Enumerating factors in intervals Given $1<a<N-N^{1/\alpha}$ where $\alpha\geq2$, denote the number of distinct factors of $N$ in  $[a,a+N^{1/\alpha}]$ as $\sigma_{0,a}(N,\alpha)$ denote $\beta(N,\alpha)=\max_a\sigma_{0,a}(N,\alpha)$ which is an non-monotone increasing function of $N$. How fast does $\beta(N,\alpha)$ grow? What I am looking for is given $N_0$, what is $$\max_{N<N_0}\beta(N,\alpha)?$$
 A: Let $$\gamma(N,\alpha):=\max_{n< N} \max_{a<n-n^{1/\alpha}}\sigma_{0,a}(n,\alpha)$$ as in the question. Let also $\sigma_0(n)$ denote the number of all divisors of $n$ and $$\gamma(N):=\max_{n< N} \sigma(n).$$
We have that $\gamma(N) = N^{\Theta(1/\log\log N)}$ by the divisor bound (Wigert). This growth is essentially achieved by numbers of the form $n_T=\prod_{p<T} p$, because $n_t = e^{\Theta(T)}$ and $\sigma_0(n_T)= 2^{\Theta(T/\log T)}$ by the prime number theorem.  
(Recall the Big Theta notation: $f(n) = \Theta(g(n))$ if asymptotically $f$ is bounded above and below by $g$ up to constants.)
So $\gamma(N) = O(N^\epsilon)$, and since the linear measure of an interval $[a,a+N^{1/\alpha}]$ is only $N^{-\frac{\alpha-1}{\alpha}}$ of the full $[1,N]$, one might naively expect a much better bound on $\gamma(N,\alpha)$. However the divisors of a number spread better on a \emph{logarithmic scale}, so some such intervals do not even get ``infinitesimally smaller" in comparison to $[1,N]$. For these matters and much more see the very good [1].
Answer: in fact, we have, for fixed $\alpha\geq 2$ and for $N\to\infty$: 
$$
\gamma(N,\alpha) = N^{\Theta(1/\log\log N)}.
$$
Proof. The upper bound is 
$$\gamma(N,\alpha)\leq\gamma(N)= N^{\Theta(1/\log\log N)}.$$ For the lower bound, consider, say, numbers of the form $N=n_Tm$ with $m\sim n_T^{\alpha-1}$. So $$\gamma(N,\alpha) \gtrsim \sigma_0(n_T) = n_T^{\Theta(1/\log\log n_T)} = N^{\Theta(1/\log\log N)}.$$
[1]: Hall, R. R., & Tenenbaum, G. (1988). Divisors. Cambridge University Press
