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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 $$\max_{N<N_0}\beta(N,\alpha)?$$

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    $\begingroup$ This question is not too well formulated. What kind of answer do you expect? An upper bound by some elementary function of $N$ (and possible of $a$) like $(\log N)^{17}$? Otherwise we can say a tautology: the quantity in the question is equal to itself, period. $\endgroup$
    – GH from MO
    Sep 29, 2015 at 18:15
  • $\begingroup$ For every $N$, there an $a$ such that interval $[a,a+N^{1/\alpha}]$ holds maximum number of factors. What is this maximum as a function of $N$? $\endgroup$
    – user76479
    Sep 29, 2015 at 18:25
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    $\begingroup$ @GHfromMO I would interpret this question as seeking the asymptotics, for fixed $x$, for the maximum number of divisors of $n$ in $[a,a+n^x]$ over all all $a$ and all $n\leq N$. For $x\geq 1/2$, this is question about the extremal order of the divisor function (which has a well-known answer). My strong guess is that the maximum in question is approximately attained when $a=1$, which would mean that one would be able to answer this question by referring to the known results on the distribution of the smooth numbers. However, I do not know how to prove this. $\endgroup$
    – Boris Bukh
    Sep 29, 2015 at 18:56
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    $\begingroup$ @Arul: I don't think there is any explicit formula for this function of $N$ other than the definition itself. Boris Bukh's interpretation (see his comment above) makes more sense, and this harmonizes with my comment. At any rate, you should clarify your question, because as it stands it makes little sense. $\endgroup$
    – GH from MO
    Sep 29, 2015 at 19:16
  • $\begingroup$ @GHfromMO why is the post unclear? Do you agree there an $a\in[1,N-\delta]$ such that $[a,a+\delta]$ has maximum number of factors? Do you agree that this maximum value changes with $N$? Then it is a function of $N$? $\endgroup$
    – user76479
    Sep 29, 2015 at 21:23

1 Answer 1

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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

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