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The Masked Avenger's user avatar
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The Masked Avenger
  • Member for 11 years, 5 months
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General Euclid-Fermat sequences
The idea is that for a certain class of recurrences, you may be able similarly to show that simplicity plus slow growth plus some other condition means not coprime.
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General Euclid-Fermat sequences
I'm thinking of a result that for polynomials says that if p divides one value of f(n) then it also divides f(n+kp) for p a prime. It might be possible to replace polynomial by a slightly larger class of functions.
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General Euclid-Fermat sequences
Not much. You can find a version in Crandall and Pomerance. Rather than give code, I'll give a partial output of difference arrays. From input array 1 I get output 2; using 2 I get 4,2; I use this to generate 6,4,2,4,2,4,6,2, which I use to get an array of 48 numbers corresponding to differences between consecutive coprimes of 210. If you haven't done it, it is fun to code. Besides assignment, array access and index increment, the operations are two additions to accumulate, a mod by p, and a test for nonzero. Use subtract if you hate mod.
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An easy-to-state elusive combinatorial problem (revisited)
I'm not convinced. For the values allowed for r, less than half might work for x, and a separate less than half might work for y. You need to argue in the main case that that does not happen (there is at least one r that works for both x and y), and I do not see that argument yet.
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General Euclid-Fermat sequences
You might be able to show for some recurrences that slow growing means existence of a p that divides more than one term. You might also enjoy wheel sieving, which is a sped-up Eratosthenes sieve; I have four lines of obfuscated code which generate such sieves available upon request.
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Mathematics of privacy?
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Mathematics of privacy?
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bounded partitions and bounded signed partitions of integers
I don't know, except that you have a prism instead of a cube. I remember a web page of Kevin Brown (formerly at seanet.com) that attempted this problem. You might find some asymptotics there.
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Algorithm for load balancing during exchange between pairs
I don't know how it can be done. For seriously unbalanced cases, you might try a greedy approach: find vertices i and j such that t_ij is maximal. If there are several, find those forming a cycle. Choose direction(s) to minimize the imbalance, do the transfer, and set the tij to zero. Except for cycle finding, this can be done pretty quickly.
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Counting factors: is this approach in the literature on multiperfect numbers?
The intent is not to check the approach. The intent is to check the collective memory to see if someone remembers an approach similar to or even mildly resembling this. If someone says "this reminds me of paper X", that remark may prove useful. I will do the research regardless, but if someone has a useful pointer, that will speed things up for me. Or is what I am asking not a reference request?
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An easy-to-state elusive combinatorial problem
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An easy-to-state elusive combinatorial problem
You can imagine a union of a parameterized set of squares. Imagine drawing with a square shaped pen tip in a paint program, for example.