Done.

The brent code doesn't have a bug, it's impossible to get that wrong

and from what I can see neither does my wire code

it's all integers, so no floating point shenanigans

so I think it's really what's happening

that's really bizarre

@El'endiaStarman did you generate that gif?

So many easy questions today...

@orlp Yes! :D

@El'endiaStarman with what?

Don't ask me to generate the gif for n=36...

I wrote the code myself, actually.

;(

Caching.

no, it dropped back down

Aye, but we posted a good number of questions today.

It'll go back up. :P

15 today.

And stop compulsively refreshing that page.

hahaha how did you know

I just posted my third question

:P

I think my electrons in a wire question might be ready to post.

but but n=36

Really throws a wrench into the works, eh? :P

Oh hey @orlp, think you could modify your code to use threading?

That way, it won't be eternally stuck on the really high ones.

could you also try skipping 46 to see if any more bizarreness pops up

or does everything past 46 get too slow

It's not exactly "everything", but more like it's unpredictable.

I wonder if there's a connection to primes...

@El'endiaStarman I don't believe so

this is cycle-finding

which means that it's a single thread of data dependency

46 16777214
47 32
48 4194302
49 4092
50 510
51 504

using pypy

I mean with respect to the n values.

Oooh!

wow

16777214.

At least that one sticks to the pattern, eh? :P

23/1

And the one for 48 is 21/1.

52 taking looong

I'll just keep it running

             1         2         3         4         5
    123456789 123456789 123456789 123456789 123456789 1
1s:  1 213 3251634 439265C1A693E45 54C3?9C2A675CBN1LA86
0s:  1 121 1213121 121312141213121 1213?214121312151213

why are there question marks?

That's n=36.

oh right

Really, really threw a wrench into the works. :P

I mean, feel free to triple check my code

I mean, it breaks even the abacaba pattern.

and run it yourself for n = 36

but it's really what I get out of it

@El'endiaStarman well, not really

it still ends with 1 zero

it's just that... the rest doesn't make sense

Hmm, true.

it doesn't follow 1*0*, is the problem

52 134217726
53 2044
54 2097150
55 112
56 1022
57 65532

@El'endiaStarman want to do n=52 by hand?

26/1, wow

you could be here for a while

@orlp HA HA NO.

I'm pretty happy I'm using a proper cycle finding algorithm, though

haha, definitely

was initially thinking to just cheat a bit using sets

but with a cycle length of 134217726 I'm glad I didn't do that

             1         2         3         4         5         6
    123456789 123456789 123456789 123456789 123456789 123456789 12
1s:  1 213 3251634 439265C1A693E45 54C3?9C2A675CBN1LA86Q9K49E
0s:  1 121 1213121 121312141213121 1213121412131215121312141213121

(the current algorithm is actually pretty much constant memory, apart from the fact that the integers storing the cycle lengths can grow unbounded)

I wonder... if the abacaba is symmetrical, will there be another really weird outlier at n=59?

@Doorknob Don't put a ? for n=36 for the zeroes

Definitely a sweet algorithm.

@orlp well, it doesn't even fit the pattern of "ones then zeroes"

but sure

it doesn't?

well

seriously weird

I'd just say the 0s should be 'trailing zeroes'

okay

then the pattern doesn't get destroyed, at least for 0

Doesn't help very much with predicting the next ones though.

which is what we're really after :P

@somebody Try checking the prime factorization of the periods.

[1, 2, 1, 6, 4, 14, 1, 14, 12, 62, 8, 126, 28, 30, 1, 30, 28, 1022, 24, 126, 124, 4094, 16, 2046, 252, 1022, 56, 32766, 60, 62, 1, 62, 60, 8190, 56, 174762, 2044, 8190, 48, 2046, 252, 254, 248, 8190, 8188, 16777214, 32, 4194302, 4092, 510, 504, 134217726, 2044, 2097150, 112, 1022, 65532]

starting from n = 1

no new numbers, just nice reference

Hahahaha, I plopped that list into Wolfram|Alpha with "factor" in front, and it couldn't understand my query, so it showed me some information about one number: 174762, the oddball.

wait

found something interesting

there's always 2^n-1 as a factor (where n is random number)

(does that mean i write 15 instead of 3*5?

@El'endiaStarman Are you sure that for n=36 the outcome is 174762 and not 524286?

@Adnan I didn't come up with that, orlp did.

Oh

My results agree with his where I can/did test them.

For the others, every even n is A160657

How interesting.

Both are GoL-style oscillator periods.

@RikerW mwahahaha

@CᴏɴᴏʀO'Bʀɪᴇɴ brainwashing -> success

I literally just spent the last 5 hours doing math homework

Fun but tiring? :D

Yes yes

Especially since I spent about 8 hours yesterday working on a lab

wow

And as I was doing so, I realized that the time was about 1 am...

eval"s.split($/).map(&:chars)#{(".map{|x|x.join.gsub(/(?<=o) +(?=o)/){|x|x.tr' ',?|}.chars}.transpose"*2).sub'|}','-}'}.map(&:join)*$/"

any Ruby golfers?

this is horrible

That's got to be golfed better

What's it for?

:27542868 Good prediction!

got any more data?

@El'endiaStarman lol, I'm a bit slow haha

@Doorknob Me? No. :P

factorisation:

um

are the CS:GO matchmaking servers down?

@Doorknob Ugh. Not productive after 5 hours of math homework. Sorry

haha, it's okay

@Cyoce Gaben is too busy bathing in money from the sale. forgot to turn them back on

9.9 q/day still

you should make a challenge to create an average questions per day figure

for n=36 the outcome is probably 524286

that's exactly what I did an hour ago, no? ;)

             1         2         3         4         5         6
    123456789 123456789 123456789 123456789 123456789 123456789 12
1s:  1 213 3251634 439265C1A693E45 54C3?9C2A675CBN1LA86Q9K49E
0s:  1 121 1213121 121312141213121 1213121412131215121312141213121

D: but it was unclear

It's funny how strong the desire to squash n=36 into the pattern is. :P

but... i'm sure there's a reason for the pattern

Certainly. But the rule has to come from the observations, not the other way around. :P

Fit the theory to the data, not the data to the theory.

@Doorknob you should probably include 0 1 for 1, 3 etc

the pattern is easier to see without them

you mean the blocks of length 2^n-1?

I mean the abacaba pattern at the bottom

Once @orlp's program gets 58, you'll have the rest up to 63 and beyond.

I should graph these.

Gimme three minutes.

back to 10 q/day, by the way

Values were log'd.

no discernible pattern...

@Doorknob For the sake of your fingers and internet connection........stop refreshing the page! :P

hahaha

Do we allow functions to return the string 3.14 for the number 3.14?

Okay, need to shower and eat. Be back in a bit.