Suppose $\alpha$ is a fixed given irrational number with $\alpha\in [A, B]\subset [0,1]$, are there any (efficient) methods to compute the least integer $n$ such that the decimal part of $n\alpha$ lies in $[0,1]\setminus [A,B]$?



(efficient) method to test $\{n\alpha\}\not\in [A, B]\subset [0,1]$
The multipliers which could be candidates for what you want form a very sparse subset of the integers. They can be found efficiently by


* Finding the (start of the) continued fraction expansion of $\alpha$
* Using it to find certain especially good rational approximants to $\alpha$
* Taking the denominators of these fractions and finding the first number from that brief list which works.


I'll illustrate with an example: 


I have in mind a certain real number with $0.4414961245 \le \alpha \le 0.4414961265$ 
Suppose too that I have $A \lt B$ and want the smallest positive $n=n(\alpha,A,B)$ so that the fractional part $\lbrace n\alpha \rbrace \in(0,A) \cup (B,1).$ 


A few calculations will show: If $n(\alpha,A,B) \lt 10000$ then it is one of the $22$ values $1,2,3,5,7,9,16,25,34,43,77,111,188,265,$$453,641,829,1470,2299,3128,3957,7085$ . 


By quick calculations it is easily found that the continued fraction expansion of $\alpha$ begins 


$$\alpha=0+\frac1{2+}\frac1{3+}\frac1{1+}\frac1{3+}\frac1{2+}\frac1{2+}\frac1{3+}\frac1{1+}\frac1{3+}\frac1{2+}\frac1{2+}\frac1{?+\cdots}$$ where the unknown $12$th and later denominators require more precise bounds on the value of $\alpha.$


I will use the notation $\frac{a}{b} \vee \frac{c}{d}=\frac{a+c}{b+d}$ for the ["mediant addition"](http://en.wikipedia.org/wiki/Mediant_%28mathematics%29) of fractions.


* The $0$ and $2$ give $\mathbf{\frac01}$ and $\mathbf{\frac12}$ as approximants to $\alpha$ as well as the first two potential multipliers.


The following $9$ denominators $3,1,3,2,2,3,1,3,2$, with sum $20$, give us $20$ more approximants of which $10$ are convergents, the $20$ denominators are the rest of the potential multiplier list:


* $3 \Rightarrow \frac01 \vee \frac12=\frac13,\ \frac13 \vee \frac12=\frac 25,\ \frac25 \vee \frac12=\LARGE{\frac37}$
* $1 \Rightarrow \frac12 \vee \frac37=\LARGE{\frac49}$
* $3 \Rightarrow \frac37 \vee \frac49=\frac7{16},\ \frac7{16} \vee \frac49=\frac{11}{25},\ \frac{11}{25}\vee \frac49=\LARGE{\frac{15}{34}}$
* $2 \Rightarrow \frac49 \vee \frac{15}{34}=\frac{19}{43},\ \frac{19}{43} \vee \frac{15}{34}=\LARGE{\frac{34}{77}}$
* $2 \Rightarrow \frac{49}{111},\LARGE{\frac{83}{188}}$
* $3 \Rightarrow \frac{117}{265},\frac{200}{453},\LARGE{\frac{283}{641}}$
* $1 \Rightarrow \LARGE{\frac{366}{829}}$
* $3 \Rightarrow \frac{649}{1470},\frac{1015}{2299},\LARGE{\frac{1381}{3128}}$
* $2 \Rightarrow \frac{1747}{3957},\LARGE{\frac{3128}{7085}}$


Here is a list of the pairs $[k,\lbrace k\alpha \rbrace]$ arising from the denominators 
$$[1,.44], [2, .88], [3, .32], [5, .20], [7, 0.09], [9, .97], [16, 0.06], [25, 0.03], [34, 0.01],$$
$$ [43, .98], [77, .995], [111, 0.006], [188, 0.001], [265, .996], [453, .997], [641, .99902], $$$$[829, 0.0002], [1470, .9993], [2299, .9995], [3128, .9998], [3957, 0.0001], [7085, 0.00005]$$ 


One can check that these are precisely the pairs up to $k=10000$ with $\lbrace k\alpha \rbrace$ closer to $0$ or to $1$ than for any previous value. Hence they are indeed the only possible values of $n(\alpha,A,B) \lt 10000$


In the event that **both** $A \lt 0.00005$ and $0.9998 \lt B,$ we would require a multiplier larger than $10000.$ This would require being more specific about the value of $\alpha$, extending the continued fraction, and finding a few more denominators.


