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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 …
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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 …
Dave Benson's user avatar
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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 …
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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 …
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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 …
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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 …
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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 …
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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 …
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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.
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