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Given $\alpha\in\Bbb N$, can there be more than $(\log N)^4$ divisors (composites allowed) of $N$ in $\big[\frac\alpha2,\alpha\big]$ when $\sqrt N\in\big[\frac\alpha2,\alpha\big]$?

What is the maximum number of divisors (composites allowed) asymptotically?

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This is certainly possible. I will construct an example with $\alpha=\sqrt{N}$. That is, I will exhibit a square number $N$ with more than $(\log N)^4$ divisors lying in $[\tfrac{1}{2}\sqrt{N},\sqrt{N}]$.

Let $x>2$ be a large parameter. Consider $$ M:=\prod_{p\leq x}p \qquad\text{and}\qquad N:=M^2\lfloor\tfrac{2}{3}\log M\rfloor^{10}, $$ where the product is over the primes up to $x$. Note that $x\sim\log M$ by the prime number theorem. Hence, for large $x$, $M$ is divisible by any product $p_1p_2p_3p_4p_5$, where $p_i$ are distinct primes with $$ \tfrac{3}{5}\log M < p_1 < p_2 < p_3 < p_4 < p_5 < \tfrac{2}{3}\log M. $$ The number of such divisors $p_1p_2p_3p_4p_5$ of $M$ is $\gg(\log M/\log\log M)^5$ by the prime number theorem, i.e. more than $(\log N)^4$ when $x$ is large. Each of these divisors uniquely determines the divisor $Mp_1p_2p_3p_4p_5$ of $N$, and this divisor also satisfies $$ \tfrac{1}{2}\sqrt{N} < M\lceil\tfrac{3}{5}\log M\rceil^5 < Mp_1p_2p_3p_4p_5 < M\lfloor\tfrac{2}{3}\log M\rfloor^{5} =\sqrt{N}.$$ Done.

Added. Here I address the OP's second question which was added after my response above.

I claim that, for any $c<(\log 2)/3$, there are infinitely many square numbers $N$ with more than $\exp(c\log N/\log\log N)$ divisors lying in $[\tfrac{1}{2}\sqrt{N},\sqrt{N}]$. This is best possible up to the constant, because $c>\log 2$ would contradict Wigert's classical bound for the total number of divisors $\tau(N)$.

For the proof we present a variant of the construction above. We take $x$ and $M$ as before, and we decompose the interval $[1,\lceil\sqrt{M}\rceil]$ into $\ll\log M$ intervals of the form $[K,L]$ with integers $1\leq K < L \leq 2K$. Clearly one of these intervals, say $[K,L]$, contains $\gg\tau(M)/\log M$ divisors $d$ of $M$. Now we put $N:=M^2 L^2$, and we observe that each divisor $d$ considered above uniquely determines the divisor $Md$ of $N$. Moreover, this modified divisor $Md$ satisfies $$ \sqrt{N}/2=ML/2\leq MK\leq Md\leq ML=\sqrt{N}, $$ hence to finish the proof it suffices to remark that, for large $x$, $$ \tau(M)/\log M = \exp((\log 2+o(1))\log M/\log\log M)>\exp(c\log N/\log\log N).$$ This inequality follows from the prime number theorem combined with the bounds $N<2M^3$ and $3c<\log 2$. Done again.

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  • $\begingroup$ Yes. If $n\geq t$ and $n\in\mathbb{Z}$, then $n\geq\lceil t\rceil$. Similarly, if $n\leq t$ and $n\in\mathbb{Z}$, then $n\leq\lfloor t\rfloor$. Hence my last display is clear with $\leq$ around $Mp_1p_2p_3p_4p_5$. However, we talk about a product of different primes, so strict inequality also follows (namely for $p_2$, $p_3$, $p_4$ the intermediate inequalities are strict). Note also that $\leq$ is sufficient in the last display. $\endgroup$
    – GH from MO
    Nov 24, 2015 at 23:15
  • $\begingroup$ @Arul: There is no such bound, i.e. you can get $(\log N)^{1000}$ divisors if you wish. In fact the number of divisors in the given range can be much larger than polylogarithmic, but I have not attempted to determine the maximum order of this quantity. $\endgroup$
    – GH from MO
    Nov 25, 2015 at 3:04
  • $\begingroup$ It will be useful for my purposes if I have a good bound $\endgroup$
    – user76479
    Nov 25, 2015 at 4:33
  • $\begingroup$ @Arul: See my added section, which answers your second question up to the precise value of the best constant $c$. $\endgroup$
    – GH from MO
    Nov 25, 2015 at 6:43

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