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Feb
3
revised The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
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Feb
3
comment The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
Yes, one certainly has to use the identity $\omega F_i=F_{i-1}-(-\omega)^i$, together with bad rational approximately of $\omega$. This implies for instance $\lceil(\omega(\pm F_i+k)\rceil=\pm F_{i-1}+\lceil\omega k\rceil$ for each nonzero integer $k$ with $|k|<F_i$. Nevertheless, trying to work things out along this idea seems to be a pain. So someone probably has a better idea.
Feb
3
comment The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
@LevBorisov: You are right, there are some details missing for which I don't have the time to check.
Feb
3
revised The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
added 209 characters in body
Feb
2
awarded  Nice Answer
Feb
2
revised The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
added 720 characters in body
Feb
2
answered The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
Jan
8
comment Is this a proof of the Hardy-Littlewood inequality?
The `proof' doesn't seem to use anything specific about primes. Seems to use only that $\pi$ is the counting function of a strongly monotonous sequence $p_1,p_2,\dots$ of integers. Unless I'm overlooking something (have spent less than 30 seconds on the paper), it is utter nonsense.
Jan
7
revised Is this system always solvable in radicals by quartics, octics, $12$-ics, etc?
added 369 characters in body
Jan
6
answered Is this system always solvable in radicals by quartics, octics, $12$-ics, etc?
Jan
3
comment Number of real roots of an exponential polynomial
@FedorPetrov You are right, my comment gives a weaker bound.
Jan
3
comment Number of real roots of an exponential polynomial
I believe that math.stackexchange.com/questions/688606/… is an answer to a more general question. Probably also mathoverflow.net/questions/44443/… is relevant here.
Jan
3
revised factorization of the regular representation of the symmetric group
added 135 characters in body
Jan
3
answered factorization of the regular representation of the symmetric group
Jan
2
comment Covering of a surface of a cube $n\times n \times n$ by pieces of paper $1\times 6$
@fedja: Oh, I had stopped the programs after a few days :-(
Dec
23
comment Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
Now fedja's solution and this argument eventually produced a proof from the book ...
Dec
14
comment How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$?
@zeraouliarafik: Are you really unable to check if my examples are counter examples or not? At math.stackexchange.com/questions/1306593/… I've added a second dis-proof which does not require to compute with large integers.
Dec
11
comment How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$?
I don't care about the two ``proofs'' if there is a simple counterexample. As you don't seem to be able to understand the simple reformulation (suggested by Silverman), here is the example in the original setting: Set $x=\frac{3}{4\sqrt{3}}$, $y=\frac{1}{\sqrt{3}}$, $z=\frac{9}{7\sqrt{3}}$. Then $xy+yz+zx=1$, and with $m=7$, $n=8$, $\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}-\frac{3}{2}(\frac{1‌​}{\sqrt{3}})^{n-m}=-\frac{30373554930137225}{11449504266834352656}\sqrt{3}<0$.
Dec
9
awarded  Enlightened
Dec
9
awarded  Nice Answer