Reputation
7,333
Next privilege 10,000 Rep.
Access moderator tools
Badges
1 24 69
Newest
 Nice Answer
Impact
~108k people reached

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