Ian Mateus
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 Mar 26 reviewed Close Determining parametric equation given 3 points Feb 10 answered Problem 10 of Section 1.2 from Hatcher. Dec 21 answered A little question about convergence of sequence Dec 8 comment How did mathematicians decide on the axioms of linear algebra "(...) the early history of vectors was complicated by people wanting to have a way to multiply vectors in analogy with multiplying real numbers or complex numbers (...)" This is very reasonable, but is there a written example? Nov 2 awarded Nice Question Oct 21 revised Find the sum of all $x$ such that $y = 2 - \frac{9}{x+1}$ has an integer solution $(x, y)$ deleted 13 characters in body; edited title Oct 16 awarded Yearling Oct 7 revised Finding $y$ so that $y^2 + x^2 \equiv 1 (\mod x + y)$ added 5 characters in body Oct 7 answered Finding $y$ so that $y^2 + x^2 \equiv 1 (\mod x + y)$ Oct 1 answered Is $f$ uniformly continuous? Sep 30 comment On the number of integers whose prime factors are congruent to 1 modulo some given number. The last formula should be $(2n^2H_{2n}N^{(2n+1)/2n})/((2n+1) \log^2N)$, sorry. We should expect, then, $$C_m(N)\sim\frac{N}{\log^2 N}\sum_{n\geq 1} \frac{2n^2}{2n+1}H_n \frac{N^{1/2n}}{\varphi^n(m)}.$$ Sep 30 comment On the number of integers whose prime factors are congruent to 1 modulo some given number. In general, the $n$th sum is asymptotic to $$\frac12\sum_{k=1}^n\pi(N^{k/2n})\pi(N^{(2n+1-k)/2n}),$$ which is like $$\frac12 \sum_{k=1}^n \frac{2nN^{k/2n}}{k\log(N)} \frac{N^{1-(k/2n)+(1/2n)}}{\left(1-(k/2n) +(1/2n)\right)\log(N)} =\frac{4n^2H_{2n}N^{(2n+1)/2n}}{(2n+1) \log^2N}$$ by the prime number theorem. Sep 30 comment On the number of integers whose prime factors are congruent to 1 modulo some given number. I think this possible in principle. Write $[1,N] = [1, N^{1/2}]\cup[N^{1/4}, N^{1/2}]\cup[N^{1/2}, N^{3/4}][N^{3/4}, N]$. Then the second sum is approximately $$\frac{1}{2}\left[\pi(N^{1/2})^2 + (\pi(N^{1/2}) - \pi(N^{1/4}))(\pi(N^{3/4}) - \pi(N^{1/2}))+ \pi(N^{1/4})(\pi(N) - \pi(N^{3/4}))\right].$$ Keeping only the relevant powers and using the prime number theorem, this should be asymptotic to $$\frac{10}{3}\frac{N^{5/4}}{\log^2(N)}.$$ Sep 30 comment On the number of integers whose prime factors are congruent to 1 modulo some given number. Heuristically, I would expect $C_m(N)$ to be like \frac{1}{\varphi(m)}\sum_{p