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It is well known, that given an integer $n$ whose prime factorization is $$ n = p_1^{a_1} \times p_2^{a_2} \times \ldots \times p_n^{a_n} $$ the number of its positive divisors is given by $$ d(n) = (a_1+1) \times (a_2+1) \times \ldots \times (a_k+1) $$ Now the question is the following: How many of the above divisors of $n$ lie between given bounds $a$ and $b$? Is there a systematic method for answering this question ?

In other words: given two positive integers $a

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    of course it is well defined. – 2017-01-26
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    does checking every divisor count? – 2017-01-26
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    well, that counts for an algorithm however my question is what would that function be. – 2017-01-26
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    You might as well ask for a function $d'(n,b)$ that counts divisors below $b$. Then $d(n,a,b)=d'(n,b)-d'(n,a)$ – 2017-01-26
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    @KonKan Not only is it well-defined, it's clearly periodic in $n$ (with a period that depends on $a,b$). – 2017-01-26
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    @Erick Wong: could you elaborate further on that? Maybe i'm missing something simple but i don't really see why this should be periodic on $n$. – 2017-02-01
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    @KonKan Sure thing. Fix any $a$[a,b]$. Let $L$ be the lcm of the set $\{a,a+1,\ldots,b\}$. Then any integer $c$ with $a\le c \le b$ by definition divides $L$. In particular, $c \mid x \iff c \mid x+L$. It follows that $d(n,a,b) = d(n+L,a,b)$, so $d(\cdot,a,b)$ is periodic with period at most $L$. – 2017-02-02
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    ah .. ok, now i see what you mean. thank you for noticing that! – 2017-02-03

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Here is an approximation that works well when $n$ is enormous, has few factors, and $b$ is large but small compared with $n$. We will look for a function $d'(n,b)$ that counts divisors of $n$ that are below $b.$ As an example, we will consider the case where $n=2^k3^m$ and assume that $2^k, 3^m \gt b.$ At each of the lattice points $(x,y)$ in the first quadrant we can associate the log of one potential factor, so at $(x,y)$ we associate $x \log 2 + y \log 3$. The factors less than $b$ are the lattice points below the line $x \log 2 + y \log 3 =\log b$. The area of the triangle is $\frac {(\log b)^2}{2 \log 2 \log 3}$ so we would expect that many factors less than $b$. The extension to more factors is clear. If there were prime factors $2,3,5$ the number of factors less than $b$ would be $\frac {(\log b)^3}{3! \log 2 \log 3 \log 5}$ because we would have a tetrahedron in a $3D$ lattice.

The limitations are easy to see. If there are not enough factors of $2$ or $3$ to exceed $b$ our triangular region can become a rectangle with the corner cut off. The line or plane dividing the factors below $b$ from those above will pass between lattice points, so there is uncertainty at the boundary. As the boundary is one dimension lower than the bulk of the lattice points this error reduces as $n$ gets large. A feel for the error comes from the fact that for $b=10^6$ the area formula gives $125$ factors while there really are $142$. For $b=10^9$ the area formula gives $282$ compared to $306$.

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    thank you for providing this. Although it does not provide a definite answer to the question, it is certainly a very interesting approximation and nicely presented. I think that the initial problem virtually "breaks" into a system of simultaneous diophantine-like inequalities and what you are doing here is a kind of graphical approximation to the solution of such a problem. +1. Do you know of some further references on similar methods, for this or related problems? – 2017-02-01
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    On the other hand, I have discussed this problem elsewhere but I have not received significant feedback (not even close). So, i think it is fair -after a few days have passed- to accept your answer. – 2017-02-01
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Several mathematicians have done research on this very topic. A good place to start might be this paper of Kevin Ford: "The distribution of integers with divisors in a given interval."

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    thank you for the reference. However, after taking a quick look through it, i think that it discusses a similar but different problem. It is interesting in any case. – 2017-02-01