TheSimpliFire's Chatroom - 2019-07-10 (page 1 of 2)

Wow, that's actually a good question

$(n+1)^{n+2}=(n+1)^n(n^2+2n+1)\equiv 0, 1\pmod 4$

Therefore we need a value $n$ such that $n^{n+1}\equiv n^{n+1}, 1+n^{n+1}\pmod 4\equiv \pm 1 \pmod 6$

8:53 AM

@MrPie why not 3 mod 4?

@TheSimpliFire what do you mean? I think I'm also doing something wrong, idk

you say congruent to 0,1 mod 4

How did you exclude 3 mod 4?

If n is odd = 2k + 1 then we have $(2k+2)^{2k+3}$ which is clearly divisible by 4

If n is even = 2k then we have $(2k+1)^{2k+2}=(4k^2+4k+1)^{k+1}$

Got you @MrPie :) (my brain's a bit slow today)

@MrPie there are also composites of the form 6k+1 or 6k-1

this follows by Dirichlet's theorem on A.P

@TheSimpliFire I was just gonna do that xD

I was gonna let $n$ be even and $n$ be odd

guess my brain is slower (because I just woke up and it's... uh.. 7:01pm)

it's 0959 here

What day is it for you?

9:00 AM

Wed

9 hour time difference

yup

It's Wednesday for me too

Geez louise

(at least we're in the same year ;)

or are we

i am in Year 10

You are in... university? ;)

9:02 AM

year 13 technically

Well if I'm 15 then that makes you 18 turning 19 (because I'm turning 16) or 19

...unless you are way beyond your years (which I presume)

In the UK a new school year starts in June instead of August

@TheSimpliFire I know, I'm just messing around ;)

My memory isn't that bad

Wait

What are we talking about again?

heheh

I'm starting a bounty in 15 minutes

@MrPie Note that considering each term separately means you'll need to do more work at the end :P

@TheSimpliFire is there a badge for handing out bounties, or not?

Just wondering, because I don't do them for the badges

9:07 AM

Altruist, Investor, Benefactor, and one other that I've forgotten

They can only be earned once so you don't get more badges by repeatedly bountying something

Ah, Promoter.

Ok, I should be experienced enough to know this, but...

How do you check the badges you have earnt?

number-theory prime-numbers conjectures algebra-precalculus proof-verification proof-writing elementary-number-theory

math.stackexchange.com/users/477343/mr-pie?tab=badges

Mr Pie, Melbourne VIC, Australia

3.9k 3 14 46

Oh wait, nvm, ah bonkers

That was easy to find xD

Thanks anyways :P

@TheSimpliFire what question are you going to start the bounty on?

The prime number algorithm question

It surprisingly got 226 views despite not being on the HNQ...

@TheSimpliFire I can put it up if you want

9:17 AM

nah, you need your rep :P

Well, my last bounty will be awarded in a week's time (so there aren't so many bounties on the homepage)

1600/(3947+1600)=28.8%

that's how much of your total rep you've given to bounties :o

@TheSimpliFire wow!

Q: Can we remove any prime number from this strange process?

19

This is a little algorithm I made today, which may appear to be quite complex, so I will start with an example. Questions are at the end of the post. The process goes as follows: Start with the first prime number, $2$. From $2$, add the next prime number ($3$) to get $2+3=5$. There are ...

number-theory prime-numbers recreational-mathematics recursive-algorithms

started

I need to reach 30% XD

9:20 AM

For me it's 17.8%

I should have 3039 rep

You want to solve $$\frac {1600+x}{(3947-x)+(1600+x)}=0.3$$

so x = 64.1

I just did $2360(1+28/100)$

wait, on Puzzling or MSE?

Oh, I was doing Puzzling.

You were doing MSE, right?

9:25 AM

yes

(I was wondering where 3947 came from...)

I did give a +500 rep bounty once on Math.SE but I haven't given many bounties on it

Which means.... for Puzzling... uh oh

the heck?!

you have 2360 rep but you've given 4850 worth of bounties

HAHAHAA

I am stopping at 5000. Holy crap

(at least you earned 75)

I did not expect that

How sweet

I would have given 5000rep worth of bounties before my 100th post

9:34 AM

:)

Unfortunately, to keep it there, I probably would never put another bounty ever again xD

Nah, I'll just reach 10k.... but very slowly next time

what if you could transfer rep between SE sites :P

Can you do that?

You can't, right? Aw

nope

10:00 AM

Getting close to the 50000th message...

48353 messages (including this one)

I already had the honours of the 20000th (thanks again @TheSimpliFire :P) so who will have the honours this time round?

Omega perhaps?

ping @OmegaKrypton

Hello @Haran, I just bountied this

?

@OmegaKrypton the 50000th message is getting near in this room, just thought if you wanted to have it reserved for you :)

this is your chatroom, you should have it :) thanks for asking tho

btw, where is the message count?

10:07 AM

We have a different person every 10000th message

I think it was me -> MrPie -> Holo -> (who had the 40000th msg?)

@OmegaKrypton go to 'info' and look at the bottom right of the largest box

oh i see thanks

It was actually Nobody -> me -> Holo -> MrPie

Am going through the list of starred msgs and I just realised we haven't seen KevinL, Frpzzd or Abcd in a while...

Oh yes, KevinL, I remember him!

Idk about Frpzzd and Abcd

@MrPie Frpzzd was the first person you talked to in this room apart from me

chat.stackexchange.com/transcript/82585/2018/9/1

Ooohh, thanks @TheSimpliFire

I remember Kevin L used to like Riley riddles

10:14 AM

sometimes I look back at old messages in this room (hasn't even been a year old) and think that they are already ancient ;)

@TheSimpliFire hahaha

Oh ho ho ho

I remember another chatroom

That is probably dead by now

That anagram chatroom of mine XD

it's strange, because sometimes this room has more traffic than the main Mathematics site ... guess it mainly due to me concocting weird stuff

How can we see the most active chatroom?

@MrPie RIP, I should've added the 'dot' messages to that room to stop if being frozen

@MrPie chat.stackexchange.com/?tab=site&sort=active

chat.stackexchange.com/rooms/36/mathematics this one and the spinx's lair seem to have the most users on board

10:20 AM

You forgot Charcoal HQ ;)

typically there are around 30 users in that room

Where is that?

Got it: chat.stackexchange.com/rooms/11540/charcoal-hq

HOLY MOLY

A lot of them are mods/bots/SE employees

I see

I couldn't upload anything

And from the looks of it, they try to keep the Stack Exchange clean and healthy

So I won't bother them ;)

@MrPie how come?

@TheSimpliFire idk. I typed in "hello...?" and it wouldn't allow me to upload

The upload button was originally orange, and then as soon as I typed, it went grey and I clicked upload and it wouldn't.

oh wait what am I doing

[Brain Processing Error]

11:58 AM

@Grešnik hello

What is it now, you would like that I solve something for you?

how do I make the integral of (ln x) ²?

sorry for the inconvenience

=/

@YODA Integration by parts?

@TheSimpliFire Explain that to YODA.

Right

12:01 PM

@TheSimpliFire yes

u* v - integrate u * du

Write (ln x)^2 = (ln x)^2 * 1

We have u = (ln x)^2 so u' = (2 ln x)/x

And v' = 1 so v = x

Can you proceed? @YODA

I did not understand u '= (2 ln x) / x

By the chain rule

f'(g(x))*g'(x) ?

yes

Let g = ln x

and f = x^2

12:04 PM

@TheSimpliFire thank you!!!!!

@YODA If you have u=f^2 then u´=2*f*f´, by the chain rule, it´s easy

You saved my day!

No problem :)

Have fun with integration :)

@TheSimpliFire The trick f=f*1 is helpful sometimes, it can be that some do not know about that trick.

Indeed, especially for integrals involving ln x since its derivative cancels out $\int 1$

12:08 PM

@TheSimpliFire f'=2x ?

correct

2xlnx 1/x correct?

2xlnx/x ?

You shouldn't have the x on the numerator

No, just 2lnx/x

@YODA f'(x) = 2x, but remember to do f'(g(x)) = 2 ln x

and of course, g'(x) = 1/x

12:12 PM

@TheSimpliFire where was the x of the expression f?

g=lnx and f=x^2 so f(g(x))=(lnx)^2 so (f(g(x))´=f´(g(x))*g´(x)=2lnx*(1/x) @YODA

We replace x by ln x

You plug g(x) instead of that x

@Grešnik ahhhh

thank you!

I'm happy to learn!

No problem, @TheSimpliFire did most of the job, as I knew he surely knows how to do those integrals. I also know, but am not in a mood for much typing now since I am drinking the "morning" coffee.

12:17 PM

haha

Where do you live?

I am Brazilian

=)

I´m in Europe, under suitable definition of Europe.

I can write you which country, but that won´t help you much.

kazakhstan? just guessing

turkey?

or russia

I like how none of those countries are in 'proper' Europe xD

Hint: the 'š' in Grešnik should give you a clue

@TheSimpliFire Traitor! :D

12:23 PM

Hehe, I know that as I used to learn Czech

I had a lot of trouble correctly pronouncing ř

You would have also much trouble with Ž, i think.

That's easy, it's just like zh-

ř = r + Ž

(roughly)

But you almost have no words that contain "zh", right?

There are in the English language

pleasure = plezhur

and a lot of French words have that sound too

Could be, the Ž is spoken as G in Girondins, most probably.

12:32 PM

Then there's words like čtvrt ....

@Grešnik How similar are Czech and Croatian?

Haha, what a traitor! :D I do not know exactly, they have similarities, at least in pronounciations and structure, but I am not an expert.

@Grešnik Hey, you used to write that in your profile :P

Yes, but do you see it now? And why not?

@Grešnik I guess. Though it is indeed possible to Google Translate your username...

anyway ... :)

brb

@TheSimpliFire
I was seeing the exercise response but it's different.
In the integration by parts u * v - integrate u * du. It appears as x (ln x) ² - 2 integrate lnx dx. But the du = 2 lnx * 1 / x where did the rest go?

method of substitution?

to my view would be x (ln x) ² - integral x lnx / x dx

12:40 PM

And when you cancel x´s in xlnx/x dx?

ahhh

=)

thank you

ahahha

@TheSimpliFire I don't have a clear understanding of your question. Can you write about 15-20 of the first few steps of your process?

I will try your problem and @Grešnik 's problem today

hello @Haran

@TheSimpliFire For n = 1 , we get 9, which is not prime. For n = 392, we get another prime.

@Peter What problem is this?

12:47 PM

@Haran Is that the question on deleting primes from the lists?

Yes

The function $$f(n)=n^{n+1}+(n+1)^{n+2}$$

What exactly you do not understand? @Haran

The process

@Haran Let me explain

We start with 2

12:48 PM

I have a vague understanding, but it will get better if you write about 10-15 more steps

2

2, 3

2, 3, 5

3

3, 5

3, 5, 7

7

7, 11

7, 11, 13

7, 11, 13, 17

7, 11, 13, 17, 19

7, 11, 13, 17, 19, 23

7, 11, 13, 17, 19, 23, 29

11, 13, 19, 23, 29

11, 13, 19, 23, 29, 31

Dude

Looks like your problem is simple

You did that "by heart", without paper and pencil and computations, I think.

You cannot reach $p_{n+1}$ without $p_n$ coming in your sequence

Due to the way you have defined it

@Haran But will every prime be cancelled?

12:52 PM

Oh that is what you are asking alright

@TheSimpliFire As soon as the sum is divisible by the first prime, this prime is removed. I am pretty sure this must be the case eventually.

What makes the problem complicated is that I do not see a way to write down a closed form for the formula

Compute the lengths l(n) of the lists at the n-th step, just to have a feeling about what happens in some reasonable range, if that could help.

What is the command in Python to generate the length of an array?

@Grešnik A slight complication is that we have to handle possible loops. But the rule then simply is to add the next higher prime.

12:56 PM

@TheSimpliFire I do not know that.

Guess it's just len()

would make sense

@Haran The link to the code in Python is starred (third one down)

I guess that the lists are "almost-always" strictly increasing.

@TheSimpliFire A loop could also involve more than two steps ? At which point do we then add the next higher prime ?

I guess when the first repetition of a vector occurs

1:02 PM

Yes

e.g. if we have (83, 89, 97) as a loop then just add 101

Alternatively, we could still add 97 but remove 89

Do we till now at least have the result that $lim_{n \to + \infty} l(n)=+ \infty$?

But in this example, 101 would have caused the loop, so we have to add 103, right ?

@Grešnik yes

@Peter Ah yes

Anyway, I think we can easily show that a loop can only occur at once because if we add two more primes we cannot fall back to an earlier array because the product gets to large hence other entries have to be removed as well.

Did someone find a second "cyclic prime" ?

They were 269 and 94793

1:09 PM

Wow, and what was the smallest prime in this array ?

Before overflow, all primes up to 16903 were eliminated

if that's what you mean

yes, that is what I mean

Another question would be to determine the maximum number of primes that the nth step can remove

We can ask a multitude of questions about that problem, but I think that we need to "focus" on one by one, if we were to settle something.

i.e. If $\sum\limits_{i\in I} p_i=k$ then a good starting point would be to solve $\min\limits_{j}\prod\limits_{i\in I}^jp_i\le k$

@Grešnik yes

@Peter Did you manage to program something in PARI that allows for loops to occur?

hello @SimplyBeautifulArt

Any luck with the partial Beta derivatives?

1:17 PM

How large was your reputation was when you opened this chat-room? @TheSimpliFire

I can check, why?

So something's definitely wrong with this. I tried writing out $S_{5,3}$ in terms of $J(b,c)$ and I got everything cancelling into $S_{5,3}=0$.

Because I do not know can all users open chat-rooms, no matter of how large or small their reputation is?

@Grešnik math.stackexchange.com/help/privileges/chat-rooms

100 rep

@SimplyBeautifulArt Yes, $\int_0^1\frac{\left(\ln x\right)^4\left(\ln\left(1-zx\right)\right)^3}{x}dx>0$ for $z<0$ and vice versa

From my experience, getting a zero integral is usually the consequence of missing a minus sign or something

@SimplyBeautifulArt Are you sure you didn't miss an $x$ term in $I(a,b,c)=(-1)^aI(a,c,b)$

I'm mostly sure?

that part isn't even really relevant to the issue

Key part is relating $I$ to $S$ and also $J$

1:26 PM

The substitution $1/x$ brings it to $$I(a,b,c)=\int_0^\infty x(-1)^a\ln^a x\ln^b\left(1+\frac1x\right)\ln^c(1+x)\,dx$$

Since $u=1/x\implies du=-\frac1{x^2}dx$ and you need an extra $x$ to balance $dx/x$

you missed the $-1/x^2$ part and the bounds swap

uh what

What

$\frac{dx}x=-x\cdot-\frac1{x^2}\,dx$

forget what I said, it works

cdn.discordapp.com/emojis/366999788803325952.png?v=1

indeed :P

@SimplyBeautifulArt "3669997888" that's a weird sequence

shrugs

1:37 PM

But if $\lim_{n \to + \infty}l(n)=+ \infty$ isn´t it necessary that there exists some natural $M$ such that cardinality of the set $\bigcap_{n=M}^{+ \infty}l(n)$ equals AT LEAST 1?

Yes

We can't have an empty set

But if that is a yes then there is AT LEAST one prime which is not cancelled?

Not necessarily

As more primes are added to the sequence, there is a point at which all primes less than k are eliminated

Perhaps this is more of an issue with $\infty$

Can primes be cancelled and then again return to the list later?

No

1:41 PM

hello my friends

hello again @YODA

hello YODA, you surely have some question for us

hehe, that's what they come here for :P

integrate 3x/3x-2

How can I solve it?

hint: 3x = (3x-2) + 2

1:42 PM

@TheSimpliFire =)

3x/(3x-2)=((3x-2)+2)/(3x-2)=1+2/(3x-2)

if you're still talking about that prime sequence thing, can't you cancel some primes and the last prime, and then add back that last prime without getting into a loop?

e.g. 2+3+5 cancels the 5, but the 5 comes back

Then that's a loop

Unless you're doing 2+5

Though it accelerates the prime cancelling process

At the start you have 2

then you have 2+3, and nothing cancels

then you have 2+3+5, and the 2 and 5 both cancel

but the 5 still comes back on the next step with 3+5

@Grešnik ((3x-2) +2) / (3x-2) ? How did you integrate it?

1:47 PM

@SimplyBeautifulArt but that's what I did

yeah, and then you said that primes that get cancelled don't return to the list

from the post we have [3], [3 + 5], [3 + 5 + 7], [7], [7, 11], ...

What do you must differentiate to get 1/(3x+2)? @YODA

@SimplyBeautifulArt I thought Gresnik meant if a product contains a prime which was removed then it will be added, which is ofc a no

otherwise you add the next prime regardless of its previous state, unless it is a loop

well I'm just throwing out a comment without reading the full context :P

1:49 PM

@Grešnik ln (3x+2) ?

Perhaps I should draw out a flow diagram of the process

@YODA missed out a factor of 1/3

I just meant "if some prime gets cancelled at some step can it return to the list". What is not precise about this question?

@Grešnik It was a misunderstanding from my part

@TheSimpliFire I did not program this yet. Do you want a doublecheck ?

OK, I checked Euxhen's code yesterday and it worked fine

though yours was much more efficient in terms of length

1:53 PM

But then we have relative and absolute cancelling, relative is when prime gets cancelled and is able to return and absolute is when prime gets cancelled and is not able to return?

@TheSimpliFire I did not understand it right. =( expression = integrate ln(3x-2) dx .... u= ln(3x-2) ;dv= dx; v=x; du = 3/3x-2; ok ? u *v - integrate (v * du) = xln(3x-2)- integrate (3x/3x-2)

right?

@Grešnik An absolute prime is one that is smaller than any of the primes that will be added to the sum

Aaaah you cannot solve everything with just one method. @YODA

A relative prime is then one of the primes that will be added to the sum (maybe call this a future prime)

@YODA $$\frac d{dx}\left[\frac13\ln(3x-2)\right]=\frac13\cdot\frac{d\ln(3x-2)}{d(3x-2)}\cdot\frac{d(3x-2)}{dx}=\frac13\cdot\frac1{3x-2}\cdot3=\frac1{3x-2}$$

@TheSimpliFire translate please =/