# Greg Martin

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 Name Greg Martin Member for 3 years Seen 2 hours ago Website Location Age 43
 2h revised Order type of the smallest set containing the identity function and closed under exponentiationedited tags 2d accepted Are sums of the inverses of prime siblings finite? 2d answered Are sums of the inverses of prime siblings finite? May20 comment 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. May19 comment Sequences equidistributed modulo 1I 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? May17 comment What is most current greatest lower bound on gaps between P2 almost primesI 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. May15 comment Applications of the Chinese remainder theoremI 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 $p$). May14 comment p such that p+1 has a large prime factor, effectivelyWe 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. May14 comment Proof of the weak Goldbach ConjectureI 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. May13 comment Another colored balls puzzle (part II)Navid, have you run simulations? May13 comment Another colored balls puzzleI 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. May13 answered Another colored balls puzzle May13 comment Another colored balls puzzleNote 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 $m(n-m)$. May9 comment Maximum number of Vertices of Hypercube covered by Ball of radius RThat 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 $\mathbb R^n$. May8 comment 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. May7 comment estimate for i-th smooth number, gap between consecutive smooth numbersThe 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". May7 comment average involving phi functionAgreed. I think the constant is $|\zeta'(2)|/2\zeta(2)$, though I might have missed something. May7 answered average involving phi function May2 comment Is it true that $p^2+1$ is square free if $p>7$ is a Mersenne primeIndeed, 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 $4$). Apr29 accepted Least non primitive root Apr29 comment Least non primitive rootYes, 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). Apr29 awarded ● Civic Duty Apr29 answered Least non primitive root Apr29 revised Least non primitive rootadded 15 characters in body Apr28 accepted Uniform convergence of $g(x) = \sum_{n=1}^\infty (1-x) a_n x^n$ Apr28 answered Uniform convergence of $g(x) = \sum_{n=1}^\infty (1-x) a_n x^n$ Apr28 comment Bounds for the largest divisor of n less than n^0.5From 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. Apr28 comment A divergent series related to the number of divisors of of p-1Presumably 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 $x(\log x)^{-3/2}$. Apr25 comment A divergent series related to the number of divisors of of p-1FYI: 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 $x(\log x)^{-3/2}$. Apr25 comment A divergent series related to the number of divisors of of p-1Nitpick: the prime number theorem has as an easy consequence the fact that $p_n-1\le n^2$ for all sufficiently large $n$. Apr22 comment Intersection of two arithmetic progressionsAnd 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. Apr22 comment 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 Apr22 comment 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. Apr20 comment A “bit” of primesI'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. Apr18 comment Giving a general term of a recursive function, and upper bound for itYour 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. Apr18 awarded ● Nice Answer Apr17 answered Is rigour just a ritual that most mathematicians wish to get rid of if they could? Apr11 accepted Transcendency of numbers of a special form. Apr11 answered Transcendency of numbers of a special form. Apr10 comment 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. Apr10 comment Asymptotics of the Number of Non-Isomorphic Equivalence Relations and the Number of Non-Isomorphic RelationsThis seems like a strange use of the term "kernel" to me. Apr5 comment Impossibility of certain carmichael numbersIs your first sentence at all related to your question? Apr4 comment Ordinary Generating Function for MobiusFair enough, although on the first page it cites unconditional results as well. Apr3 answered Ordinary Generating Function for Mobius Apr3 awarded ● Citizen Patrol Apr2 answered Two different definitions of Erdos-Rényi random graph Apr2 comment product 1+1/p in terms of Chebyshev’s theta or psi functionAsymptotically, 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. Apr2 comment 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 $3$). Apr1 awarded ● Yearling Mar30 answered product 1+1/p in terms of Chebyshev’s theta or psi function