Greg Martin
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Registered User
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2h |
revised |
Order type of the smallest set containing the identity function and closed under exponentiation edited tags |
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2d |
accepted | Are sums of the inverses of prime siblings finite? |
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2d |
answered | Are sums of the inverses of prime siblings finite? |
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May 20 |
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What is barycentric simplicial subdivision? I think you have some misconceptions about the objects being discussed. A simplex is a pretty general object - the specific angles and side lengths aren't relevant. The barycentric subdivision of a 2-dimensional simplex (triangle) is a set of six triangles; yes, these new triangles can be right triangles, but that's not a problem. |
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May 19 |
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Sequences equidistributed modulo 1 I still don't think you've nailed down exactly what you mean by "nontrivial". For example, in Weyl's and Vinogradov's results, you have an assumption on $\alpha$, namely that is irrational. Why is "irrational" okay but "normal in base 2" not okay? |
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May 17 |
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What is most current greatest lower bound on gaps between P2 almost primes I imagine the question being asked is: what is the greatest lower bound on the largest gap between P2s up to $x$, as a function of $x$? In other words, an analogue for P2s of Rankin's result on large gaps between primes. |
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May 15 |
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Applications of the Chinese remainder theorem I agree that any $k+1$ people can compute $f$, using, say, Lagarange interpolation ... but, using the Chinese remainder theorem? I don't see $k+1$ different moduli here - only (mod $p$). |
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May 14 |
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p such that p+1 has a large prime factor, effectively We would love to have $\theta$ larger - I don't know how large "large" needs to be in the paper you're citing. Also I'm pretty sure the proof would go through with the additional restriction $p\equiv 2\pmod 3$ if needed. |
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May 14 |
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Proof of the weak Goldbach Conjecture I for one certainly believe the proof is correct. It was already known modulo a large finite computation; the new ideas are aimed (I believe) towards lowering that modulo-computation, to the point where it has been done already. |
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May 13 |
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Another colored balls puzzle (part II) Navid, have you run simulations? |
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May 13 |
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Another colored balls puzzle I don't fully understand this answer. For the variable $X_i$, are you pretending that there are only two colors, $i$ and not-$i$? If so, there are steps in the real game, involving two different not-$i$ colors, that don't affect the pretend game. Furthermore, why would the real game's expectation be the sum of the pretend expectations? - it seems to me that it should be the max of the pretend expectations. |
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May 13 |
answered | Another colored balls puzzle |
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May 13 |
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Another colored balls puzzle Note that once the balls have been reduced to two colors, say $m$ red balls and $n-m$ green balls, this is exactly the Gambler's Ruin problem; the expected number of turns remaining at this point is exactly $m(n-m)$. |
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May 9 |
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Maximum number of Vertices of Hypercube covered by Ball of radius R That seems like a good guess to me. You could try using the fact that any set of more than $2^k$ vertices of $H^n$ cannot be contained in a $k$-dimensional affine subspace of $\mathbb R^n$. |
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May 8 |
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What is known about the area of the symmetric Pythagorean tree? Definitely the question should be edited to include a precise definition of the Pythagorean tree, at least. |
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May 7 |
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estimate for i-th smooth number, gap between consecutive smooth numbers The terminology "smooth" is admittedly entrenched in the (English) literature on this topic, but it's a poorly chosen term. The more modern term, which hopefully will oust the older term, is "friable". |
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May 7 |
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average involving phi function Agreed. I think the constant is $|\zeta'(2)|/2\zeta(2)$, though I might have missed something. |
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May 7 |
answered | average involving phi function |
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May 2 |
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Is it true that $p^2+1$ is square free if $p>7$ is a Mersenne prime Indeed, given any $q\equiv1$ (mod $4$), the condition that $q^2 \mid (p^2+1)$ means $p^2\equiv-1$ (mod $q^2$), which is equivalent to $p$ lying in one of two residue classes modulo $q^2$. For example, when $q=5$, those residue classes are $7$ and $18$ (mod $25$). Solving $2^a-1\equiv7$ or $18$ (mod $25$) gives $a\equiv3$ or $18$ (mod $20$). Therefore $(2^a-1)^2+1$ is divisible by $5^2$ if and only if $a\equiv3$ or $18$ (mod $20$). While Mersenne primes are rare, nothing seems to keep them out of the residue class $3$ (mod $20$). Similar calculations hold for any $q\equiv1$ (mod $4$). |
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Apr 29 |
accepted | Least non primitive root |
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Apr 29 |
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Least non primitive root Yes, the lower bound would apply only for infinitely many $q$. That's the best one can hope for: the only lower bound that would apply for all prime $q$, or even almost all prime $q$, would be $n(q)\ge2$ (assuming Artin's conjecture). |
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Apr 29 |
awarded | ● Civic Duty |
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Apr 29 |
answered | Least non primitive root |
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Apr 29 |
revised |
Least non primitive root added 15 characters in body |
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Apr 28 |
accepted | Uniform convergence of $g(x) = \sum_{n=1}^\infty (1-x) a_n x^n$ |
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Apr 28 |
answered | Uniform convergence of $g(x) = \sum_{n=1}^\infty (1-x) a_n x^n$ |
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Apr 28 |
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Bounds for the largest divisor of n less than n^0.5 From Ford's paper, cited above, Theorem 1(v): when $\epsilon$ is sufficiently small, you still don't get density 1 from $f(n) = n^{1/2-\epsilon}$. You do get density 1 from $f(n) = n^{1/4}$, although I can't immediately tell whether that's the best exponent or not. |
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Apr 28 |
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A divergent series related to the number of divisors of of p-1 Presumably your $N$ and $x$ are the same. Note that $e^{a+b/x+c\ln x} = e^a x^c e^{b/x} \sim e^a x^c$. So your model is simply predicting that the sum is asymptotic to a constant times a power of $x$. This is almost surely false: the sum is bounded above by $cx/\log x$ by the prime number theorem, and bounded below by $cx/(\log x)^2$ if there are infinitely many Sophie Germaine primes. As I commented on the original post, the sum is probably asymptotic to a constant times $x(\log x)^{-3/2}$. |
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Apr 25 |
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A divergent series related to the number of divisors of of p-1 FYI: Since $1/d(n)$ is a multiplicative function that is $1/2$ on every prime, standard heuristics predict that the summatory function $\sum_{n\le x} 1/d(n)$ should have order of magnitude $x(\log x)^{1/2-1}$. (One could probably even write down the leading constant if careful.) On the further heuristic that random shifted primes $p-1$ aren't all that different from random $n$, one predicts that $\sum_{p\le x} 1/d(p-1)$ should have order of magnitude $x(\log x)^{-3/2}$. |
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Apr 25 |
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A divergent series related to the number of divisors of of p-1 Nitpick: the prime number theorem has as an easy consequence the fact that $p_n-1\le n^2$ for all sufficiently large $n$. |
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Apr 22 |
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Intersection of two arithmetic progressions And when it is nonempty, the intersection is a single arithmetic progression modulo $ac/\gcd(a,c)$. What you have written is equivalent, I think, but much more complicated. |
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Apr 22 |
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On the proof of Modified Vitali Lemma. Come to think of it, I wouldn't be surprised if a Smith-Volterra-Cantor set were a counterexample to this. en.wikipedia.org/wiki/Smith–Volterra–Cantor_set |
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Apr 22 |
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On the proof of Modified Vitali Lemma. When you say "decreasing", you allow equality, right? And do you mean decreasing for $r$ sufficiently small in terms of $x$? Otherwise there are trivial counterexamples, such as $C$ being the unit disk. |
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Apr 20 |
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A “bit” of primes I'll add: starting at a specific prime $p_n$, there is a slight tendency for the bit-1 of the next prime $p_{n+1}$ to be the opposite of the bit-1 of $p_n$; basically, this is because $p_n+2$ gets first crack at being prime, before $p_n+4$ (and $p_n+6$ also gets its chance before $p_n+8$, etc.). However, this slight bias gets smaller and smaller as the primes involve increase. |
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Apr 18 |
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Giving a general term of a recursive function, and upper bound for it Your question (2) is a real question; but for the life of my I can't see how you didn't answer question (1) even before you asked it. |
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Apr 18 |
awarded | ● Nice Answer |
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Apr 17 |
answered | Is rigour just a ritual that most mathematicians wish to get rid of if they could? |
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Apr 11 |
accepted | Transcendency of numbers of a special form. |
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Apr 11 |
answered | Transcendency of numbers of a special form. |
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Apr 10 |
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Poles of products of Gamma functions Are you missing a square root in the definition of your product? Otherwise I can't imagine why it should matter that the argument of $\Gamma$ is near a square. |
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Apr 10 |
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Asymptotics of the Number of Non-Isomorphic Equivalence Relations and the Number of Non-Isomorphic Relations This seems like a strange use of the term "kernel" to me. |
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Apr 5 |
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Impossibility of certain carmichael numbers Is your first sentence at all related to your question? |
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Apr 4 |
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Ordinary Generating Function for Mobius Fair enough, although on the first page it cites unconditional results as well. |
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Apr 3 |
answered | Ordinary Generating Function for Mobius |
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Apr 3 |
awarded | ● Citizen Patrol |
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Apr 2 |
answered | Two different definitions of Erdos-Rényi random graph |
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Apr 2 |
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product 1+1/p in terms of Chebyshev’s theta or psi function Asymptotically, your product is $\log \theta(y)/\log\theta(x)$, by my answer below. But this won't have that good an error term. Taking $\log\theta(\cdot)$ seems like a pretty unnatural operation, since $\theta(\cdot)$ is a sum. |
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Apr 2 |
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A Diophantine equation involving prime powers. The equation $p^x-1=3^y$ forces $p$ to be odd, so you're really looking at $2^x-3^y=1$. So again Catalan's conjecture/Mihailescu's theorem says there are no solutions (and similarly for any odd prime in place of $3$). |
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Apr 1 |
awarded | ● Yearling |
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Mar 30 |
answered | product 1+1/p in terms of Chebyshev’s theta or psi function |
