_{n+1}-p_n$can be where$p_n$is the$n^{th}$prime number. This is a well known problem, and there There are many results regarding the size of the largest prime gap. Unconditional: The work of Baker, Harman and Pintz shows that $$p_{n+1}-p_n \ll p_n^{0.525}$$ for some computable constant. This means that your zipper function may be taken to be$f(n)=Cn^{40/19}$for some constant$C$. We get The$\frac{40}{19}$appears in the exponent because$\frac{40}{19}=\frac{1}{1-0.525}$. Conditional: If we assume the Riemann Hypothesis, then we have $$p_{n+1}-p_n \ll \sqrt {p_n}\log p_n,$$ and we may take$f(n)=n^2 \log n$. Assuming Cramer's conjecture, which says that $$p_{n+1}-p_n =O\left((\log p_n)^2\right),$$ would allows us to take$f(n)=Cn(\log n)^2$for some constant$C$. Also see this Wikipedia article on prime gaps. Remark: Note that finding a prime zipper which grows slower than$f(n)=Cn^{40/19}$would imply better bounds on the largest prime gap, so your question is equivalent to asking what is the largest prime gap. **By your definition, if$p_n$is the$n^{th}$prime number, then$f(n)=p_n+1$will be dthe the slowest growing prime zipper, but that is not a meaningful statement. 2 added 311 characters in body; added 19 characters in body; added 209 characters in body If you avoid trivial things**, then your zipper depends on how large$p_{n+1}-p_n$can be where$p_n$is the$n^{th}$prime number. This is a well known problem, and there are many results regarding the size of the largest prime gap. Unconditionally Unconditional: The work of Baker, it has been proven Harman and Pintz shows that $$p_{n+1}-p_n \ll C p_n^{0.525}$$ for some computable constant. This means that your zipper function may be taken to be$f(n)=Cn^{40/19}$for some constant$C$. We get$\frac{40}{19}$in the exponent because$\frac{40}{19}=\frac{1}{1-0.525}$. Conditional: If we assume the Riemann Hypothesis, then we have $$p_{n+1}-p_n \ll \sqrt {p_n}\log p_n,$$ and it is conjectured we may take$f(n)=n^2 \log n$. Assuming Cramer's conjecture, which says that $$p_{n+1}-p_n =O\left((\log p_n)^2\right).$$ See p_n)^2\right),$$would allows us to take$f(n)=Cn(\log n)^2$for some constant$C$. Also see this Wikipedia Articlearticle on prime gaps. **By your definition, if$p_n$is the$n^{th}$prime number, then$f(n)=p_n+1\$ will be dthe slowest growing prime zipper.

1