Search Results
| Search type | Search syntax | 
|---|---|
| Tags | [tag] | 
| Exact | "words here" | 
| Author | 
                    user:1234 user:me (yours)  | 
            
| Score | 
                    score:3 (3+) score:0 (none)  | 
            
| Answers | 
                    answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234  | 
            
| Views | views:250 | 
| Code | code:"if (foo != bar)" | 
| Sections | 
                    title:apples body:"apples oranges"  | 
            
| URL | url:"*.example.com" | 
| Saves | in:saves | 
| Status | 
                    closed:yes duplicate:no migrated:no wiki:no  | 
            
| Types | 
                    is:question is:answer  | 
            
| Exclude | 
                    -[tag] -apples  | 
            
| For more details on advanced search visit our help page | |
Results tagged with nt.number-theory 
                Search options                                 answers only
                                 not deleted
                                 user 460592
    Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
            9
            votes
        
    Unusual clump of small prime numbers?
                This happens earlier at 18444 to 18450, and again from 21109 to 21115. It doesn't seem very special. If you want 37 as well, try 138411 to 138417.
This seems very much in line with Wojowu's naïve esti … 
            
        
       
    
            18
            votes
        
    111...11 base p = 111...11 base q
                I'm guessing that $1+5+25=31=1+2+4+8+16$ is the only example. There are certainly no more small examples, and probabilistically they get rare very quickly. But I only checked the first 80 primes to th … 
            
        
       
    
            5
            votes
        
            
                
                Accepted
            
    Primes above the distant prime neighbors
                This is probably true, and will almost certainly be hard to prove. The prime neighbours satisfying your hypothesis grow more quickly than one might expect, and the $p$ are listed in A002386. The corre … 
            
        
       
    
            10
            votes
        
    On the number of distinct prime factors of $p^2+p+1$
                There is the following theorem of Halberstam, "On the distribution of additive number-theoretic functions. III." Let $\omega(n)$ be the number of prime factors of $n$. Given any irreducible polynomial … 
            
        
       
    
            19
            votes
        
            
                
                Accepted
            
    Trivial homomorphism from a non-abelian group to an abelian group
                $SL_2(\mathbb{Z})$ is generated by $U=\left(\begin{smallmatrix}0&1\\-1&1\end{smallmatrix}\right)$ of order $6$ and $V=\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$ of order $4$, subject  … 
            
        
       
    
            6
            votes
        
            
                
                Accepted
            
    Closed formula for number of ones in a proper factor tree
                This seems to be Sloane's A002033, namely the number of perfect partitions of $n-1$. Since the OEIS doesn't give any closed formula, there probably isn't one, but it's probably worth checking the refe … 
            
        
       
    
            5
            votes
        
    Schur multiplier of a Chevalley group of type $D_5$
                According to Theorem 5.3 of the first paper of Mike Stein that you mention, the universal central extension of the Chevalley group of type $D_5$ over the integers is given by the Steinberg generators  … 
            
        
       
    
            3
            votes
        
            
                
                Accepted
            
    Product/quotient of factorials beget dyadic powers
                This is easy. Start with $$n!.2^n=(2n)(2n-2)\dots 2.$$ This implies that $$((2n-1)!(2n-3)!\dots 1!).n!.2^n=(2n)!(2n-2)!\dots 2!.$$ Then $$((2n)!(2n-1)!\dots 1!).n!.2^n=((2n)!(2n-2)!\dots 2!)^2.$$ Thus … 
            
        
       
    
            7
            votes
        
            
                
                Accepted
            
    Learning Inverse Galois Theory
                I would like to suggest that a good place to start is with John Thompson's work on the subject. He initiated the modern approach using the notion of "rigid" tuples of conjugacy classes.