Consider a continuous time doubling branching random walk on the nonnegative integers. It starts with a particle at $0$. The initial particle dies after producing two offspring at $1$, each after an independent exponential$(1)$-distributed amount of time. In the same fashion, a particle born at $k$ will die upon producing two offspring at $k+1$ after, each after an independent exponential$(1)$-distributed amount of time. Suppose the last of the $2^n$ particles born at $n$ occurs at time $T$. This is known as the "Last-Birth Problem" and is fairly well studied. We are curious how many particles reach $n$ just before $T$. 


To make this concrete let $X\_n$ be the number of particles that reach $n$ at times in $[T-1,T)$. Is there a function $f(n)\to \infty$ such that $P(X\_n > f(n)) \to 1$? 


Perhaps this is false and only $O(1)$ particles arrive in the final second. However, there may be a huge wave of particles just in front of the last one. 



Almost last births in branching random walk
Consider the binary tree of ancestry (thus, the vertices at depth $n$ are the particles that reach location $n$). Write on each edge $e$ between level $n-1$ to level $n$ the time it takes to generate the particle at level $n$, and call this variable $X\_e$. If I understand correctly your model, these times are iid exponential (1). For any vertex $v$, let $S\_v$ be the partial sum of $X\_e$ for $e$ on the geodesic connecting the root to $v$. If I understand correctly,
$T=\max\_{v: |v|=n} S\_v$. The question you are asking is then a question on the extremal process of this branching random walk (how many particles in an interval of size $1$ to the left of $T$). By a result of Madaule, <https://arxiv.org/pdf/1107.2543.pdf>, the extremal process converges to 
a decorated Poisson point process (with intensity $e^{-cx}$, some explicit $c$) viewed from its tip, and therefore $X\_n$
is a tight sequence. So $f(n)=O(1)$.


