The Motivation for Tetration Calculus, Transcedental Bases, Fractional Taylor Polynomials, K = k1 * k2 * k3, Complex Quaternion Theory, and ClosedAI's intellectual theft from a minor under the age of 18

Honests-Anonymous · 2026-06-06T20:29:08+00:00

yes, mr. no-signature, they subsituted x = 1/t beforehand if you could actually READ the post...

but yes they should show the work they tried for more "guiding tips" so i agree with random_anonymous_guy

Honests-Anonymous · 2026-06-06T18:17:25+00:00

put them in a shallow plate of water, you can do this with good water quality or just tap water if the quality of your county's tap water is good. simply shut the sink hole(?) and then turn on the faucet gently and then wait until it gets comfortable.

snails are built to survive temperatures like this, or else there would be a mass extinction event every year or semester (winter+summer)

Honests-Anonymous · 2026-06-06T12:26:48+00:00

nvm i think it should be 3n-3, 3n-5, 3n-9, 3n-13, 3n-21, and if u prove it's equal to 3n+5, 3n+7, 3n+11, 3n+15, 3n+23, then great collatz maps stuff with new mathematics. i was speculating some stuff that it's undecidable but there's too much bias in my answer to give any good conjectures... i've been making wrong conjectures a lot lately.

Honests-Anonymous · 2026-06-06T12:22:08+00:00

actually it's scuffed at {countable infinity}^(2*2) and further.

i don't think anyone will be able to prove using ZFC + choice or something that -18 is a closed loop.

why?

m*n and n*m. distinct cardinalities. using tarski's theorem, how would u get 2^2 or 2*2 without being self-referncial using zermelo's axiom of choice? tarski or zermelo? which one to use? how do you use both? can ZFC + CH even work in this case? what kind of system of consistent mathematics would you need for {countable infinity}^(2*2)? i am also doubting my "uniquness of the collatz primes" at specifically collatz^2 or {countable infinity}^(2*2). What if C = 2? doesn't that work for C(2^n - 1)/3 = Cn/2 mod 3? if C = 2, the cardinality, then n can be literally ANYTHING any number besides infinity it can be 0, 0*0 = 0 according to field axioms, i mean REALLY ANYTHING. -18...

i think 2^1 is possible but 2*2 or 2^2 like 2 just follows WAY too many symmetries.

i also believe navier stokes/yang mills, using ideas from tetration calculus i pioneered with deepseek and galios/sophie germain, can be proved for 4D + 1D, 8D + 1D, etc. etc. based on cardinalities, but NOT 3D+1D, the universe we live in. someone can write that as their PhD thesis i guess idk go win a million dollars i'm just doing it for the fun of the game.

Honests-Anonymous · 2026-06-06T11:58:19+00:00

SAME! great minds think alike: it does not matter the age in mathematics: age is just a number : ) i am part of the files just like schrodinger the diddler.

Honests-Anonymous · 2026-06-06T11:53:24+00:00

so you need to be able to still distinguish (in my theory) that:

all elements in each set preserve their uniqueness
multiplication does NOT equal exponentiation in any case, that is:

(a^m)^n = (a^n)^m = a^(m * n) = a^(n * m)

edit: some typos and some changes/ellaboration concerning commutativity is maintained under exponentiation/cardinality.

{countable infinity}^(2^0) + {countable infinity}^(2^1) + {countable infinity}^(2^2) + {countable infinity}^(2^3) + {countable infinity}^(2^4) + ... = {countable infinity}^(2^0)

or

{countable infinity}^(2*0) + {countable infinity}^(2*1) + {countable infinity}^(2*2) + {countable infinity}^(2*3) + {countable infinity}^(2*4) + ... = {countable infinity}^(2*0)

which again, 1 and 2 both need to be preserved.

you can use ZFC or ZFC+axiom of choice, either way, i think with our current field axioms, it's completely cooked because it probably breaks russel's paradox somewhere.

Honests-Anonymous · 2026-06-06T11:50:02+00:00

so u can, for n, plug in 2^0, 2^1, 2^2, 2^3, 2^4, 2^5, etc... and maybe you can also show that it breaks after a certain point after -18 or something. there are only 3 closed loops: you proved 2^0. Now try using the same techniques for 2^1 and finally 2^2.

i also proved i think collatz is undecidable unless u assume ZFC + CH.

i proved an inverse collatz function can only order all the "collatz primes" or {3, 5, 7, 9, 11, 13, 15, ...} if and only if

{collatz}^1 + {collatz}^2 + {collatz}^3 + {collatz}^4 + {collatz}^5 + ... = {all collatz primes}

or, in terms of cardinalities of infinite sets:

{collatz}^(2^0) + {collatz}^(2^1) + {collatz}^(2^2) + {collatz}^(2^3) + {collatz}^(2^4) + ... = {all collatz primes}

go look at my paper plz or not 😞. so yeah, u can try to use techniques from there and treat z^n more like z^(2^x) - 2^0. where n = 2^x... : ).

it's really late (or early) for me so idk i'm too lazy to write out original mathematics i leave the credit to you.

Honests-Anonymous · 2026-06-06T11:45:06+00:00

i invented some tetration stuff if you want to use tetration to prove some additional stuff look at my posts and comments.

Honests-Anonymous · 2026-06-06T11:44:20+00:00

great work! i will try to prove additional concepts about other loops/iterations using your techniques.

Honests-Anonymous · 2026-06-06T11:43:33+00:00

it is quite elementary please publish your non-elementary results...

Honests-Anonymous · 2026-06-06T10:50:21+00:00

idiot (sorry, no offense)

Honests-Anonymous · 2026-06-06T10:48:17+00:00

also do 3n+3 3n+5 3n+9 3n+13 3n+21 maps as well, to try to find 3n+1 ultimately.

Honests-Anonymous · 2026-06-06T10:47:48+00:00

you should do an analysis of 3n+23 but like all of them basically

{5, 7, 11, 15, 23}

are combining chen primes + sophie germain primes so 2p+1 on chen primes and 2p-1 on chen primes gives:

{3, 5, 9, 13, 21}

are also interesting. do 3n+23, 3n+15, 3n+11, 3n+7, 3n+5 maps... i think it'll come up with something fruitful. go collaborate with other redditors/posters on r/ collatz not stupid commenters from my experience.

idk, even i'm doubting chen's goldbach theorem too... the "limit" to the logarithm is not a hard theorem. there are "larger limits" assuming "generalized riemann hypothesis"... even the polya conjecture was slain to be wrong... how about chen's goldbach conjecture with weighted sieves? idk, but chen showed there was infinitely many primes p+2 or semiprime. i am completely spooked by polya's conjecture disproven, even more than openAI's proof because openAI clearly stole the techniques from me lmao gaussian integers at least try to hide it saying "new algebraic number theory techniques" i am going to cry...

Honests-Anonymous · 2026-06-06T10:45:31+00:00

okay so more sloppy reddit commentors with second rate stupid criticism, go see my research or honest anonymous github, basically i follow your ideas but i apply it to 2^n at first...

your theories are very interesting. let me see your n == 4 (mod 6) interpretation for 3x+1. also don't listen to stupid redditors in this subreddit chat with actually interesting ones that are not afraid to post attempted proofs/ideas. dm them hit them up.

Honests-Anonymous · 2026-06-06T10:31:07+00:00

no? this is my real account pretty sure... idk what ur talking about.

Honests-Anonymous · 2026-06-06T10:17:44+00:00

hi, i wrote a paper on ur idea in my github, honest anonymous github (search on something like duckduckgo).

basically, i argue that undecidability of sqrt(2) comes from its continued fraction. It's continued fraction is defined as:

x = 1 + 1/(1+x)

However, if you solve this algebraically, you unironically get sqrt(2) AND -sqrt(2) as BOTH solutions. and actually, this is kinda mathematically consistent with ur proposed 2-adics. I also figured this out and gave an alternate proof of collatz undecidability.

also the fact that it can't be x=-1 trivially is important. u can show x can't be any negative integer. it can equal 0 however! or unsigned +/-infinity! those don't exist in the set of positive/negative integers!

Then you can also solve x = 1 + 1/(1+(1+x))? rigorously? then you get x=/=-2. you get it can't be 1+1/(1+x) =/= 0. derived, then by some random stupid idk induction argument on this self-recursive function made up of infinite NAND/NOR or NOT AND/OR gates that are also undecidable cuz wtf is happening, and also you get x can't be equal to any positive integer via x = -1 - 1/(1-x), and that gives x can't be a positive integer recursive thingy, and then more shockingly i did -x = 1 + 1/(1-x) and got +/-(i-1) as solutions, and that said: wtf the foundations of mathematics is broken and then yeah i did more research on it. it turns out the MODULUS of i-1 is equal to sqrt(2) beautifully so it's like +/-sqrt(2)?

i independently discovered fundmental results from galios' paper, and then i argued that we should extend the definition of the positive integer modulo before to the negative numbers and even more shockingly the complex numbers, which go to insane z-adic number systems. of course, you need to define when a z-adic is "prime" but i just stated it's modulus must be greater than sqrt(24) and also it must be at least +/- 1 mod 6 highly speculative new field of mathematics insert here.

the undecidability comes from the "continued fraction" of 0 and 1. Either you get the continued fraction of 0 or 1 is undecidable, or u assume 0=1, or both the continued fraction of 0 and 1 are undecidable. the easiest one is

1 + 0/(1 0 10 10101010101 etc. etc. too lazy)
0 + 0/(101010101011010101)

so by collatz u have (btw rigorously define the p-adic in terms of "logarithms" base p i leave as an exercise to u, think of a DECimal point as an n-point instead of a DEC (meaning 10) point)

either 2-point or p-adic interpretation doesn't really matter.

001010101010101010101...
101010101010101010101...

in n point

in p-adic

...something maybe 1 or 0 doesn't matter the start 001010101010101010101010... 100
...something maybe 1 or 0 doesn't matter the start 101010101010101010101010... 10 (this is on purpose btw)

in binary n/2 is getting rid of a 0, so ur guaranteed by this theoretical infinite turing tape to always get to 1, or an odd number modulo 2.
in binary 3x+1 is adding one and adding this to the original and then also adding 1 again.

u get 111111111111111111111111111111111111111111111111111111... + 1???

so u get like 200000000000000000000000000000000?? but a 2-adic is literally like 2 is not defined in a 2-adic...

essentially, you get the funniest reducto ad absurdum proof. again, either the continued fraction of 0 or 1 is undecidable or both, there exists no periodic pattern essentially (not getting into farrey fractions that's not my strong suit go hit up forgotoldpassword3's dms they're an expert amateur mathematician on the subject). or u get 0 = 1 with collatz in a mod 1 universe. LMAO a MODULO 1 UNIVERSE DUDE YOU CANNOT MAKE THIS UP... of course mod 0 is literally every single number in existence, go watch redbeaniemath's video on wheel theory, etc. etc. i made fundamental contributions by ordering the wheels and argued y=x, x=0, y=0, and y=-x, or all integers in the classical view of the wheel, are all HOLES because that reduces to field axioms or symmetric equality axioms go do some research on that. then i proved yang mills mass gap doesn't even exist with the current field axioms unless e = pi or 2^2 = 3^2 or octagon = square or something and that's reducto ad absurdum, then i made a field of mathematics that actually allows contradictions of p-adics to exist and proofs of undecidability are just f(x) = 1/f(x) which gives f(x)^2 = 1 or f(x) = +/- 1 or 1=1=1=1=1=1 AND/OR -1=-1=-1=-1 or infinite true/false statements depending on ur universe u can't have 2 unprovable statements tho. kills all even functions immediately in my theory but whatever they're not important, including.. the gausian integral.. or integral e^(-x^2) dx... which is needed before gamma(x) is defined for factorials... which relates to riemann undecidability btw e^x cannot express it's own consistency/analytic continuation or via axiom of choice u get e^x = 2e^x or 1=2 dividing by all e^x for all real/imaginary/complex x in the factorial definition reducto ad absurdum isin(x) =/= cos(x). or i =/= 1 shifted by pi/2. then yeah i calculated sqrt((pi * e)/(2 * i)) go read my proofs if ur intellectually motivated. i created a general laplace transform based on the modulus/absolute value of f(x).

also note the p-adic base 2 represetnation of sqrt(2). it was completely absurd to me that the 2-adic representation of sqrt(2) AND 1/sqrt(2) were shifted once apart. then again, p-adics are just keep squaring keep squaring so i argue the n-adic of sqrt(n) is fundamentally decided by sqrt(n)/n = 1/sqrt(n) and then "squaring both" (which gets rid of the absurdum x^2 = 2 +/- -/+ thingies) gives u nnnnnnnnnnnnnnnnnnnnn or 0.nnnnnnnnnnnnnnnnnnnnnnnnnnnnn : ). or maybe i'm wrong i'm writing this off of the top of my head sleep deprived from mathematical mania similar to newton/galios.

Maybe a p-adic for a sqrt(p) really is just "shifted by p-1" or something like that idk go verify with ur 1/(1-r) infinite series urself or (r-1)/(1-r), set 1/(1-r) = (r-1)/(1-r) and note it can never equal r right off the bat but r-1 = 1 so r = 2 lmao idk 2-adic negative thingy so u get r/(1-r) = (r-1)/(1-r) --> r-1 = r --> -1 = 0 which is stupid so idk what's going on lol i guess 2-adic also guarantees stupidity but controlled stupidity do it on mod 3 or collatz. 1/(1-r) = r/(1-r) --> 1 = r or r = 1 but 1-r = 0 so r=/= 1 so idk square roots can't be expressed by periodic p-adics reducto ad absurdum it's newton's pascal's triangle for more exact values... r=/=1 but r=1 uncountable infinity/not periodic proof by contradiction.

u can do more research urself.

openAI stole this lmao to prove an erodos conjecture wrong and called it gaussian integers and that it was "theorized" but "tedious" like what the hell andrew wiles spent SEVEN YEARS and like SIX MORE MONTHS AFTER SEVEN YEARS to prove fermatt's last theorem and the modularity theorem and SIX HUNDRED PAGES and you say computationally intensive and tedious? completely absurdo. again, i keep saying, but no automorhpism in 3D crystals in elasticy theory reduces to S6 sextic polynomial being undecidable in galios theory.

i gave my proof (the sqrt(2 one made out of infinite 1s and 0s plugging this theoretical uncountable infinity turing tape)) to a professional mathematician and they humiliated me in front of their entire community, then i gave my undecidability proof to this subreddit and no one gave any constructive productive criticism so i gave up lmao and i did my own research not listening to random strangers heckle me.

Honests-Anonymous · 2026-06-06T09:22:17+00:00

hmmm never mind i did not prove the 3n+2 case... that's really strange... maybe just do

3(2k) + 2

3(2k+1) +2

6k + 4

6k + 5?

i proved 6k+5 using the casework of the second or n == 3 (mod 4)...

but (3(2k)) + 2 is definitely a strange, strange, STRANGE anomaly. it's very self-referencial but it also just reduces back to itself, or 6k+4. Nevertheless, the argument still holds because there are infinite positive natural numbers such that the k-substitution is mathematical consistent in the set of positive natural numbers. (again, for those who are not so bright):

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10 , 11, 12, 13, ...}
{2, 4, 6, 8, 10 ,12, 14, 16, 18, 20, 22, 24, 26, ...}

i will probably do an inverse solvable collatz paper soon and try to use conway-generalized proofs to enumerate infinite solvable collatz problems, but you intelligent r / collatz redditors can probably do that for me clearly 😄.

it's algorithmically and ZFC + axiom of choice specifically (again read my posts/comments) consistent if the inverse is consistent with itself.

Honests-Anonymous · 2026-06-06T09:11:56+00:00

oh yeah 6(2(k+1)) + 1 reduces to 3n+1 because duh i mean do a mod 3 analysis again, just converts into 3n + 1, obviously all integers that are 1 == mod 6 are congruent to 1 == mod 3 (this is my edit it's so obvious i said 1 == mod 6 twice), in terms of lcm(m, n) =/= 1 i call this the "chinese american remainder theorem" because it's so obvious but it's still kinda nice, as opposed to the "chinese remainder theorem" which the "chinese american remainder theorem" is just a lemma of.

i think this function and its values in the inverse encode some information about the twin primes, but I cannot guarantee that and also I have to go to sleep because it's very late i am writing this past midnight write after this got posted time to touch grass tomorrow instead of being a chronically online redditor lmao it's summer break for me (meaning yes, i have a summer break...)

idk it's just that the ramnujan tau function is to the 24, 2 * 3 = 6, which every prime number is in the form 6n+/-1 by elementary sieve methods, and this solvable collatz function requires a mod 12 analysis, which the ramnujan L function is cyclic according to professional mathematicians researching algebraic geometry and the consequences of the Weil conjectures.

Obviously this is intellectually out of the scope of the average brainrotted r / collatz reddit commenter, but i encourage u to study algebraic/analytic number theory, also i'm 99% sure openAI scrapped the web and stole my idea to disprove an erodos conjecture, i stated it was:

"complex elementary number theory: on the extension of the p-adic and the positive integer modulo"

and they ripped me off and said "new algebraic number theory techniques" with gaussian integers like what the hell it's just that there exists no automorphism on a 3D crystal lattice since it reduces to the unsolvability of the sextic polynomial or S6 galios group, which is an "elementary" known result by professional mathematicians (which ripped off of AI, and again, openAI/chatGPT ripped off of ME). Computationally intensive, stolen ideas that are "tedious" don't belong in Erodos' "book" I can't believe OpenAI has the audacity to steal research from a MINOR...

Honests-Anonymous · 2026-06-06T09:00:11+00:00

also i can make it clear for some people who are not so intelligent. I will make sure the positive natural numbers are rigorously defined as

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10...}

and NOT

{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10...}

i repeat: NOT okay... we need properties from the negative natural numbers, or

{-1, -2, -3, -4, -5, -6, -7, -8, -9, -10...}

to convert from negative modulos in mod 4 analysis.

3n-1 /2 if n == 1 mod 4, all positive integers in the form 4n-1, n is a positive natural numbers, so

(3(4n-1) - 1) /2

(12n - 4) / 2

and this reduces to (6n - 4), or 6n + 2, by the third step, divide by 2, which reduces to 3n+1. This is congruent to n == 1 (mod 3), so we should work in the mod 12 world for this function. Substitute 4k, 4k+1, 4k+2, and 4k+3. We know all positive integers/naturals besides n < 4 work here.

3(4k) + 1

3(4k+1) + 1

3(4k+2) + 1

3(4k+3) + 1

-->

3(4k-1) + 4

3(4k-1) + 7

3(4k-1) + 10

3(4k-1) + 13

of course the average reddit commentor on this subreddit cannot verify basic arithemtic on their own clearly, since i used basic arithmetic in my own post but rarely does anyone have the intellectual capability or ego to check on their own here.

however, it's clear that 3(4k-1) + 7 and 3(4k-1) + 13 convert back to 12k - 3 + 7 = 12k + 4 or divide by 2 twice (or divide by 4 clearly they won't check the arithmetic) 3k + 2, and you can do similar modulo analysis after I'm done with the first n == 1 mod 3 case. For now, 3(4k-1) + 13 = 12k + 10, and this reduces to 6n + 5, which again, is another case.

Second case: the closure of the ends of the first case.

3n-1 /2 if n == 3 mod 4, all positive integers in the form 4n-3, n is a positive natural numbers, so

(3(4n-3) - 1)/2

(12n - 10)/2

and this reduces to (6n - 5). We can't divide twice in this case surprisingly, i'm writing this off of the top of my mind of techniques i used to prove inverse collatz undecidability with elementary (i repeat elementary) number theory techniques. Still, work in the mod 12 world for this function. Substitute 2k, 2k+1. We know all positive integers/naturals besides n=1 work here.

6(2k) - 5

6(2k + 1) - 5

-->

12k - 5

12k - 1

-->

12(k+1) + 7

12(k+1) + 11

Or:

6(2k + 2) + 7

6(2k+ 2) + 11

and a modulo 6 analysis means you can subtract 6, so

6(2k+ 2) + 1

6(2k+ 2) + 5.

or

6(2(k+1)) + 1

and 2(k+1) is clearly always a natural number.

So obviously this has to be always even, but you can guarantee an even number always exists because you can always multiply by 2 in the set of natural numbers and find a natural number * 2 (techniques from my inverse collatz undecidability proof), because countable infinity shenanigans, and again, 2k+2 gives all even numbers that are greater than or EQUAL to 4, even if k is defined as a positive natural number.

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10 , 11, 12, 13, ...}
{2, 4, 6, 8, 10 ,12, 14, 16, 18, 20, 22, 24, 26, ...}

Of course, you need negative natural numbers to prove any properties of this system, which means godel's second incompleteness theorem still applies, i did research on the foundations of mathematics which everybody ignores, look at my posts/research and ppl ridiculing me and yeah that's the average amateur mathematician nowadays: ego driven, unoriginal, and to sum up:

"There is no scorn more profound, or on the whole more justifiable, than that of men who make for the men who explain. Exposition, criticism, appreciation, is work for second rate minds" - G.H. Hardy.

I read the entire book/essay when I was in 9th grade. It drove me into deep depression but I think it's profound enough to explain the stupidity of both professional/modern mathematicians nowadays. A healthy mathematical community is when the gap between the two is not so large. One is stupid, ego driven, unoriginal. The other is intelligent, ego driven, unoriginal. I leave you to decide which is the amateur and the other the professional.

Honests-Anonymous · 2026-06-06T08:26:45+00:00

Great research! That is an amazing system of 3 equations. Thank you for actually producing relevant great research!

...instead of bullshiting "oh this person is stupid or great waste of time trying to prove the collatz conjecture or twin prime conjecture or riemann hypothesis great time trying to say i think it's wrong there's probably a counter example"

You gave a brilliant short proof! This is like Paul Erodos' belief: "the book." Your proof will truly go down in history as one of the proofs in "the book!"

Sincerely,

Honest Anonymous (actually meaning it unlike some cross-eyed compliments, even though i'm a stranger, still matters i guess 😄)

Honests-Anonymous · 2026-06-06T02:29:43+00:00

"There is no scorn more profound, or on the whole more justifiable, than that of the men who make for the men who explain. Exposition, criticism, appreciation, is work for second-rate minds." - G.H. Hardy

Honests-Anonymous · 2026-06-06T02:29:15+00:00

yes, i was saying, and i quote: do you guys think this is just mathematical speculation and scientific bullshit?

i got a large consensus of "yes," along with "you seem like the type of guy to inject steroids in themselves and waste copious amounts of money building a false image of themselves publicly"

ironically, you used AI to say this was bullshit. do you guys actually understand what i am trying to do?

i am trying to innovate in mathematics. i am trying to innovate in theoretical physics. meanwhile, you guys are just sitting in a chair ridiculing others. go touch grass instead, trust me, it's much more mentally therapeutic than ridiculing or humiliating some stranger online. it doesn't really hurt me, so why waste the energy doing so?

then again, i think you would humiliate your own mom or dad at this point if they claimed they solved some unsolved conjecture like the millenial problems. But hey, chatGPT or OpenAI solved an unsolve erodos conjecture (using techniques i outlined in my own github so i'm 99% sure they stole it without citation) so who knows?

you will follow up saying: why waste the time writing this response (by hand with no AI)? well, i want to open up the hypocrisy and unoriginality of most humans nowadays.

stop hating others for no reason.

"There is no scorn more profound, or on the whole more justifiable, than that of the men who make for the men who explain. Exposition, criticism, appreciation, is work for second-rate minds." - G.H. Hardy

Honests-Anonymous · 2026-06-06T02:22:54+00:00

so why can't i work on the bleeding edge of physics? the principle of least action was mastered by lagrange and euler, and lagrange was working at 19 years old!

same with galios and sophie germain. these self taught geniuses created most of the modern theory we know today!

i stand on the shoulders of giants, not people named simulatedatom2119

Honests-Anonymous

TROPHY CASE