bio | website | grpaseman.wordpress.com |
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location | S.F. Bay Area | |
age | ||
visits | member for | 5 years, 4 months |
seen | 2 days ago | |
stats | profile views | 4,652 |
The sdfae.* websites are not up (technical difficulties); until they rise, I invite you to check out http://grpaseman.wordpress.com for the month of October and your dose of System Design.
Ask Me About System Design. I am willing to do email correspondence on the subject.
Social Networking Data
__Location: (Headed to) Back from Seoul for ICM2014. See MathOverflow @ ICM2014 : We Want You! for detail, including email address
__Interests: General Algebra, Computability, Enumeration, Prime Gaps
__Project: Bounds on Jacobsthal's Function
__Project: Combinatorics of P's Ring Toss
__Have: copy of Erik Westzynthius' (Only?) Paper
__Want: symbolic dynamics on infinite directed sets, esp. forests
__Contact: through Will Jagy, or guess the following Hangman dro_d__r_ard__ma_l.com
(Please do not merge keep me, or even merge me: merge keep 3402 instead)
May 24 |
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The most number of points that realize only $k$ distinct distances
It is related, but not exactly the same. In Joseph's version, k (number of distinct distances) is fixed and n is wanted, whereas the literature you mention seems to me to have n fixed and estimates k given n and d. However, that entry a good place to start. Also, the book titled something like "Unsolved problems in geometry" might have a discrete portion that gets closer to Joseph's question. Gerhard "Looking From The Other End" Paseman, 2015.05.23 |
May 22 |
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How did Cole factor $2^{67}-1$ in 1903
It seems he uses most of the devices David Speyer lists in the posts, plus some other observations about representing $2^{67} -1$ as $((u+v)/2)^2 - ((u-v)/2)^2$. Anyway, he did do some sifting and a lot of modular arithmetic. Gerhard "Maybe It Was A Casio" Paseman, 2015.05.22 |
May 22 |
answered | How did Cole factor $2^{67}-1$ in 1903 |
May 20 |
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Find the intersection between two convex hulls, in this specific case
It looks like the solution set is invariant under permutations of the indices. It looks like it may be enough to consider finding l such that d(lhat) <e < d((l+1)hat), where hat of a number l has the first l coordinates 1 and the rest 0. With l in hand, you can now consider linear combinations of the right number of vectors and do it up to permutation of coordinates. Gerhard "Use Symmetry To Your Advantage" Paseman, 2015.05.20 |
May 19 |
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Looking for reference or proof to some facts stated on Anand Pillay's book
It is my hope that things like "that" are encouraged to happen (authoritative answers from such sources) when we keep this forum's quality level high. Unceasing vigilance and compassion, I say. Gerhard "Expect Quality To Follow Quality" Paseman, 2015.05.19 |
May 15 |
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Smallest constant so that there are at least $n/\log_2{n}$ primes between $n$ and a constant multiple of $n$
Indeed, such a computation has been outlined by GH from MO, who has even checked smaller values and found that $c=13/6$ does not quite cut it. Gerhard "Missed It By That Much" Paseman, 2015.05.15 |
May 15 |
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Smallest constant so that there are at least $n/\log_2{n}$ primes between $n$ and a constant multiple of $n$
You can probably use older inequalities from Rosser/Schoenfeld which apply for all $n \gt 286$ to do a proof by hand. Just check that $\epsilon$ is small enough to pull the result through. Gerhard "Rewriting Terms Really Helps Here" Paseman, 2015.05.15 |
May 15 |
answered | Smallest constant so that there are at least $n/\log_2{n}$ primes between $n$ and a constant multiple of $n$ |
May 15 |
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Smallest constant so that there are at least $n/\log_2{n}$ primes between $n$ and a constant multiple of $n$
No reason to be embarrassed. If it turns out that you had a thought like "Hey, maybe every even number $n$ can be a sum of two primes both at least $n/c$ in size by this easy counting argument...", you could reveal that and we could say "Yes, but ..." or "No, because ...", and answer that speculation for future readers. If the motivation is "mature enough" (it doesn't have to be "technical enough"), it's OK for MathOverflow. Just don't use too many words for it. Gerhard "Like To Know Your Thoughts" Paseman, 2015.05.15 |
May 15 |
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Smallest constant so that there are at least $n/\log_2{n}$ primes between $n$ and a constant multiple of $n$
Iterated log and a messed up version of $(\log n)^2$. It would be nice for you to edit the question not only to clarify but to include some motivation. Gerhard "If You Would, Pretty Please" Paseman, 2015.05.15 |
May 15 |
revised |
Important open problems that have already been reduced to a finite but infeasible amount of computation
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May 15 |
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Important open problems that have already been reduced to a finite but infeasible amount of computation
Thanks. Will fix. Apologies to both Kolyvagin and Konyagin Gerhard "Do Not Call Me Passman" Paseman, 2015.05.15 |
May 15 |
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Smallest constant so that there are at least $n/\log_2{n}$ primes between $n$ and a constant multiple of $n$
I can think of three interpretations for $\log_2 n$. Which one do you have in mind? Gerhard "Perhaps You Mean Base Two?" Paseman, 2015.05.15 |
May 15 |
answered | Important open problems that have already been reduced to a finite but infeasible amount of computation |
May 15 |
revised |
Important open problems that have already been reduced to a finite but infeasible amount of computation
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May 15 |
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Second differences of primes determined by increasing first differences: every positive even integer?
Another way to look at this is to consider "generalized power sequences", where one considers a sequence such that each kth difference is positive, and then one asks how bounded such a sequence is (or what range the kth difference has). I see this as not far from asking if any "positive squarish" subsequence contained in the primes must have unbounded coefficient. Note that if the second difference were constant for long, one would have a successful prime-producing quadratic polynomial. I have a related question on squares here. Gerhard "Giving Another Avenue Of Research" Paseman, 2015.05.15 |
May 15 |
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Long gaps between primes
You might note that they removed a factor of logloglogn from the denominator. That may inspire more click throughs to the abstract. Gerhard "Think Of It As Teaser" Paseman, 2015.05.14 |
May 14 |
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A question on the bounds of the $n$-th composite $c_n$
It is an additional parameter such that the kth primorial is less than half of $c_{p_n - n}$. So if k =4, then we should be talking about composites larger than 420. I think it is not too hard to show the inequality holds for $m$ such that $c_{m-n}$ is less than the $k$th primorial, since the interval from 1 to the first primorial tends to have more primes in it than any successive interval of integers of that length. You might even extend it to cases where $c_{m-n}$ is less than twice the $k$th primorial. Gerhard "It's The Middle That's Toughest" Paseman, 2015.05.14 |
May 13 |
answered | A question on the bounds of the $n$-th composite $c_n$ |
May 13 |
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A question on the bounds of the $n$-th composite $c_n$
I ask because $m$ is not restricted in the inequality above. Even taking $c_0=1$, I get $c_{p_n -n} +c_0=p_n +2$ for $m=n$ or $m=p_n$. When $m$ is not restricted, $c_{m-n}$ can have a negative index, and it makes sense to ask if there are other failures of the inequality. Gerhard "Trying To Get It Clear" Paseman, 2015.05.13 |