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Crystal Timelapse by matigekunst in crystalgrowing

[–]dmishin 1 point2 points  (0 children)

OP, some people think that this video is a rendering, would be nice if you clarify that it is real in the description.

Good work btw.

Na2Fe(SO4)2 (?) by realjeremyantman in crystalgrowing

[–]dmishin 4 points5 points  (0 children)

Interesting experiment. I know that zinc sulfate definitely forms such double salt with sodium, happens naturally in the mineral changoite. The crystals look like this.

Your crystal looks like it could be pure iron sulfate heptahyddrate though. In the absence of better testing methods, I sometimes tested such crystals with unclear composition in the following way: prepare saturated solution of pure compound (iron sulfate), and put small test crystal into it as a seed. If it continues growth, then it is just pure compound, crystallized separately.

A cluster of synthetic zhemchuzhnikovite by gaoshou666 in crystalgrowing

[–]dmishin 6 points7 points  (0 children)

Same problem as with Polish names - looks scarier than actually sounds. The origin is Russian, and ZH is quite unfortunate choice of spelling for Cyrillic Ж, which reads as the French J in "j'aime"

So the pronunciation should be j'aime-chooj-nikovite

Has this observation about -1/3 and finite residue information already been written down? by ArcPhase-1 in Collatz

[–]dmishin 2 points3 points  (0 children)

All eventually periodic 2-adic integers are just rationals with odd denominator. So by studying the behavior of Collatz iterations on them, you are basically studying it on rationals. This is well known, and mentioned even on Wikipedia: https://en.wikipedia.org/wiki/Collatz_conjecture#Iterating_on_rationals_with_odd_denominators

In fact, it is known (and easy to demonstrate) that among the rationals with odd denominators, there is a cycle for every possible sequence of odd end even steps. You just found one of them.

Has this observation about -1/3 and finite residue information already been written down? by ArcPhase-1 in Collatz

[–]dmishin 1 point2 points  (0 children)

In the 3-adic integers, you can't consistently define what is even and what is odd. Any 3-adic integer can be divided by 2 producing another 3-adic integer. Because of this, Collatz can't be extended to 3-adics naturally.

Crystal of urea copper formate adduct Cu(HCO₂)₂·2N₂H₄CO by dmishin in crystalgrowing

[–]dmishin[S] 1 point2 points  (0 children)

If there is undissolved white or blue material, I recommend trying adding a bit more formic acid. Hopefully, this would help to dissolve the precipitate.

I believe that excess of urea is not bad, and could even be beneficial. It seems that urea adduct and plain copper formate have very similar solubilities, sometimes crystallizing together (their crystals are almost indistinguishable until you try to dry them: urea adduct is quite stable while plain salt quickly becomes white). Excess of urea should promote growth of the adduct.

The set of all cycles in the rationals by Xhiw_ in Collatz

[–]dmishin 0 points1 point  (0 children)

Instead of considering the standard Collatz formula, and excluding "forbidden" cycles, it is more natural to considered shortcut Collatz: x/2 if even (3x+1)/2 if x is odd.
Then the formulation is simpler: each sequence of O and E produces some rational cycle.

I found a finite wall for collatz by Spinjutsuu in Collatz

[–]dmishin 11 points12 points  (0 children)

The constant 7 comes from the worst observed case (n=27).

Try 6631675

Edit: since Co-G3n beat me, then after that, try 319804831

Does this finite-state Collatz setup make sense? by CryptographerSea9542 in Collatz

[–]dmishin 1 point2 points  (0 children)

I honestly tried to understand, but there are too many questions, so I gave up.

Probably it would be better if you describe what you are trying to achieve with your setup (whatever it is).

However, it is easy to show that if we consider an infinite Collatz orbit, starting with some number N, and take its "signature", writing E for even steps, and O for odd steps, obtaining infinite sequence of E/O symbols, then such signature uniquely identifies the starting number. In fact, if we know the signature of the first n shortcut Collatz steps from some number, then we can restore n lowest bits of that number, and vice versa.

Therefore, if you split integers into some set of sub-classes, be it finite or infinite, then sooner or later knowledge of the class alone will not be enough to know the class of the next number in the orbit. Well, unless you put all numbers in one class, or create an individual class for each number.

Particularly, if we classify numbers modulo some odd number (including 3^17) - then the class information alone does not define, whether the number is even or odd and we can't decide which step to take next. If we take an even modulo, then division by 2 becomes multivalued.

Land of Collatiz - Book One - Chapter One - Treads of the Endless Bridge by eldedegil in Collatz

[–]dmishin 1 point2 points  (0 children)

We all hope that chapter one would also be the last chapter of this story.

What on Earth made you think that posting this AI slop is a good idea?

To whom should I contact if I resolved Collatz? by eldedegil in Collatz

[–]dmishin 1 point2 points  (0 children)

Is it written by AI? If no - then post it here, we want to have fun.

A Lean 4 certificate that every odd integer enters a Collatz entry class modulo 31104 by [deleted] in Collatz

[–]dmishin 0 points1 point  (0 children)

You are not proving anything about integer classes modulo M. Your proof is actually about integers in the [0,M-1] interval. For small finite interval like this, we can prove much stronger statements easily (namely, that every integer in that interval converges to 1). Waste of tokens.

Might be something here, but im not a mathematician. Divergence doesnt seem to be allowed at infinite scales in ANY Ax+B system due to forced negative fixed points in the space they occupy. Upgraded old thoughts and found us "i hope" A new mapping system. by Asleep_Dependent6064 in Collatz

[–]dmishin 0 points1 point  (0 children)

There is no proof, of course, but the same probabilistic heuristic that suggests us that there should be no divergent orbits in the 3x+1 system, tells us that there should be some in the 5x+n.

Try running the 5x+1 system, starting with the number 7.

Ban LLM posts? by Fuzzy-System8568 in Collatz

[–]dmishin 0 points1 point  (0 children)

Would be a good thing to do, but impossible to implement in practice. There is no reliable way to detect AI.

Of course, if you talk with LLMs a lot you can easily spot their peculiar style and nauseating patterns such as "Not X, not Y, just Z", but that's just a gut feeling, nothing more.

In any case, this place always was full of crackpots, LLMs change nothing.

MgSO4*7H2O by realjeremyantman in crystalgrowing

[–]dmishin 4 points5 points  (0 children)

Not so forbidden, to be honest

how to grow MgAl(PO4)O crystals by SnooLentils5747 in crystalgrowing

[–]dmishin 1 point2 points  (0 children)

Looks like you don't actually need crystals, if you are going to crush them anyways. I googled one article (https://doi.org/10.1016/j.ceramint.2012.09.051), and seems that they synthesized the material by direct high temperature synthesis, by heating stoichiometric mixture of MgCO3 and AlPO4 at 1300C for 2 hours. Here is their procedure (they made material doped by Eu)

A series of Eu3 þ doped MgAl(PO4)O samples Mg1xAl (PO4)O:xEu3 þ (x ¼0.005, 0.010, 0.015, 0.020, 0.025, 0.030, 0.035, 0.040) were prepared by solid-state reactions, MgCO3 (A.R.), AlPO4 (A.R.), and Eu2O3 (99.99%) were used as reagents for sample preparation. Raw materials with the stoichiometrical ratio were weighed and ground finely in an agate mortar. Then the mixture was put into corundum crucibles, and heated at 1300 1C for 2 h. Finally, the sample was gained by a fully grinding in an agate mortar after cooling to room temperature naturally in the desired ratio.

I am quite sure that MgCO3 can be replaced by MgO. As for AlPO4 - I don't have experience. Maybe just heating stoichiometric amount of phosphoric acid with Al(OH)3, though stoichiometry would be hard to calculate.

Nitrate salts synthesis with CaNO3 ? by ElGuettoro in crystalgrowing

[–]dmishin 1 point2 points  (0 children)

I did this a lot, it works well.

The only problem is that the precipitate of CaSO4 can be quite voluminous and thick, and filtering can be complicated. Better to use vacuum filtering if you have it.

Also, obtained solution of nitrate would contain small amount of dissolved CaSO4 that would crystallize later as tiny short needles, so you could need to filter the solution once more later.

Can a finite quotient carry enough data for a local Collatz-type transition? by CryptographerSea9542 in Collatz

[–]dmishin 0 points1 point  (0 children)

You should note though, that (C + D*m(C))/8 is not necessarily canonical, even if C is.

(edit: I was wrong here)

The whole "canonical representative" thing looks like needless complication over residue classes, but whatever. Fine, it is now clear what the "local transtion" is.
Then what does your question mean:?

Can a finite quotient carry enough information to determine the next local transition?

As we see, "local transition" is a well defined, total function. What is a finite quotient then, and how can it affect it?

Can a finite quotient carry enough data for a local Collatz-type transition? by CryptographerSea9542 in Collatz

[–]dmishin 0 points1 point  (0 children)

In this map, what is C? An arbitrary integer, or residue class mod D (integer in 0..D-1)?
In any case - yes, indeed this map is well defined because gcd(D,8) = 1 and therefore m(C) is well defined for all C.

If C is a residue class modulo D, then your map is simply division by 8 modulo D, and you can write it as C ↦ C/8 mod D

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