@mike yes

@PedroTamaroff I need a generating function?

I don't know generaing function+

s

I doubt it's ever true.

that the sum is a multiple of 2027?

yes

except for $k=0$ ofc

why?

....

i want to prove it :p

hence the silence

Put down 0?

Some context might be helpful, but 0 is an integer.

:)

i assume this is a contest problem

@AlexanderGruber The probability that a group is a non-abelian 2-group is 1. =O

it's on Brilliant.org

@PedroTamaroff the expected probability, afaik.

@PedroTamaroff conjectured.

@Mike Sorry, I don't know the difference. Never studied probablity.

I WIN

it's supposed to be a simple one(lvl 4), but I can't crack it.

@AlexanderGruber Oh, some asstwat didn't clarify that. =P

@Pedro 'expected' meaning conjectured. :P

Here.

@PedroTamaroff allow me

@AlexanderGruber It is clear that if $\Phi(G)=G'G^p$ for finite $p$-groups, then $\Phi(G)$ is the least normal subgroup of $G$ for which $G/N$ is elementary abelian, yes?

@AlexanderGruber it seems that a proof should come from asymptotic analysis now, no further group theory necessary

@Mike counting groups is not easy.

@AlexanderGruber I've seen some stuff, kiddo.

@AlexanderGruber Do you know the Janko groups? The twisted Dickson groups?

twisted Dickson. CHUCKLE.

@AlexanderGruber those are some pleasant-ass asymptotics

@user4140: Hint: 2 times 1013 < 2027.

Not sure how much of a hint is usually given for Brilliant problems, so I'll stop mentioning things now.

@GeorgeV.Williams That hint seems kinda worthless to me :/ It's the absolute first thing that occured to me, for instance.

I'm trying to describe exactly what the sum can be mod 2027 and it's not going well.

@AlexanderGruber Why are there so many groups of order 1024? =O

@PedroTamaroff there aren't that many

@PedroTamaroff good name for a band.

@PedroTamaroff you should see 2048.

@AlexanderGruber that's nothing.

@Mike ¬¬ Rephrase: why are there so many groups of order 1024 among those of order 2000?

@GeorgeV.Williams I allready knew that...

@PedroTamaroff because you're so smelly

@PedroTamaroff intuitively it makes sense that $2$-groups would be the fastest growing type of isomorphism class

@AlexanderGruber Why?

@PedroTamaroff think about all the ways to extend groups by quotients

$2$-groups have "small components"

@PedroTamaroff the more prime factors a group has, the more structure is forced on it; so p-groups will be most groups. the larger p is, the more structure is forced on it again.

or at least the smallest among the $p$-groups

p-groups have basically nothing forced on them.

this is sort of a weenie answer, of course.

@GeorgeV.Williams did you allready figure it out or are you trolling me?

@Mike if anybody knew a non-weenie answer it'd be publishable

Beg pardon?

but that is also how i think of it

It's your problem, not mine.

Did you solve it??

weenie is a great word.

i'm a fan.

@AlexanderGruber what's a generating function for 2n choose n?

no idea

@Mike Some trigonometric crazyness.

surely there's something better than that

You can actually consider $$f(x)=\sum_{n\geqslant 1}\frac{1}{4^n}\binom {2n}n x^n$$

That's nicer.

It think you get somethink like an arctangent.

Let me check.

It is not nice.

then that idea wont work

I know.

i've got the idea to solve this stupid thing though

then i can move on with my life

Sorry, the generating function of $\frac{1}{4^n}\binom{2n}n^{-1}$ is awful.

hahaha

$$\frac{1}{1-t}\sqrt{\frac{t}{1-t}}\arctan \sqrt{\frac{t}{1-t}}+{\frac{t}{1-t}}$$

That's the GF.

It's not that hard to find, though.

kill me

Since we have a nice relation between $x_n=\frac{1}{4^n}\binom{2n}n^{-1}$ and $x_{n+1}$:

no stop

Namely, $(2n+1)x_{n+1}=2(n+1)x_n$.

Then we get a diffie.

And that's it.

=)

so are you in combinatorics now?

To wit: $2t(1-t)f'(t)-(2t+1)f(t)+1=0$.

@Mike I always liked it.

If I could, I'd be all over the place.

so that's going to be yet another class that you already know the content of...

No, I don't know all the content.

Just some of it.

I think it has Ramsey theory.

over half of it

I don't know. I don't care that much!

:P

That's how you upset me. Heh.,

I knew I'd win.

Hey guys

@FernandoMartin YO.

I'm stuck on a silly exercise

ORLY.

Ya rly

Don't drop crazy stuff like Alex.

Nah, this is tame stuff

I'm stuck on a silly exercise: "Let $A$ be a coherent...."

@FernandoMartin OK.

I'm pretty sure it's obvious but I don't see it

OK

Suppose $f$ is a smooth real-valued function such that $f''+f=0$

It's easy to see that $f'(x) \cos x + f(x) \sin x = f(\pi/2)$

This somehow implies that $\frac{f(x)-f(\pi/2)\sin x}{\cos x}$ is constant on $(-\pi/2, \pi/2)$

(this is part of a linear algebra exercise)

@FernandoMartin Oh. There's a slickest solution to that man.

$f''f'+f'f=0$.

Thus $(f')^2+f^2=K$.

?

that is slick

@FernandoMartin $f''+f=0\implies f''f'+ff'=0\implies (f')^2+f^2=\rm const.$.

Yes, I understand that. I just don't see how that's connected to the problem.

@FernandoMartin Differentiate that man.

You should get $0$.

@Pedro: what are you proving?

@FernandoMartin You want to show that thing is constant.

Mhm

Well differentiate the sucker.

But that derivative is awful

¬¬

$\sec^2(x)(f(x) \sin x + f(\pi/2))$

$$\frac{d}{{dx}}\left[ {\frac{{f(x) - f(\pi /2)\sin x}}{{\cos x}}} \right] = \frac{{f'(x)\cos x - f(\pi /2){{\cos }^2}x + f\left( x \right)\sin x - f(\pi /2){{\sin }^2}x}}{{{{\cos }^2}x}}$$

$$ = \frac{{f'(x)\cos x - f(\pi /2) + f\left( x \right)\sin x}}{{{{\cos }^2}x}} = 0$$

Ahh, writing it that way makes it easier to see it.

I should stop using Wolfram.

Yes, Wolfram sucks camel d**k.