What is the best known upper bound for G(k) in Waring's Problem? Waring's problem: for fixed $k, s$ write natural numbers $n$ in the form $n=x_1^k+\dots+x_s^k$.
$$
G(k):= \min\{s:\text{ all sufficiently large $n$ can be written as above}\}
$$
A quick search / Wikipedia / Vaughan-Wooley's survey paper all suggest that the best upper bound for $G(k)$ is (Wooley)
$$
G(k)\leq k\log k+k\log\log k+O(k).
$$
Recent work of Bourgain, Demeter, Guth (https://arxiv.org/abs/1512.01565) and Wooley (https://arxiv.org/abs/1708.01220) on Vinogradov's theorem seems to improve several quantities related to Waring's problem and similar questions (eg. $g(k)$, which is as $G(k)$ but for all $n$). But I can't find any evidence of improvements in $G(k)$ following from this result. Does anyone know if such improvements are possible, and if so where to find them?
 A: While not exactly the classical quantity $G(k)$, there are partial improvements on asymptotic behavior for the Waring-Goldbach problem, namely for $H(k)$, defined to be the least integer $s$ such that every sufficiently large positive integer
congruent to $s$ modulo $K(k)$ may be written as $$p_1^k + p^k_2 + ··· + p^k
_s = n,$$
where $p_1,\dots,p_s$ are prime numbers and where $K(k)=\Pi_{(p-1)|k} p^{\gamma}$, and $\gamma$ is a parameter defined below. Shortly after Bourgain, Demeter, and Guth's resolving of the Vinogradov's conjecture Kumcev and Wooley used their results to improve Hua's estimate 
$$H(k)\leq k(4 \log k + 2 \log \log k + O(1)) \text{ as }k\rightarrow\infty, \text{ to }$$
$$H(k)\leq (4k − 2) \log k + k − 4 \text{ for } k\geq 3.$$
This was improved again later by Kumcev and Wooley to 
$$H(k) ≤ (4k − 2)\log k − (2 \log 2 − 1)k − 3$$
holding for large $k$. To compute the value $K(k)$, let natural $k$ and prime number $p$ be given, and define $\theta = \theta(k, p)$ to be the integer with $p^\theta|k$ but $p^{\theta+1}\nmid k$, and $\gamma=\gamma(k, p)$ by
$$ γ(k, p) =\begin{cases}
\theta + 2,\ \text{ when } p = 2 \text{ and }\theta > 0,\\
\theta + 1, \text{ otherwise.}
\end{cases}$$
