Henry Cohn
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 Sep 22 awarded Enlightened Sep 22 awarded Nice Answer Sep 17 comment electron configuration on manifolds Nope, it's generally far from unique, even on spheres. The first case seems to be 16 points on $S^2$, for which there are (at least) two local optima. As the number of points grows, the number of local minima seems to increase exponentially, but no proof is known. In higher dimensions there are cases with arbitrarily large numbers of non-isometric global minima (the configurations in the last line of Table 1 in arxiv.org/abs/math/0607446). Aug 27 awarded Nice Answer Aug 24 awarded Good Answer Jul 3 awarded Good Answer Jun 28 awarded Nice Answer Jun 27 comment How to explain the concentration-of-measure phenomenon intuitively? That's a good way of putting it. Making one component small forces the others to be large in aggregate, but when there are many of them this is still compatible with each one being small individually. Jun 27 answered How to explain the concentration-of-measure phenomenon intuitively? May 15 awarded Enlightened May 15 awarded Nice Answer Apr 15 awarded Nice Answer Apr 14 comment Question on the irrationality of $e$ Incidentally, $\int_0^1 (1-x)^k e^x dx = k!(e - \sum_{0 \le i \le k} 1/i!)$, so if you use $(1-x)^k$ instead of $x^k$ you get the usual approximation to $e$ through Taylor series. Apr 4 comment Open problems in continued fractions theory What's the question here? The title and answers look like you are after a list of open problems or conjectures on continued fractions, but the body of the question focuses 100% on one conjecture. If you want a list, it would be clearer to ask for that in the body of the question and move the current content to an answer. Apr 1 comment Existence of a “quasi-uniform” probablility distribution on $\mathbb{Z}$ Oops, I was being silly (and had somehow convinced myself this wasn't enough to get full independence for more than two). Apr 1 comment Existence of a “quasi-uniform” probablility distribution on $\mathbb{Z}$ How do you get independence for different primes? They are certainly pairwise independent, but I don't see how to deduce mutual independence for more than two primes. Mar 18 awarded Yearling Mar 9 comment Factorization when a factor is partially known (I edited the answer to try to clarify this, since the original version did make it sound like the assumption of equal-sized factors might be essential.) Mar 9 revised Factorization when a factor is partially known added 148 characters in body Mar 9 comment Factorization when a factor is partially known Knowing the first 75 digits of $a$ is essentially the same as knowing them for $b$ (since you know all the digits of the product $ab$ and can do approximate division), so you can still apply Coppersmith's algorithm to $b$. It's a little less efficient if you don't know how big the numbers are: the setup I have in mind requires knowing the number of digits, but you can brute force this since there are only 125 possibilities.