0}}$ is any sequence of connected graphs such that the
number of vertices in $H_i$ is strictly increasing with $i$, then the sequence has
some pebbling threshold $t(n)$ which is $\Omega(\sqrt{n})$ and
$O(2^{\sqrt{2 \log_2 n}} n/\sqrt{\log_2 n})$, where $n$ is the number of vertices in a
graph in the sequence.
\end{corollary}
\begin{proof} By Theorems \ref{geoub} and \ref{geolb}, for sufficiently large $i$,
the geometric pebbling threshold $T_i$ of $H_i$ satisfies
$$\sqrt{n_i \log 2}\le T_i \le
\frac{2^{\sqrt{2\log_2 n_i}} e}{2\sqrt{\log_2 n_i}}(1 - (\log_2 n_i)^{-1/4})^{-1}n_i,$$ where
$n_i$ is the number of vertices in $H_i$. Define the function $t(n)$ by
$t(n_i):=T_i$ for each $i\in\ZZ_{>0}$ and $t(n):=n$ if $n$ is not equal to any $n_i$.
Now apply \cite[Theorem 11]{moewxxxx} and
argue as in the proof of \cite[Theorem 1.3]{bek2003} to prove that $t(n)$ is a pebbling threshold
for $(H_i)_{i\in\ZZ_{>0}}$.
\end{proof}
\begin{theorem} \label{tbouquet3}
There is some constant $K>1$ such that,
if $n\ge 2$ is an integer and
$$\sqrt{n} \le t \le \frac{2^{\sqrt{2\log_2 n}}}{\sqrt{\log_2 n}}n,$$
then there is some connected graph $H$ with $n$ vertices whose geometric pebbling
threshold is between $t/K$ and $K t$.
\end{theorem}
\begin{proof}
Set $L_0:=L_0(\frac 12)$.
We are free to choose an arbitrary connected graph for $H$
for a finite number of values of $n$,
at the cost of worsening $K$, so we can assume that $n\ge 2^{2 L_0}$, $n/(4\log_2 n)\ge G_0$,
and $\sqrt{2 \log_2 n} +1\le \log_2 n$.
Then, we will always choose $H$ to be some ${\cal B}_{n,g,L}$. Set $\beta:=t/n$
and $\beta_c:=2^{\sqrt{2 \log_2 G_0}}e/(2 \sqrt{\log_2 G_0})$.
\begin{enumerate}
\item If $\beta<\beta_c$, $g$ will always be 1. Let ${\hat L}:=1+\log_2(\Phi(\beta)n)$.
Then
we let $L$ be $L_0$ if ${\hat L}< L_0$, $\floor{\hat L}$ if $L_0\le {\hat L}\le (\log_2 n)-L_0$,
and $\floor{\log_2 n}-L_0$ if ${\hat L}> (\log_2 n)-L_0$.
\item If $\beta\ge \beta_c$, let $g$ be the maximal integer in $G_0$, $G_0+1$, \dots,
$\floor{n/(4\log_2 n)}$ with $2^{\sqrt{2\log_2 g}}e/(2 \sqrt{\log_2 g})\le \beta$.
Let $L$ be $\ceil{(\log_2 n)+\sqrt{2 \log_2 g}}$.
\end{enumerate}
It is straightforward to verify that, regardless of $t$ or $n$,
the geometric pebbling threshold $t'$ of $H$ can then be computed with Theorem \ref{tbouquet1} or
Theorem \ref{tbouquet2}, and that there is some absolute constant $K>1$ such that
$t'/t$ is always in $[1/K, K]$.
\end{proof}
\begin{corollary} If $t(n)$ is any function of integral $n\ge 1$ such that $\sqrt{n} \le t(n)
\le 2^{\sqrt{2\log_2 n}}n/\sqrt{\log_2 n}$ for all $n\ge 2$, then there is some sequence of
connected graphs $(H_n)_{n\in\ZZ_{>0}}$ with pebbling threshold $t(n)$
such that $H_n$ has $n$ vertices for each $n$.
\end{corollary}
\begin{proof} Let $H_1$ be the 1-vertex graph, and, for each $n\ge 2$, let $H_n$ be the
connected graph
given by Theorem \ref{tbouquet3} which has $n$ vertices and geometric pebbling threshold between
$t(n)/K$ and $K t(n)$. Then, apply \cite[Theorem 11]{moewxxxx} and \cite[Theorem 1.3]{bek2003}
to prove that $t(n)$ is a pebbling threshold of $(H_n)_{n\in\ZZ_{>0}}$.
\end{proof}
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{\tt