7:07 AM
15

If $z_n < 2y_n$ then $y_{n+1} = 4y_n - 2z_n$ $z_{n+1} = 2z_n + 3$ Else $y_{n+1} = 4y_n$ $z_{n+1} = 2 z_n - 1$ Consider the following limit: $$\lim_{n\rightarrow\infty} \frac{1}{n}\left(z_{n+1} - \sum_{k=1}^n z_k\right) \tag{\star}$$ The limit may or may not exist, and it may or may n...

4 hours later…
10:55 AM
@TheSimpliFire For you prime sums problem, at each given point, you have a state:

$$p+\sum_{q\in\mathbb A} q$$

Where $p$ is the smallest prime not removed and $A$ is the set of all other primes not removed so far. We can write that as $p+S_q$ which is $p+S_q = m \pmod p,m\ne 0$ since $p$ is by definition not removed yet.

Notice $m$ has $p-1$ possible states. By adding the next $q$ in line, then $m$ changes to a different state $\in\{0,1,\dots,p-1\}$ of $p$ possible states.

Your question "Do we remove every prime" is equivalent to "Does $m$ eventually reach $0$ in this sequence of state
That is define the sequence similarly but remove the prime $p$ if the sum is $=M \pmod p$, where $M=0$ always is your sequence.
If the proof to your question is not specific to $M=0$, but seems to hold for a more general case, it should be easier to work out?
For example, we can try running the python code but removing condition is $M=p-1$ instead of $M=0$.

2 hours later…
1:14 PM
@Feeds You are the only non-bot user in the SE network named Feeds :P
I can now perform a super-ping :)
45

I've seen the word 'superping' be used across SE meta sites. What does it mean? Is it just a synonym of the word 'ping' (ie. using @ followed by a user's name to give them a notification in their inbox).

1:37 PM
@TheSimpliFire ahem excuse me??
Look at my profile!
;)
I've never heard of a super-ping

@Feeds I just got told this by another mod on OR :P

Well, I was able to prove something useful that you might have already proven regarding the $p^p+2$ conjecture
I showed my teacher who said it was correct
So I can legitimately say, as messy as it might be, it's 100% true :D
...or else my teacher will probably never teach maths again XD

Let's see...

@TheSimpliFire i'm sick of spamming the chat, actually, so I'll write it with Mathjax and send you a photo

Whatever suits you best

1:44 PM
Do you have trouble accessing photos here?

Nope
We're surprisingly close to 60K msgs now :)

Welp, in the meantime, I can give you a question that I couldn't solve today
I understand it, but don't know the process to solving it (unless I had a thousand years spare - you'll know what I mean after you see the question)
@TheSimpliFire I think you'll solve it pretty quickly - you ready? :P

@Feeds That's me when I look at a geometry question
any time

@TheSimpliFire it's not geometry, lucky for you ;)
Ok here goes. Pretty short, actually:
A sequence $a_1,a_2,a_3,\ldots$ is defined by $$a_{n+2}=\frac{1+a_{n+1}}{a_n}\quad\text{for}\quad n\geq 1.$$
Given that $a_1=2$ and $a_2=5$, what is the value of $a_{2002}$?
Good luck, and I won't see what you write until later, once I finish writing this $p^p+2$ thing :P

The recurrence is cyclic, so $a_{n+k}=a_{n+k-5}$ for $n+k>5$. As $2002\equiv2\pmod5$, the answer is $a_{2002}=5$.
@Feeds

1:58 PM
@TheSimpliFire how did you get that?
I mean, I kind of reiterated the sequence a bit and I thought it was cyclic too, but couldn't prove it
(I shoulda went with my gut - dangit!!)

@Feeds Because $a_3=3$, $a_4=4/5$, $a_5=3/5$, $a_6=2=a_1$ and $a_7=5=a_2$

@Feeds You want to say you solved the $p^p+2$-conjecture ?

@Peter no, I just made some progress, but it looks familiar to @TheSimpliFire's previous work I've seen
And glad you are here for me to tell you the bad news... which I think I already told you... but I don't fully remember... so... I'm gonna tell you again... :/
Basically my computer broke. The whole screen went kaput when I had it in my schoolbag and dropped it because it was heavy

@Feeds Wait... so are you typing MathJax on your phone??

So, uuhh... I'm using my dad's laptop now, and I'm not allowed to download anything, he said.

2:01 PM
Ah ok

I'm a bit slow at typing, today :)

That's alright

Bloody chatjax is harmless, why dad, why... mumble mumble...

@Peter point is, the progress of the trial division thing....?
:/ I think you know what happened

2:04 PM
ChatJax is just an online script

@Peter (however the technicians at school say it might not have stopped, and the screen is probably just stuffed up.... omg let's hope that's the case)
@TheSimpliFire wait, so that means.... ;)

Chatjax: oh no, not him again. RUN!!!
MrPie: HEHEHEH >:D

Just remember to remove the bookmark after you use it though

Hahaha, yes, thanks for that tip :P
As for @Peter, however, sorry about what happened to my computer. I was being careless, really. I hope I can do better for any future projects you might need help with.
For now, though, I can't do much :/
$\rm At \; least \; I \; have \; Mathjax \; rendered$

2:10 PM
hooray

ok, dokey, hang on a sec

... and I have now cast exactly 1000 votes on OR.SE :D

@TheSimpliFire hooray! :D
I can't work out how to screenshot on this computer, so... here goes a long message
$p^p+2$ must be prime for $p>3$.

$\therefore p^p\equiv -1\pmod 6$ and $p\equiv -1\pmod 6$

$\implies p^p-p\equiv 0\pmod 6$.

Note: $p^p-p=p\underbrace{(p-1)}_{\rm even}\underbrace{(p^{p-2}+p^{p-3}+\cdots + 1)}_{\rm even}.$

$\therefore p^p-p\equiv 0\pmod 4$. But since $p^p-p\equiv 0\pmod 6$ as well, we have $p^p-p\equiv 0\pmod {12}$.

$p^p-p=p(p^{p-1}-1)\implies p^{p-1}\equiv 1\pmod {12}$.

Recall: $p\equiv -1\pmod 6$

$\therefore p^{p-1}-2p\equiv 3\pmod {12}\implies p^{p-2}\equiv 5\pmod {12}$
@TheSimpliFire The first time I wrote this proof was on my shower wall xD

Aug 10 at 9:44, by TheSimpliFire
As $p^{p-1}\equiv1\pmod{12}$ we have $p^p=12a+5=(12b+1)(6c-1)$

2:26 PM
@Feeds I doublechecked the range upto $p\le 30\ 000$ , the only prime is still $29$

Thanks @Peter
@TheSimpliFire right
$p^p=6(12bc-2b+c)-1$
So $c$ is odd.

@Feeds It's not always true that $p^{p-2}\equiv5\pmod{12}$
Actually, $p^{p-2}$ alternates between $5$ and $11$

Ok, that's it

But $p^{p-3}\equiv1\pmod{12}$ is always true

Mr Jusen, I'm personally firing you

2:30 PM
Heh

@TheSimpliFire (for the fiftieth time now...)
Where did I stuff up...

@Feeds They should've realised:
Consider $p=11\equiv-1\pmod{12}$. Clearly, $$p^{p-2}\equiv11^9\equiv(-1)^9\not\equiv5\pmod{12}$$

Ok so something went wrong with the algebra, then?

" $\therefore p^{p-1}-2p\equiv 3\pmod {12}\implies p^{p-2}\equiv 5\pmod {12}$ "
What did you do?

$2p\equiv -2\pmod {12}$
And $p^{p-1}\equiv 1\pmod {12}$

2:33 PM
yes...
How did you divide by $p$?

So $p^{p-1}-2p\equiv 1-(-2)\equiv 3\pmod {12}$... or so i thought...?

Even if you divide, you'd get $p^{p-2}-2\equiv3p^{-1}\pmod{12}$ which is, of course, ...

@TheSimpliFire but $p$ is prime!
$p^{p-1}-2p=p(p^{p-2}-2)\equiv 3\pmod {12}$

Right
then?

Clearly since $p$ is prime and $p>3$, then the only option left is that $p^{p-2}-2\equiv 3\pmod {12}$?
Beacuse otherwise $3\mid p$

2:36 PM
No!

...

What about $p\equiv-1\pmod{12}$ and $p^{p-2}-2\equiv-3\pmod{12}$?

...Do I need some cancellation law revision?

You considered 3 = 1 * 3 but not 3 = (-1) * (-3)

OH
HOT DAYM

2:37 PM
:P

@TheSimpliFire sorry if I sounded like I was doubting you, I was just confused is all

I knew that, that's why I was asking you questions along the way

The Socratic method, also known as method of Elenchus, elenctic method, or Socratic debate, is a form of cooperative argumentative dialogue between individuals, based on asking and answering questions to stimulate critical thinking and to draw out ideas and underlying presuppositions. It is a dialectical method, involving a discussion in which the defense of one point of view is questioned; one participant may lead another to contradict themselves in some way, thus weakening the defender's point. This method is named after the Classical Greek philosopher Socrates and is introduced by him in Plato...

Anyway, your $p^{p-3}$ case is fine

$p$hew

2:40 PM
But not to worry

@TheSimpliFire thanks for the chat. I'll still work on this, but tomorrow
I need to get off now, it's 12:40am
Cya

Ofc
Bye

:P

@Feeds Remove the Chatjax!
(guess I was too late :P)

I still search another prime of the form $$n^{n^k}+(n+1)^{(n+1)^k}$$ with positive integers $n$ and $k$. Still, the largest prime I found of this form is $2^{16}+1$ , the $4$ th Fermat number.

3 hours later…
6:03 PM
@TheSimpliFire Seems that upto $p\le 35\ 000$ there still is no further prime $p^p+2$ with prime $p$

6:17 PM
@Peter Which means $p^p+2$ must have over $159\,000$ digits

yes, assuming pfgw worked correctly.
In fact, it is very unlikely that a prime of this form different from $29$ exists. According to the heuristic, at most about $5$%
I would like to extend the search limit from primes of the form $$n^n+(n+1)^{n+1}$$ but my computer is too slow to get significant further.

6:38 PM
@Peter Perhaps you could try booking a supercomputer :)

@TheSimpliFire Any new project ?

Nothing new, but I bountied my prime number algorithm question again

7:21 PM
@TheSimpliFire regarding your prime algorithm, if it true that adding and/or removing some primes from a sum of some primes has a positive chance to result sum being any $m=0,\dots,p-1$ modulo $p$, then we are done? Since then every $p$ gets removed eventually, assuming the sum does not repeat values (which is guaranteed due to not adding cyclic-primes). Even more, algorithm should work more generally regardless of your "removing case" which is currently $m=0$.
(Sum repeating values, I actually meant a cyclic sequence of remainders repeating)

@Vepir In other words, we want to show that as the number of iterations tends to infinity, the smallest prime in the sum can be arbitrarily large. I don't know if it is the case for m > 0.

I think if it works for $m=0$ it should be weird to not work for other arbitrary remainder.

It would be an interesting experiment to test for m > 0, EuxhenH (the scriptwrite of the Python code) actually sent the link in this room

We actually need to understand how adding primes to a sum of primes affects factorization. This reminds me of a similar unsolved problem.

Jul 8 at 13:17, by EuxhenH
https://codeshare.io/5gERew

7:27 PM
Or removing them.
If it does not work for arbitrary remainders, then adding primes to a sum of primes will have a pattern in resulting factorizations, since probability of reaching some $m$ modulo $p$ will be zero?

another downvote :/

On the bountied question? I'm not sure for a reason other than a random troll passing by.

Then again the $p^p+2$ conjecture has 2 downvotes

There will always be downvotes on posts that reach enough people.

fair point
gtg, see you tomorrow

7:33 PM
I wonder what's the expected amount of downvotes given exposure in views and interactions with the post.
Okay
I'm not sure how to exactly modify the code to hunt for different $m$.