^ this is my last one for today. The movement grows more and more hectic the more segments you have.

@ATaco Symmetrical because [90,180] is the mirror of [270,180].

I am aware.

But it's neat.

Also, is that Primes raising..?

No, just changing the angle for the primes below 10^5

But that could be interesting too.

Tomorrow let's work on these:

(good night everyone, it was fun=)

I'm glad this programming math chat exists

@ATaco Me too. You can thank @NathanMerrill for creating it.

@ATaco Is that a cardioid?

Nope.

Although it does resemble one.

It's just primes being stupid and pretty like.

I mean i know it isn't but I thought i might point out the resemblance

lovefiddle.com is down, so I can't send working snippets of my code :(

Well, in a sense, ATaco's diagram traces out the endpoint of a bunch of circles rolling around each other, which kinda explains why the outermost edge looks cardioid-y.

You'd get a similar diagram if you simply took the end point of a progressively tightening spiral.

Yeah, basically.

Because well, that's what it is. Just one with a bit of randomness to it.

@PhiNotPi And yet, with your method, you calculate the acceleration and velocity exactly and then update the position?

I'm playing chess against stockfish 3: the first time I won, but I must be getting worse, as I haven't beat it since

It's learning

This has convinced me to join Math.SE

Drawing the Prime Spiral.

Fascinating.

Sudden idea I just had: run a minimal spanning tree on the points.

I was just working on my work-project, and the thought came to me that what I was making was fairly complex, but that was so it could be reduced elsewhere. I realized that this is in a way analogous to Heisenberg's uncertainty principle, which is that you can't measure both position and velocity absolutely precisely since there will always be some error shared by them. This key word, shared, means that you can e.g. measure position super precisely if you shove all the error into velocity.

Likewise, there must be some sum total of complexity present in a program for it to do complex things. That complexity can be moved around, though. A language that is very simple internally requires the programmer to handle all of the complexity. A language that is very complex makes it possible for the programmer to write simple programs that accomplish complex tasks.

Using the same language, though, the way you structure your program's architecture, classes, objects, etc. factors into whether the rest of the program is complex.

That's why libraries are so useful. They package a large amount of complexity so you don't need as much on your end to do what you want.

True. In OOp for example, you shove complexity and functionality into a class, and just simply call methods

Exactly

I believe this is also why functions are so useful. You trade complexity in many places for complexity in one place. Net benefit!

Well, that isn't really the same thing since it's a net reduction in complexity.

In this case, I feel it is a net reduction in complexity.

A better example would be trading validation complexity around calls to a constructor for validation complexity in the constructor.

Any abstract representation of a code that has Functions or similar, simply refer to the calling of such as a single entity, one unit of complexity. You've removed unnecessary complexity, rather than simply removed complexity.

@ATaco I find it neat that you manage to constrain the drawing to the two points

It's rather simple, really. First I get the angle between the start and and point, and for each point counter rotate it by that amount, then, I get the distance from the start and end, and divided each points distance by that amount. Then simply move and scale it where I want it.

Id get you the code, but I'm I on my phone right now.

@flawr yay, Eisenstein primes :)

I thought I'd just kick this off with a larger section of the primes (up to a = 200). I couldn't be bothered to add the associates and conjugates to the plot yet:

there are some curious vertical sections where there are several primes in a row in steps of 1+2ω.

@MartinEnder Yaay=)

@MartinEnder What did you use to plot it?

Mathematica, of course ;)

I need to get into mathematica at some point.

Probably the best if I put that off till after my exams:)

The code is pretty concise actually:

ω = Exp[2 Pi*I/3];
maxA = 200;
eisensteinInts = Catenate@Table[{a, b}, {a, 0, maxA}, {b, 0, a/2}];
eisensteinPrimes = Select[
  eisensteinInts,
  #2 == 0 && Mod[#, 3] == 2 && PrimeQ@# || PrimeQ[#^2 - ## + #2^2] & @@ # &
];
Graphics[{
    Gray,
    [email protected],
        Point[ReIm[# + #2 ω] & @@@ eisensteinInts],
    [email protected],
    Black,
        Point[ReIm[# + #2 ω] & @@@ eisensteinPrimes]
}]

Oh thanks for sharing, I'll look into that later, I have to do some work now :)

random question are there "rings" (small empty hexagons) of Eisenstein primes other than the ones surrounding 4+2ω (and associates). I'm seeing clusters with 5 of those 6 primes higher up, but up to 400 I can't see another full ring.

@MartinEnder How hard would it be to color those almost-rings in red or something?

harder than I'm willing to figure out right now :D ... I might give it a try tonight, depending on when I get home

it also seems that outside of the first small ring there are never three primes that are mutually adjacent. it seems like there is a hex grid like this over the triangular grid, where primes can only appear in the # positions:

 # # . # #
# . # # . #
 # # . # #
# . # # . #
 # # . # #
# . # # . #

I'll bet that's because there's an equivalent of 2 - the "parity", so to speak.

(the only exception to this being 2 + ω)

@El'endiaStarman yeah, modulo 3 appears in both characterisations for being an Eisenstein prime

random question: how far can you get from the origin by only jumping between primes, using a certain jump (e.g. 2+ω and rotations thereof)

okay, with that jump you'd have to start on 2 to get anywhere

@MartinEnder Using the same direction each time?

no, you're allowed to make turns

I can't seem to get very far by hand. 19 + 6ω if I didn't miscount.

hm, I think I did make a mistake though

nah okay, I can still get there

it's a fairly arbitrary question though.

@MartinEnder For that you would need some number x where x+1, x-1, x+w, x-w, x+w^2, and x-w^2 are all primes.

@El'endiaStarman actually, yes. 2 + ω is that equivalent, because you can walk the grid with that and its associates (such that every such point ends up being a multiple of 2 + ω).

@El'endiaStarman yep. no clue how to go about proving whether or not that exists (apart from the first one)

For starters, such a ring cannot be on an "axis" (the unit multiples) since both x-1 and x+1 (picking the 0-degree axis w.l.o.g.) would have to be integers of the form 3n-1, which is impossible since x-1 and x+1 are 2 units apart.

yes, although I'm seeing several 5-out-of-6 rings on the axis

Yeah, but we're looking for 6-rings, aren't we?

yeah of course.

I might just write a quick search script for these.

I was thinking that the next pieces of the proof would be to show that if x+w and x-w are both primes, then this condition holds, and likewise for x+w^2 and x-w^2. Finally show that those two conditions cannot both be true at the same time (unless we're close to the origin).

I have a feeling that the prime that gets in the way is one of those in the first small ring. i.e. after we've poked holes in the grid with 2 + ω to get the hex grid, if we then poke more holes with one of those and its associates, we'll end up removing at least one prime from every ring, except for that inner one.

they do exist

{4,2}
{392,196}
{476,238}
{532,14}
{616,308}
{3808,308}
{3892,1946}
{5768,1414}
{9100,4550}
{10640,5320}

so most of these end up up being on the b == a/2 diagonal

Curiously, all of these components are divisble by 14.

And the as are divisible 28 even

(Except the first one, of course.)

That is quite curious.

Oh, I forgot that I left it running in the background. It's up to 40k now:

{12768,6384}, {12922,4844}, {13132,6566}, {13286,1918}, {13286,3304},
{13440,6720}, {14644,6188}, {15008,3724}, {15428,3472}, {16744,8372},
{17024,8512}, {17052,3150}, {18088,7826}, {18452,9226}, {18606,4788},
{19012,1316}, {19376,9142}, {19712,2128}, {22778,9478}, {23842,8834},
{23842,10556}, {23856,6174}, {24122,4144}, {25088,3430}, {25088,12544},
{25956,4788}, {26600,13300}, {26908,10346}, {27202,10640}, {28014,42},
{28154,11998}, {28182,8904}, {28630,9212}, {28728,13986}, {29078,4102},
{29512,2912}, {29526,2058}, {30450,4200}, {30772,3878}, {30842,6916},

Huh, more common than expected?

well, I think it's just that they have a low density that doesn't fall very quickly

btw, here the a aren't necessarily divisible by 28 any more, but still all components are divisible by 14

hmmm, their density seems to be increasing actually

if I compute N^2/#{a<N,b} for N as 10000, 20000, 30000, 40000 (and then drop a few zeros), I get 1.11111, 1.42857, 1.91489, 2.22222

Same thing for N in increments of 1000:

also interesting: out of the 72 rings we found, only 18 are on that diagonal that seemed to be more common initially:

{4, 2}, {392, 196}, {476, 238}, {616, 308}, {3892, 1946}, {9100, 4550},
{10640, 5320}, {12768, 6384}, {13132, 6566}, {13440, 6720}, {16744, 8372}, {17024, 8512},
{18452, 9226}, {25088, 12544}, {26600, 13300}, {32032, 16016}, {37940, 18970}, {39648, 19824}

Sorry to interject, but is this room for all kinds of math and programming stuff, or some specific prime number project of yours?

I had the impression that it's all kinds of (mostly recreational) maths and programming stuff.

Ok, nice.

@MartinEnder I'm stupid, that's the reciprocal of the density...

Okay, so the density is falling slowly. Curiously, the density without the rings on the diagonal seems remain almost constant after N = 15k.

These pairs or rings have exact integer distances:

{616, 308}, {3808, 308}
{13286, 1918}, {13286, 3304}
{18606, 4788}, {25956, 4788}
{23842, 8834}, {23842, 10556}
{25088, 3430}, {25088, 12544}

{16744, 8372}, {34762, 5348}
{26908, 10346}, {35728, 7322}
{28014, 42}, {29526, 2058}
{28182, 8904}, {39270, 12138}
{28182, 8904}, {39648, 19824}
{32032, 16016}, {36232, 11606}
{34720, 10682}, {38752, 18242}

I haven't checked yet for the second half whether they lie on a diagonal.

Doesn't seem to be the case.

@MartinEnder Rational integer or gaussian integer distances? :)

Real integer :P

Ah crap.

I didn't compute the distances correctly.

sqrt(a^2-a*b+b^2) should work

the norm on the gaussian integer induces an actual norm

nah, the problem was that I was simply computing euclidean distances on the (a,b) pairs.

okay, here we go:

{616, 308}, {3808, 308}
{13286, 1918}, {13286, 3304}
{15008, 3724}, {15428, 3472}
{18606, 4788}, {25956, 4788}
{23842, 8834}, {23842, 10556}
{25088, 3430}, {25088, 12544}

{22778, 9478}, {26600, 13300}
{26908, 10346}, {27202, 10640}

{19376, 9142}, {29078, 4102}
{32032, 16016}, {34384, 16898}
{34720, 10682}, {38752, 18242}

the first set is obviously along a diagonal (they share one component). the second set is along the the ω^2 diagonal. the third set is not on a diagonal.

(so those are the fun ones, I guess)

@MartinEnder So that is a whole new category of primes XD

whole new category?

well these aren't even primes :P

But hexagons of primes?

@MartinEnder well funny primes as in sexy primes and twin primes , friendly numbers etc=)

yeah, I think they are comparable to twin primes, I guess

it's just that I've listed them by their centre which itself isn't prime.

Not sure why I didn't think of this earlier, but here is a plot of the prime rings up to a = 40k:

Is the density distribution of Eisenstein primes known?

I think you could derive bounds by generalizing the derivation of the bounds of the distribution of the rational primes.

> In fact, though, this is true for general number fields.

Haha, I love how formal the comments on that answer are.

The trolls are probably too intimidated by all the math:)

Talking about primes and squares, I made something:

red is prime, blue (not really visible) is square, black is the rest

(click on image and view it enlarged for best results). So basically I start in the centre and go around in a spiral. If the index of the pixel is a prime or a square, I change the colour accordingly

You can almost see an X is the middle

I forgot to include blue: i.sstatic.net/Hgyqg.png

If you look closely, the squares (blue) seem to form a spiral

The animation of the rotating spiral formed by the squares is pretty cool

It's like a galaxy :)

@KritixiLithos That's neat.

Math patterns are fun to visualise :)

That they are. :D

Huh, I've noticed that it takes longer to draw a point than to draw a rectangle is Processing. The pattern renders in a bit over 1 second when I use rectangles, but it takes over 5 seconds when I use points.

Maybe it's doing some optimization with rectangles and not points?

That doesn't make much sense though.

Maybe because for rectangles I use fill() to set their fill colour (and I've ensured they have no stroke). But for points, their colour is specified using stroke().

So I guess it's something to do with the difference between fill and stroke (but it is still weird)

*almost continuous deformation of z to 1/z

That's cool.

also it is using half a complex dimension more=)

so, I was looking at this challenge:

I'm trying to figure out if there's an OEIS sequence for it

@NathanMerrill gonna flag this as spam, since it is about PPCG :D

stuff can relate to PPCG, just don't talk about the site (like policies and such)

:P

anyways, I've got 1,2,4,10 so far by hand

but of course, there are a gazillion sequences on OEIS for those 4 digits

Those aren't the lowest possible values, are they? 1,2,4,5 should work?

oh, I did make another assumption

a list of N values consists of 1,2,3,...N

?

so, its basically permutations of 1..N

1,2,4,10 is not a permutation of 1..10

ok, as an example for N=3

123 is not allowed, because 2 is the average of 1 and 3, but 132, 213,231,312 are allowed

(321 is the other one not allowed)\

so, for N=3, there are 4 total permutations that work

@El'endiaStarman LucasVB is an amazing animator when it comes to techical stuff on wikipedia, he said he's starting a YT channel:

@NathanMerrill I thought you were looking for oeis.org/A003278

@flawr Ermagherd!

Probably one of the channels without videos with the most subscriptions.

Yeah, no kidding; it was at 2,346 when I clicked.

@NathanMerrill oeis.org/A003407

that's it\

thanks :)

> AUTHOR N. J. A. Sloane, Apr 30 1991

Before I was born. Sheesh.

Must've been one of the sequences he had on cards.

you know, math existed before you were born

FYI

huh, there doesn't appear to be a direct formula

Math would be boring otherwise.

@flawr Well, can you imagine if Archimedes had access to a computer?

ok, another challenge: given N, generate a permutation of 1..N such that it has no 3-term progression, in O(N)

Haha, imagine Euler had access to a computer! We would not have any unsolved math problems anymore XD

the problem with the above, is that most patterns generate arithmetic progressions

slightly harder: given N, generate a permutation of 1..N such that it has no 2-term progression, in O(log(log(N))

)

you can't generate a permutation in less than O(N)

you need to consider N terms

In my case you actually do not XD

wait, 2-term progression?

yay:)

doesn't only N=1 and N=0 work for that?

Yep, it was not a very serious challenge XD

But I like how you noticed the O(log(log(N))) before you noticed the 2-term restriction :)

I read from right to left, duh :P

@flawr More likely, we would be saying "Eulerlogic" or something instead of "mathematics".

Eulogy

I hope you're aware that "Eulogy" already has a meaning in English. :P ...though it could actually work here. Maybe.

clearly the mathematical version of it would overrule and create a different word for eulogy

At this point I'm not even sure whether mathematicians haven't already abused that word.

But that doesn't work for talking about math Eulogy with normal people.

if eulogy really replaced "math", "math" is a much more common word than the current eulogy

even for normal people

anyways, I cannot think of a way to generate a permutation

One that fulfills the no-average-in-middle rule?

yeah

@NathanMerrill So as soon as there is a three termprogression as a subsequence of the permutation it is impossible, right?

right

@El'endiaStarman finally finished that BFS to figure out how far one can move across the Eisenstein integers with fixed jumps (up to rotations and reflections). my by-hand solution of (19,6) for jumps of (2,1) was correct.

So it is probably not too dificult to prove that this is always the case for N>=4

now trying more fun stuff

(3,1) gets you to (48, 7)

wait, I think I misunderstood you

N=4 is definitely possible

1,3,2,4 is a possible permutation

every N has one

1

1,2

1,3,2

(5,1) is taking a while

oeis.org/A003407

the issue is generating them in a time-efficient manner

Ah so I did not understand it!

@El'endiaStarman With (5,1) you can go to at least (1087,529). (aborted there for now to try some larger steps)

@MartinEnder Wow, impressive.

The number of reachable spots from one location is the same regardless of the step size, right?

@MartinEnder Did you plot the size of the jumps to the maximal distance?

I only have like 5 data points so far :P

CMC: It would be fun to see the primes of the integers of Q(exp(2*pi*i/5))

@El'endiaStarman Not quite. If the step size is (a,a/2) then you only have 6 possible jumps instead of 12.

same for (a,0)

Hm, this is a first for me: I just had to report an error on MathWorld

Okay, got all the distances up to a = 9 now:

    (2,0)
(2,1) => (19,6)
(3,0) => (34,7)
(3,1) => (48,7)
    (4,0)
(4,1) => (67,18)
    (4,2)
(5,0) => (10,3)
(5,1) => (1351,405)
(5,2) => (33,10)
    (6,0)
(6,1) => (53,21)
    (6,2)
(6,3) => (47,9)
(7,0) => (19,7)
(7,1) => (73,31)
(7,2) => (1183,540)
(7,3) => (58,15)
    (8,0)
(8,1) => (1450,583)
    (8,2)
(8,3) => (107,9)
    (8,4)
(9,0) => (81,11)
(9,1) => (44,9)
(9,2) => (55,27)
(9,3) => (1317,193)
(9,3) => (47,21)

the indented ones can't leave (2,0) (which is where I started for all searches)

You tend to get less far when you start from (2,1) instead:

(2,0) => (20,3)
    (2,1)
    (3,0)
(3,1) => (48,7)
(4,0) => (21,10)
(4,1) => (67,18)
    (4,2)
(5,0) => (9,1)
    (5,1)
(5,2) => (33,10)
    (6,0)
(6,1) => (53,21)
(6,2) => (479,230)
    (6,3)
(7,0) => (23,8)
(7,1) => (28,1)
    (7,2)
(7,3) => (58,15)
(8,0) => (30,7)
    (8,1)
(8,2) => (501,71)
(8,3) => (107,9)
    (8,4)
    (9,0)
(9,1) => (51,14)
(9,2) => (55,27)
    (9,3)
(9,4) => (47,21)

Very interesting.

Why are there two (9,3) in the first listing?

Because I wrote it by hand and the last line is supposed to be (9,4)

(off topic: I just found a trilogy comprising 8 books...)

:D

@El'endiaStarman 12,3 is the first one that beats 8,1 (it got to 1622,791)

starting from 2,0 again

Could you also record the number of jumps?

That would require some changes, but sure.

Okay, rerunning.

I wonder whether Eisenstein prime jumping will become an olympic discipline at some point.

distance/steps would probably be a fun quantity to look at

@flawr it'll be part of the Eisenstein/Gauss prime jumping biathlon.

    {2,0}
{2,1} => {19,6}     (10 steps)
{3,0} => {34,7}     (14 steps)
{3,1} => {48,7}     (31 steps)
    {4,0}
{4,1} => {67,18}    (21 steps)
    {4,2}
{5,0} => {10,3}     (2 steps)
{5,1} => {1351,405} (484 steps)
{5,2} => {33,10}    (16 steps)
    {6,0}
{6,1} => {53,21}    (14 steps)
    {6,2}
{6,3} => {47,9}     (12 steps)
{7,0} => {19,7}     (3 steps)
{7,1} => {73,31}    (17 steps)
{7,2} => {1183,540} (315 steps)
{7,3} => {58,15}    (15 steps)
    {8,0}
{8,1} => {1450,583} (317 steps)
    {8,2}

@El'endiaStarman ^

actually the interesting quantity isn't distance/steps but distance/steplength/steps

(13,2) has been running for ages.

I'll just leave it on overnight

It'd be neat to see the path to the furthest point and/or the tree of all reachable points.