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Oct
5 |
awarded | Nice Answer |

Jan
25 |
awarded | Yearling |

Aug
8 |
awarded | Nice Answer |

Jun
24 |
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Maximal order of Hooley's Delta function?
How close is the maximum u to log(n)/2? Is there anything in the literature on the location of u? Or can it be proved that u is closer to log(delta(n))? Gerhard "Perhaps Jacobsthal's Function Is Related" Paseman, 2013.06.24 |

Jun
24 |
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For any n and some prime p there is an elemnet in Zp* of order n
I will agree that an approach involving arithmetic progressions like that does lead to a prime p with the desired property, and may be shorter if not cleaner than an approach involving Bang or Zsigmondy. I interpret Katz's answer as saying that the "question behind the question" is about primes in such progressions; this intepretation I disagree with, and suggest for self study of this problem an alternative. However, the poster has accepted Katz's answer; perhaps a proof more than understanding is wanted. Gerhard "To Each Their Own Heaven" Paseman, 2013.06.24 |

Jun
23 |
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algebra group theory
For that matter, let A be B direct summed with countably (or more) many copies of C. I think under certain moderate conditions this can characterize such A, but I don't have chapter 5 of "Algebras, Lattices, Varieties" handy. Gerhard "Ask Me About System Design" Paseman, 2013.06.23 |

Jun
23 |
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Equality of the determinants of certain submatrices of an orthogonal matrix
Recent work (2012) of Richard Brent and Judy-ann Osborn on determinants of binary matrices uses a similar (if not identical) observation by F. Szollozi to get better bounds on such determinants. Gerhard "Ask Me About Binary Matrices" Paseman, 2013.06.23 |

Jun
21 |
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On the primitive prime divisors of $q^n-1$
Just the fact that$\pi_3(2)=\pi_3(18)$ is intriguing to me. I wonder if there is a result similar to Stormers Theorem on finitely many consecutive numbers whose prime factors all belong to a finite set $S$ that would say for a given finite set $S$ of primes that $\pi_n(m) = S$ has only finitely many positive integer solutions $m$ and $n> 1$. Gerhard "Also Relates To Odd Perfects" Paseman, 2013.06.20 |

Jun
21 |
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The functional equation of Hofstadter's Q sequence
Yes, but until one gets to the word "meromorphic" in your post, there is no hint that differentiability is a concern of yours. I thought your issue would be whether it results in a total function on the reals greater than or equal to 1. In any case, it would be of interest to see if the function needs to be defined on (0,1) in the straightline version for the functional part of the definition to work. Gerhard "With Sides Reversed Is Not" Paseman, 2013.06.20 |

Jun
20 |
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The functional equation of Hofstadter's Q sequence
Nice pictures. Suppose instead of your polynomial you use a straightline approximation: choice 1 is f=1 for (0,2] and (x-1) for [2,3], while choice 2 is f=x on (0,1) and otherwise looks like choice 1. Can you comment on the results for either of these choices? Gerhard "Maybe Make Pretty Graphs Too?" Paseman, 2013.06.20 |

Jun
19 |
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For any n and some prime p there is an elemnet in Zp* of order n
More elementarily, the poster is asking for a (positive) integer b such that b^n - 1 is a multiple of some prime p while b^m - 1 is not such a multiple for m < n. A proof by Bang was given for n > 6 (pick b=2) and for fixed b and n all the allowed cases were characterized by Zsigmondy. Gerhard "Ask Me About System Design" Paseman, 2013.06.19 |

Jun
19 |
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For any n and some prime p there is an elemnet in Zp* of order n
Start by investigating the case that n is prime, and look at factors of a^n - 1. When you handle that, try n a prime power. Gerhard "Eventually, Look Up Zsigmondy's Theorem" Paseman, 2013.06.19 |

Jun
17 |
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Mersenne Prime Sequences
Euclid proved the opposite, for a certain iterated sequence, and thus for infinitely many other iterated sequences. Later Dirichlet improved upon this by showing it held for every possible permissible instance of the iterated sequence. Gerhard "Is Talking About Iterated Adding" Paseman, 2013.06.17 |

Jun
16 |
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Functions about integers which are divisible by all numbers less than or equal to a fractional power of the integer
For numbers k which are prime powers and n that are divisible by all integers up to but not including k, one can use ceil(log_k(n)) instead of ceil(log_2(n)). Gerhard "Ask Me About System Design" Paseman, 2013.06.16 |

Jun
16 |
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Functions about integers which are divisible by all numbers less than or equal to a fractional power of the integer
For odd numbers, I think ceiling(log_2(n)) is the function he wants. Gerhard "Happy Father's Day To You" Paseman, 2013.06.16 |

Jun
16 |
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Functions about integers which are divisible by all numbers less than or equal to a fractional power of the integer
Let g(n) be 1 + ceiling (log_2 n). Then n^1/g < 2, and so your divisibility relation holds. If you wish it to hold for odd numbers, you won't do much better than g(n). Gerhard "Close Enough Sometimes Good Enough" Paseman, 2013.06.16 |

Jun
14 |
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Partitions of numbers with restrictions on repetitons
Now that I've bothered to calculate it, I see my earlier bijection of partitions isn't. Oops again. Gerhard "So Sorry About That, Chief" Paseman, 2013.06.14 |

Jun
14 |
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Partitions of numbers with restrictions on repetitons
Oops, I think I mean n is even and k is odd. Gerhard "Is Sorry About That, Chief" Paseman, 2013.06.14 |

Jun
14 |
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Partitions of numbers with restrictions on repetitons
I would suggest F(s,1)=2 (or 0) in your clarifying example, and for purposes of generating functions, things might be more cleanly stated by using F(tn,v) for integral values of t instead of F(s,v) inside the sum. Except for when n and k are both odd, I don't know the answer. Gerhard "Ask Me About System Design" Paseman, 2013.06.14 |

Jun
14 |
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Partitions of numbers with restrictions on repetitons
Also, if n is even and k is odd, I think an odd partition for s is the complement of an even partition for an s', so for a reasonable (mis?)interpretation of your question, the answer is that there are the same number of odd partitions as there are even partitions. Gerhard "Saw Someone Else In Mirror" Paseman, 2013.06.14 |