Jul 13, 2016 17:13
it is late here and I shall sleep now. Good night. :)
Jul 13, 2016 17:07
You may see the answer and point out any inappropriate points. :)
Jul 13, 2016 17:05
Briefly, because $$a_{n+1}=d_1+\cdots+d_{n+4}+d_{n+5}=a_n+a_n=2a_n.$$
Jul 13, 2016 17:02
I am writing an answer; I shall add an explanation.
Jul 13, 2016 17:00
It should be $a_3=2a_2=4\cdot28.$
Jul 13, 2016 16:58
basically $a_{n+1}=2a_n.$
Jul 13, 2016 16:57
glad to know that. :D
Jul 13, 2016 16:56
how about $28=1+2+4+7+14?$
Jul 13, 2016 16:54
yes, I mis-typed.
Jul 13, 2016 16:54
that is, $d_{n+5}=d_1+\cdots+d_{n+4}.$
Jul 13, 2016 16:54
I mean $a_{n+1}=2(d_1+\cdots+d_{n+4})$
Jul 13, 2016 16:53
hi
Jul 13, 2016 16:53
How about $d_{n+5}=d_1+\cdots+d_{n+4}?$
Jul 13, 2016 16:53
In your construction how are you sure that $d_1$ divides $a_{n+1}?$
Jul 13, 2016 16:53
Alright, so it remains to find one such $a_1?$ Also isn't $a_{n+1}$ supposed to be $\sum_{r=1}^{n+5} d_r?$
 

 Mathematics

Associated with Math.SE; for both general discussion & math qu...
May 25, 2016 08:03
2
Q: $ 1^k+2^k+3^k+...+(p-1)^k $ always a multiple of $p$?

YoungI would appreciate if somebody could help me with the following problem: Q: For any prime number $p(p\geq 3), k=1,2,3,...,p-2$, why is $$ 1^k+2^k+3^k+...+(p-1)^k $$ always a multiple of $p$ ?

May 25, 2016 08:03
Does anyone want to add some more proofs to this question, just for fun? :P
Aug 21, 2013 15:37
Does anyone know why Makoto Kato has been suspended?
Aug 10, 2013 07:07
BTW, that problem turns out to have an answer here
Aug 10, 2013 06:47
Yes, and I have no idea how to approach it.
Sorry...
Aug 10, 2013 06:45
Same here. :)
Jul 19, 2013 14:17
:D
Jul 19, 2013 14:17
Who said what?
Jul 19, 2013 14:17
XD
Jul 17, 2013 14:37
:"D
Jul 17, 2013 14:36
Hello everyone!
Jul 15, 2013 01:40
Because their differrence is infinitely small?
Jul 15, 2013 00:56
 

 Chez Cosette

Discussion pour french.stackexchange.com. Bienvenue à tous ! Y...
Sep 12, 2014 09:16
Je suis d'accord~
Sep 12, 2014 05:12
Oh, et j'ajoute quelques mots de plus. Peut-être il y a de nouvelles erreurs. ;)
Sep 12, 2014 04:58
Et merci pour tout!
Sep 12, 2014 04:55
@Laure J'ai édité mon profil en conséquence.
 
May 22, 2014 14:24
the equivalence in George's answer is a generalisation of what I am trying to prove; so I cannot take that as granted.
May 22, 2014 08:58
Sorry for the impolite tone; I was too excited to find this statement.
May 22, 2014 08:58
http://math.stackexchange.com/questions/52856/is-noetherian-condition-always-needed-when-speaking-of-a-coherent-sheaf
Check the first answer!
"The problem is that coherence is very difficult to check in general and actually for some schemes, even affine ones, the structure sheaf OX is not coherent, and in that sad case the concept coherent is essentially worthless ."
May 22, 2014 08:50
For example, in page 17 of this [pdf file](http://stacks.math.columbia.edu/download/modules.pdf), they assume that $O_X$ is a coherent $O_X$ module; this would not be needed if $O_X$ is always a coherent $O_X$-module.
Of ocurse this is an indirect evidence.
May 22, 2014 08:48
OK
I am trying to find a counter-example now. :P
May 22, 2014 08:41
...for I...
May 22, 2014 08:41
I shall ponder more upon it, or I suspect something is wrong, but I don't know where.
May 22, 2014 08:38
I mean this is the definition.
May 22, 2014 08:37
This is exactly how a professor told me. :)
May 22, 2014 08:28
I think here the morphism u is a morphism of sheaves; and I am not familiar with the notion of projective sheaf of modules. Maybe you can explain more? Thanks.
May 22, 2014 08:26
I am currently trying to read EGA; and I was confronted with a proposition that, hypotheses being as in the question, an $\mathscr O_X$-module is coherent if and only if it is $\cong \tilde M$ where $M$ is a finitely generated $A$-module. I cannot uderstand the last step of that proof, but I can complete it if I can show the statement in question. Also, it is mentioned that $\mathscr O_X$ is not necessarily a coherent $\mathscr O_X$-module.
May 22, 2014 08:26
I cannot understand the proof: Say I have a morphism $u:\mathscr O_X^n|_U\rightarrow \mathscr O_X|_U;$ how did you propose to prove that $\operatorname{Ker}(u)$ is locally finitely generated?
May 22, 2014 08:26
I don't understand what you are trying to prove here: are you trying to show that $\mathscr O_X$ is always a coherent $\mathscr O_X$-module? I think this is not true in general.
 

 The Crusade of Answers

Our menace: the Unanswered queue. Our goal: total annihilation...
Aug 8, 2013 13:13
I see. Thanks for explaining.
Aug 8, 2013 10:12
Here is an unanswered question.
Someone could help me?
http://math.stackexchange.com/questions/451565/how-to-maximize-the-volume-of-a-rectangular-parallelepiped-in-an-ellipsoid
Aug 8, 2013 10:10
Hello!