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 7h awarded Nice Answer Apr 12 awarded Popular Question Mar 2 comment An elementary inequality for a recursive double sequence BTW, as your Lemma 4 shows, we have $\sigma_n(m)\approx nm^{1-1/n}$, and I think the key to understanding $\sigma_n$ is to somehow make this approximation precise. Mar 2 comment An elementary inequality for a recursive double sequence Thanks for sharing, I'll try to see whether this (particularly Lemma 4) can be useful. Concerning monotonicity - I also believe that $\sigma_n(m)$ is strongly increasing in $n$ for any fixed $m\ge 1$, but cannot prove this. Feb 29 comment Asymptotic of a sequence related to $LCM(1,\cdots,n)$ Is it easy when $u_k$ is the zero sequence? Feb 29 revised An elementary inequality for a recursive double sequence added 235 characters in body; edited tags Feb 28 asked An elementary inequality for a recursive double sequence Feb 16 comment A large deviation / binomial coefficients bound @usul: Thank you! Feb 15 comment For a sufficiently large $a$, are there distinct (mod $a$) integers such that all powers up to the $n$-th are “close” modulo $a$? @Turbo: not sure I've got your last question right, but in case I have: you would know that at least two values of $b$ exist if you can show that the absolute value of the remainder term is does not exceed the main term less $2$. Feb 15 comment For a sufficiently large $a$, are there distinct (mod $a$) integers such that all powers up to the $n$-th are “close” modulo $a$? @Turbo: one cannot say whether it is hard without actually trying, for which unfortunately I don't have time now. Maybe, it is easy. Anyway, this is not just a "trick"; dating back to Gauss, exponential / character sums form is a rich area with many famous mathematicians involved and lots of top-rate research done on it. And, I don't understand your question starting with "also conversely". Feb 15 comment For a sufficiently large $a$, are there distinct (mod $a$) integers such that all powers up to the $n$-th are “close” modulo $a$? @Turbo: calculating the main term is a triviality, just plug in $u_1=\dotsb =u_n=0$. The non-trivial thing is to estimate the remainder term, which is the rest of the sum. If it turns out to be smaller in absolute value than the main term, then yes, it is definitely impossible to find $b$ as you want. Feb 15 revised For a sufficiently large $a$, are there distinct (mod $a$) integers such that all powers up to the $n$-th are “close” modulo $a$? typo fixed Feb 15 answered For a sufficiently large $a$, are there distinct (mod $a$) integers such that all powers up to the $n$-th are “close” modulo $a$? Feb 14 comment For a sufficiently large $a$, are there distinct (mod $a$) integers such that all powers up to the $n$-th are “close” modulo $a$? Ok, still, how about $n=2$? Feb 14 comment For a sufficiently large $a$, are there distinct (mod $a$) integers such that all powers up to the $n$-th are “close” modulo $a$? Could you prove this for $n=2$? Is it true that if $m$ is large enough, then there are at least two (distinct) elements $b\in{\mathbb Z}_m$ such that both $b$ and $b^2$ belong to $(\sqrt m,\sqrt m\log m)\pmod m$? Have you tried using exponential sums to prove this? Feb 7 comment Does $|A+A|$ concentrate near its mean? You have $|A+A|=I_0+I_1+\dotsb+I_N$, where $I_z$ is the indicator random variable of the event $z\in A+A$. It may be a little technical, but should not be difficult in principle to show that the $I_z$ are "almost independent", the exact meaning of which is that the pair correlations of $I_u$ and $I_v$ are small for $u\ne v$. As a result, $|A+A|$ should have a distribution close to the binomial. Feb 7 revised Does $|A+A|$ concentrate near its mean? edited tags Feb 4 comment If $N = qn^2$ is an odd perfect number, is it possible to have $q + 1 = \sigma(n)$? What is an "Euler prime"? What is a "quasi-Euler prime"? What is $I(n)$? Jan 23 comment Question related to Fermat curve: Does the equation $A x^n + By^n = C z^n$ have any solution in $\mathbb{N}$? ... or $z=1$ and $C=Ax^n+By^n$ ... Jan 19 awarded Nice Question