# Gerhard Paseman

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 Name Gerhard Paseman Member for 3 years Seen 57 mins ago Website Location Age
 1d comment how to proof this Stirling related equationNice 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 1d comment how to proof this Stirling related equationBy 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 1d answered how to proof this Stirling related equation 1d comment how to proof this Stirling related equationYou might fare better if you consider separately the cases 2d < m and 2d > m. Gerhard "Ask Me About Binomial Sums" Paseman, 2013.05.17 2d comment 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 2d answered General and translational Birkhoff lattices. Equational classes. 2d comment Integers n such that sigma(n)=omega(n)n and omega(n) divides nI 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 2d comment 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 2d revised Integers n such that sigma(n)=omega(n)n and omega(n) divides nmore found 2d awarded ● Fanatic 2d comment 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 May15 comment 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 May14 comment Integers n such that sigma(n)=omega(n)n and omega(n) divides nInteresting. 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 May14 comment Integers n such that sigma(n)=omega(n)n and omega(n) divides nAnd it looks like 30240 is another example. I suspect there are not many more. Gerhard "Ask Me About Premature Speaking" Paseman, 2013.05.14 May14 answered Integers n such that sigma(n)=omega(n)n and omega(n) divides n May14 comment Integers n such that sigma(n)=omega(n)n and omega(n) divides nDarn. 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 May14 comment Integers n such that sigma(n)=omega(n)n and omega(n) divides nEven 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 May14 comment Integers n such that sigma(n)=omega(n)n and omega(n) divides nNo. 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 May14 comment What is known about a^2 + b^2 = c^2 + d^2Using 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 May13 comment Another colored balls puzzleAh, 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 May13 comment Another colored balls puzzleWhy 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 May13 comment Another colored balls puzzleFor 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 May13 comment 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 May13 comment 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 May13 comment how to prove a conjecture on a “canonical equivalent” of factoringWhy 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 May11 comment A sequence based on Catalan–Mihăilescu problemWlod: 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 May11 comment A sequence based on Catalan–Mihăilescu problemOops. 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 May11 comment A sequence based on Catalan–Mihăilescu problemYou 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 May10 comment Optimal inspection path on a sphereNot 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 May9 comment 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 May9 comment 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 May9 comment 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 May9 comment Binary Operation on a Cubic SurfaceWhat 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 May8 accepted Summing ratio of ratio of partial sums of binomial coefficients May8 revised Summing ratio of ratio of partial sums of binomial coefficientsadded 3 characters in body May8 answered Summing ratio of ratio of partial sums of binomial coefficients May7 comment An ultrafilter is a set of subsets containing exactly one element of each finite partition: reference requestTom, 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 May7 comment Summing ratio of ratio of partial sums of binomial coefficientsI 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 May2 answered Modeling concurrent internet users Apr24 answered Algebras with finite essential arity Apr23 accepted Chains or Antichains slowly increasing Apr23 awarded ● Yearling Apr22 answered Chains or Antichains slowly increasing Apr22 answered smallest number of comparisons needed Apr20 answered Can every $\mathbb{Z}^2$ disk be pinball-reached? Apr17 accepted References on techniques for solving equations with discontinuous functions such as floor and ceiling? Apr16 accepted What is “Data” involved in a mathematical construction? Apr15 answered References on techniques for solving equations with discontinuous functions such as floor and ceiling? Apr15 answered What is “Data” involved in a mathematical construction? Mar11 answered Square submatrix