Gerhard Paseman
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Unregistered User
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1d |
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how to proof this Stirling related equation Nice avoidance of Stirling. To avoid confusion with misuse of the empty product, I recommend using the first line for $1 \leq i \leq d,$ and then assert the second line for $0 \leq i \leq d.$ Gerhard "Don't Want Greater Than One" Paseman, 2013.05.17 |
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1d |
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how to proof this Stirling related equation By the way, I would like some context. If this was a homework problem, I expect you to credit MathOverflow in your writeup. If it wasn't, some acknowledgement would still be appreciated. Gerhard "Or Maybe A Gift Card" Paseman, 2013.05.17 |
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1d |
answered | how to proof this Stirling related equation |
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1d |
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how to proof this Stirling related equation You might fare better if you consider separately the cases 2d < m and 2d > m. Gerhard "Ask Me About Binomial Sums" Paseman, 2013.05.17 |
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2d |
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General and translational Birkhoff lattices. Equational classes. Also William DeMeo (who has appeared on MathOverflow in the past) did some work with Ralph Freese, another lattice researcher. William might be able to say something relevant toward your questions. Gerhard "Ask Me About System Design" Paseman, 2013.05.16 |
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2d |
answered | General and translational Birkhoff lattices. Equational classes. |
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2d |
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Integers n such that sigma(n)=omega(n)n and omega(n) divides n I just noticed in the above that I did not use the second of the above conditions, only the first. If we use a weaker condition, such as which multiperfect (or other rare class of) numbers n are divisible by the number of there own prime factors, I still think there will be finitely many such, but more machinery will be needed to prove it. This exercise makes me wonder if Zsigmondy's Theorem can be used for studying odd perfects. Gerhard "Odd Perfects? How Perfectly Odd." Paseman, 2013.05.16 |
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2d |
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Cardinals without choice: interpolation (reference wanted) Butch, you missed an answer posted by Asaf Karagila, which was later deleted because of (I suspect) a nonobvious use of choice. Somehow a set which was some form of union of (images of) the a cardinals was shown to exist inside the b cardinals and that this set exhibited the desired cardinality. While it was a nice presentation, I was unsure how the set guaranteed the desired cardinality in ZF. On the matter of personal remarks, it's best to pretend you are at a televised seminar when you are making them. Gerhard "Typing This While Mostly Naked" Paseman, 2013.05.16 |
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2d |
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Integers n such that sigma(n)=omega(n)n and omega(n) divides n more found |
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2d |
awarded | ● Fanatic |
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2d |
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Constrained minimum maximal distance. x* is near the middle of C, not at the top. For more examples, pick a domain D which is sufficiently eccentric, and consider the level sets of x such max(x,y) is some constant m as y ranges over D. One can then find C intersecting D and also a "least" level set with points outside of D. Gerhard "Ask Me About System Design" Paseman, 2013.05.15 |
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May 15 |
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Constrained minimum maximal distance. I don't think it is true, even if you restrict C and D to be in a class of (pairs of) congruent triangles. It might become true if you place bounds on the eccentricity of the shapes permitted. Gerhard "Ask Me About System Design" Paseman, 2013.05.15 |
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May 14 |
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Integers n such that sigma(n)=omega(n)n and omega(n) divides n Interesting. I use arithmetic and not much more than that to get $\omega(n) \lt 5$. You might try a less analytic approach. Gerhard "Doing Much More With Less" Paseman, 2013.05.14 |
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May 14 |
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Integers n such that sigma(n)=omega(n)n and omega(n) divides n And it looks like 30240 is another example. I suspect there are not many more. Gerhard "Ask Me About Premature Speaking" Paseman, 2013.05.14 |
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May 14 |
answered | Integers n such that sigma(n)=omega(n)n and omega(n) divides n |
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May 14 |
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Integers n such that sigma(n)=omega(n)n and omega(n) divides n Darn. Looks like I spoke too soon. I will see if I can prove that there are no other examples. Gerhard "Drats! Foiled Again By Counterexample!" Paseman, 2013.05.14 |
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May 14 |
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Integers n such that sigma(n)=omega(n)n and omega(n) divides n Even if you relax the first equation to n being multiply perfect, I suspect there will still be sharp limits for divisibility, e.g. omega might need to be even and possibly abundant itself in order for it to be a factor of n and significantly greater than 2. Gerhard "Feels That's How It Is" Paseman, 2013.05.14 |
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May 14 |
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Integers n such that sigma(n)=omega(n)n and omega(n) divides n No. Consider how the product p/(p-1) grows where the product is taken over the first omega many primes. This puts a sharp upper bound on omega and should imply n is even for the first equation to hold. If omega is greater than 2, an exhaustive search should finish it off. Gerhard "Ask Me About Pi Inverse" Paseman, 2013.05.14 |
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May 14 |
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What is known about a^2 + b^2 = c^2 + d^2 Using the word "random" to describe the notes may be viewed as somewhat disparaging, as well as inaccurate. A term like "variegated" might be preferred, as I find more resemblance between Piezas's notes and a botanical garden than I do between his notes and some of the backyards I've seen lately. Gerhard "As Lovely As A Trie" Paseman, 2013.05.14 |
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May 13 |
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Another colored balls puzzle Ah, I misread it as different pairs of colors, not different pairs of colored balls. Also, I should have thought of difference in number of heads and tails being m, rather than a string of m heads, in the two color scenario with m balls of one color. Greg's analysis is looking good to me now. Gerhard "It's All About Good Looking" Paseman, 2013.05.13 |
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May 13 |
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Another colored balls puzzle Why is 5 a denominator? Isn't the chance of a transition from 211 to 22 actually 1/3? Similarly for the other transitions out of 211? Gerhard "What Does Probability Really Mean?" Paseman, 2013.05.13 |
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May 13 |
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Another colored balls puzzle For large n, I would guess no bound. To get to two colors is nontrivial, but I would expect that to be reached within an exponential in n number of tries. Gerhard "Of Course I'm Just Guessing" Paseman, 2013.05.13 |
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May 13 |
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Can a composition with itself of a universal self-map be non-universal? OK. When I was Ralph's student, he spelled his last name as McKenzie, as opposed to Mckenzie. I would be surprised if that has changed. Gerhard "More General Than Universal Now" Paseman, 2013.05.12 |
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May 13 |
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Can a composition with itself of a universal self-map be non-universal? Can you say more on why f and g in your example are universal? Also, can you say something more about McKenzie's example? Gerhard "Interested Minds Wish To Know" Paseman, 2013.05.12 |
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May 13 |
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how to prove a conjecture on a “canonical equivalent” of factoring Why do you have equality in your last step? Shouldn't the product be $(n+1)/q_0$? Gerhard "Ask Me About System Design" Paseman, 2013.05.12 |
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May 11 |
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A sequence based on Catalan–Mihăilescu problem Wlod: yes, for the problem as written, your form is appropriate. If one is looking for common differences between powers, 3^a - 2^c = 2^d - 3^b leads to a nice mod 8 restriction. I haven't taken it farther. Gerhard "Fast And Correct: Choose One" Paseman, 2013.05.10 |
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May 11 |
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A sequence based on Catalan–Mihăilescu problem Oops. Sign Error. Even so, trying the appropriate formulation with powers of 3 on one side and powers of 2 on the other side might appear in the literature. Gerhard "Let's Take Absolute Values Instead" Paseman, 2013.05.10 |
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May 11 |
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A sequence based on Catalan–Mihăilescu problem You might find it revealing to study the system 3^a + 3^b = 2^c + 2^d. That will show some constraints on the possiblities and likely answer some of your questions above. Gerhard "Ask Me About Elementary Analysis" Paseman, 2013.05.10 |
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May 10 |
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Optimal inspection path on a sphere Not quite an answer, but this variation may be worthwhile: consider an optimal (charge-seperated?) arrangement of disks on the sphere, such that traversing a steiner tree between the points gets the coverage desired. Gerhard "Ask Me About System Design" Paseman, 2013.05.10 |
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May 9 |
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Bounding a sum of binomial coefficients in terms of ‘the next one’ Also, for t far from 2, one can find decent approximations to the sum using a combination of geometric series and combining adjacent terms to express the sum as a sum of fewer terms of (n+1) choose stuff. Gerhard "Search MathOverflow For Binomial Approximation" Paseman, 2013.05.09 |
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May 9 |
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Bounding a sum of binomial coefficients in terms of ‘the next one’ My comment above would be improved by saying for large enough n, and using (1 - 1/t)/(1/t) as the target ratio. In any case, for t close to and less than 3, one can find k and then choose n accordingly. Gerhard "Still Answering The Question Affirmatively" Paseman, 2013.05.08 |
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May 9 |
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Bounding a sum of binomial coefficients in terms of ‘the next one’ Yes. For all n, the sum is minorized by a geometric series of length k and ratio close to (n - n/t -k)/(n/t - k). Now pick n very large so that the k's disappear, and so that the geometric series has sum large enough. This can be done for t arbitrarily close to 3. Gerhard "Ask Me About Sum Estimation" Paseman, 2013.05.09 |
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May 9 |
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Binary Operation on a Cubic Surface What is f(P,P)? I presume it will be P if you are serious about looking at the identities. Gerhard "Ask Me About System Design" Paseman, 2013.05.08 |
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May 8 |
accepted | Summing ratio of ratio of partial sums of binomial coefficients |
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May 8 |
revised |
Summing ratio of ratio of partial sums of binomial coefficients added 3 characters in body |
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May 8 |
answered | Summing ratio of ratio of partial sums of binomial coefficients |
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May 7 |
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An ultrafilter is a set of subsets containing exactly one element of each finite partition: reference request Tom, I've seen postings by a Butch Malahide on other fora, which may help you get in touch with him/her/it. In this day and age, referring to MathOverflow users by number should be socially acceptable. Gerhard "User 3528. Aliases 3206, 3371..." Paseman, 2013.05.07 |
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May 7 |
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Summing ratio of ratio of partial sums of binomial coefficients I really prefer my suggestion in the other post where the partial sum is written as s and is understood to depend on y and k. Rewriting ((y-1) choose k)/s as t, one gets the summand as y/(2+t), where t ranges from near 1/2^k to near (y-k)/k, giving that the sum will be something like n(k - O(log k)). Gerhard "Ask Me About System Design" Paseman, 2013.05.07 |
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May 2 |
answered | Modeling concurrent internet users |
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Apr 24 |
answered | Algebras with finite essential arity |
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Apr 23 |
accepted | Chains or Antichains slowly increasing |
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Apr 23 |
awarded | ● Yearling |
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Apr 22 |
answered | Chains or Antichains slowly increasing |
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Apr 22 |
answered | smallest number of comparisons needed |
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Apr 20 |
answered | Can every $\mathbb{Z}^2$ disk be pinball-reached? |
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Apr 17 |
accepted | References on techniques for solving equations with discontinuous functions such as floor and ceiling? |
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Apr 16 |
accepted | What is “Data” involved in a mathematical construction? |
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Apr 15 |
answered | References on techniques for solving equations with discontinuous functions such as floor and ceiling? |
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Apr 15 |
answered | What is “Data” involved in a mathematical construction? |
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Mar 11 |
answered | Square submatrix |
