1
$\begingroup$

Is there a quick way to determine which integers $D < d\leqslant 2D$ are such that $d$ has a multiple in $[X, X + H]$? Here, $H$ should be thought of as much smaller than $D$, and $X$ larger than $D$. It is easy to find this in time linear in $D$ by just checking for some $d$ if $\lfloor X/d\rfloor\ne \lfloor (X + H)/d\rfloor$. Is there any way to do better, perhaps through some structure those $d$s that do have such a multiple must satisfy ?

$\endgroup$
1
  • 1
    $\begingroup$ If, idk, $H = O(1), D = X^{3/4}$, then surely computing all divisors of all numbers in $[X,X+H]$ is quicker, no? $\endgroup$ Commented Feb 1, 2023 at 8:02

1 Answer 1

1
$\begingroup$

Depending on the value of $D$ as compared to $X$, you may want to switch from scanning the interval $[D+1,2D]$ to the interval of co-factors: $\big[ \lceil \frac{X}{2D} \rceil, \lfloor \frac{X+H}{D+1}\rfloor \big]$.

Alternatively, you can just try to factor each number in the interval $[X,X+H]$ as suggested in the comments and directly find their divisors in the interval $[D+1,2D]$, or employ particular techniques used for the integer factorization such as sieving. For example, if one finds that a prime $p$ does not divide any integer in the interval $[X,X+H]$, then multiples $p$ can be excluded from the candidates in $[D+1,2D]$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.