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Dubs theory thread

Name: Anonymous 2016-03-19 15:03

WELCOME DUBS THEORY RESEARCHERS

here you can talk about your most recent finds regarding dubs theory, I will start with the well know ``dubsless primes'':

- let к(n) be the number of bases in which n has dubs excluding base 1 and base n-1 as these are trivial, we shall call n a dubsless number if к(n)=0 and n>3

- the dubsless numbers up to 10000 are 5, 6, 29, 41, 61, 113, 317, 509, 569, 761, 797, 1213, 1229, 2617, 5297, 6221 and 8017. it turns out that all of these numbers except 6 are prime, and up to 10 million all except 6 are prime, we call these primes the ``dubsless primes'' a new kind of primes. this raises the following questions in dubs theory:

- is the set of dubsless numbers/primes infinite?
- is 6 the only non prime dubsless number?

Name: Anonymous 2016-03-19 15:07

How are 113, 1229 and 6221 supposed to be dubsless?

Name: Anonymous 2016-03-19 15:15

what about numbers that cant include same digit more than once?

Name: Anonymous 2016-03-19 15:26

I don't believe in numbers greater than 1000, that's just a myth.

Name: Anonymous 2016-03-19 15:51

>>2
Sir, this only studies repeating digits at the end of the number (dubs), generalizations for repeating digits in any position are welcome.

Name: Anonymous 2016-03-19 15:53

>>5
I see, thank you.

Name: Anonymous 2016-03-20 19:38

Bump for an actually interesting thread.

Name: Anonymous 2016-03-20 21:23

bampu pantsu

Name: Anonymous 2016-03-20 22:39

I'm not sure what your definition of a dubsless number is. It's not just a number without dubs, is it?

Name: Anonymous 2016-03-20 22:40

Oh wait I'm stupid. You were talking about bases. I was thinking of digits.

Name: Anonymous 2016-03-21 0:32

I'm enjoying these dubs

Name: Anonymous 2016-03-21 14:23

>>11
nice dubs, are you a dubs theorist?

Name: Anonymous 2016-03-22 20:52

So here's an approximate timeline:

2016, March: the Dubsless Prime Conjecture is posted on /prague/.

2016, April: as the problem remains unsolved, people who have been wracking their brains about it post it to several math forums, from which it spreads.

2016, June: a tenured math professor at a prestigious university breaks down and, instead of a boring lecture in Linear Algebra, explains to the students the significance of dubs and dubsless primes, along with her ultimately failed but otherwise fruitful approaches at cracking the problem. The video of the lecture goes viral.

2016, August: Scott Aaronson blogs about his grappling with the Dubsless Primes problem, arrogantly says that it seems to him about as hard as the Goldbach's Conjecture, despite only having spent a couple of months thinking about it day and night.

2017, February: an Australian surfer discovers a property of Dubsless Primes that links it to the various free variables in the Standard Model. The Dubs Theory disagrees with QCD about the charge of electron in the 10th digit.

2017, March: an anonymous poster on /prague/ posts a simple proof of the п(n) = P(пdubs(n)), that is, that the distribution of Dubsless Primes polynomially approximates the distribution of usual primes. The proof doesn't quite fit into the post size restriction, many mathematicians join the discussion recovering the missing parts and building upon them.

2017, March: Donald Trump pledges to build the biggest ever particle accelerator, to test the predictions of the Dubs Theory. progrider.org finally goes under the pageviews and Indian spam, gets bought from Cudder by Google for unspecified amount.

2017, still March: Chris "4Chandler" Poole goes on record saying that shutting down the textboards was his "billion dollar mistake". Meanwhile, progrider.google.com overtakes Arxiv.org as the place for mathematics and physics preprints, because anonymity gets rid of the authority bias, which obviously is liked by up and coming, lots of energy, scientists.

2017, July: LHC somehow goes above the original specifications again and verifies the Dubs Theory predictions vs QCD. Trump feels trumped, "very sad" says he in the respective Address. Someone on /prague/ speculates about the consequences of the Dubs Theory for controlled thermonuclear synthesis. Trump is reinvigorated and pledges to make America great again, again.

2020: Cold thermonuclear synthesis is achieved at last! The city of Prague unanimously (and anonymously, since by then all political decisions are discussed on specially created anonymous textboards) votes to erects the statue of Anonymous Progrider, thirty feet tall, made from solid gold, holding a Lambda in her right hand and a bundle of sticks in her left hand. She looks suspiciously like Suiseiseki.

Name: desu 2016-03-22 20:56

mailto:desu
desu

Name: Anonymous 2016-03-22 21:07

No

Name: Anonymous 2016-03-22 21:44

>>13
This is a continuation from an old /prog/ thread.

Name: Anonymous 2016-03-22 21:47

Go work on the solution of the Dubsless Prime Conjecture (DPC), it's your only chance to appear in the encyclopedia.

Name: Anonymous 2016-03-23 8:46

>>2
As an aside, repeating digits at the end of the number actually seem a lot easier to reason about to me - for example, the number of trailing zeros for a number n in base b is equal to the number of times you can fit b's prime factorization in n's prime factorization. Example: 1234772300000's prime factorization is {2, 2, 2, 2, 2, 5, 5, 5, 5, 5, 151, 81773}.

Name: Anonymous 2016-03-23 11:52

By expanding upon >>18, a number n has dubs in base k if and only if k2 divides n - k - 1.

Name: Anonymous 2016-03-23 11:54

Wait, k2 divides n - c(k - 1), for a 0 <= c < k. That's not so useful actually.

Name: Anonymous 2016-03-23 12:07

So a number n is dubsless if for all 1 < k < n - 1, and 0 <= c < k, k2 does not divide n - c(k - 1). Can it be shown a dubsless number is either 6 or prime?

If pb divides n for a b > 1, then n has dubs in base p since n will have b zeros at the end of its decimal expansion in base p. So dubsless numbers must be squarefree. That is, it's prime factorization consists of distinct primes.

Name: Anonymous 2016-03-23 17:29

I wrote a bit of code, including a decision procedure for dubs that doesn't compute its digit expansion. The way it works is, for a number n and a base b, either n has at least two trailing zeros, in which case it contains b's prime factorization twice, or it has two trailing symbols that aren't zeros, in which case it's k * (b + 1) away from a number with two or more trailing zeros for some k. This lets us test for dubs-ness in a base by knowing only its prime factorization and a few constants (k * (b + 1) for k between 1 and b - 1).

import Data.List hiding (union)
import Math.NumberTheory.Primes -- from package arithmoi

-- "multiset" operations

pull :: Eq a => a -> [a] -> Maybe [a]
pull x [] = Nothing
pull x (y:ys) | x == y = Just ys
pull x (y:ys) = pull x ys >>= \ys' -> return (y:ys')

union = (++)
difference [] ys = []
difference (x:xs) ys = case pull x ys of
Nothing -> x:(difference xs ys)
Just ys' -> difference xs ys'
intersection [] _ = []
intersection (x:xs) ys = case pull x ys of
Nothing -> intersection xs ys
Just ys' -> x:(intersection xs ys')
subset xs ys = xs `intersection` ys == xs

-- basic functions

unfactor [] = []
unfactor ((a, 0):xs) = unfactor xs
unfactor ((a, n):xs) = a:(unfactor ((a, n - 1):xs))

factorize = unfactor . factorise

-- digit expansion/prime numbers

expand :: Integer -> Integer -> [Integer]
expand n b = reverse (expand' n) where
expand' 0 = []
expand' n = (n `mod` b):(expand' (floor ((fromIntegral n) Prelude./ (fromIntegral b))))

contract :: [Integer] -> Integer -> Integer
contract ds b = contract' ds ((length ds) - 1) where
contract' [] _ = 0
contract' (d:ds) p = d * (b^p) + (contract' ds (p - 1))

rebase :: Integer -> Integer -> Integer -> Integer
rebase = undefined

decompose :: Integer -> (Integer, Integer)
decompose p = quotRem p 6

repeating :: Integer -> Integer -> Integer
repeating n b = ((genericLength . takeWhile (== (head e))) e) - 1 where
e = reverse (expand n b)

k n | n < 3 = []
k n = k' n (n - 2) where
k' n 2 =
let b = 2
ds = repeating n b
in if ds > 0
then [(b, ds)]
else []
k' n b =
let ds = repeating n b
in if ds > 0
then (b, ds):(k' n (b - 1))
else k' n (b - 1)

trailing b d n = e' !! 0 == d && e' !! 1 == d where
e' = (reverse . (flip expand) b) n

-- the trivial dubs decision procedure - it computes its digit expansion
dubs b n = e' !! 0 == e' !! 1 where
e' = (reverse . (flip expand) b) n

-- base 2

{- the set of dubs in base two is exactly the set that contains 3, those numbers
which contain {2, 2} in their prime factorization, and those numbers which
contain {2, 2} in their prime factorization when you substract 3 from them -}
dubs2_c' 3 = True
dubs2_c' n = [2, 2] `subset` (factorize n) ||
[2, 2] `subset` (factorize (n - 3))

-- base 3
dubs3_c' 4 = True
dubs3_c' 8 = True
dubs3_c' n = or [[3, 3] `subset` (factorize n),
[3, 3] `subset` (factorize (n - 4)),
[3, 3] `subset` (factorize (n - 8))]

-- base 4

{- This is wrong:
dubs4_c' 5 = True
dubs4_c' 10 = True
dubs4_c' 15 = True
dubs4_c' n = or [[4, 4] `subset` (factorize n),
[4, 4] `subset` (factorize (n - 5)),
[4, 4] `subset` (factorize (n - 10)),
[4, 4] `subset` (factorize (n - 15))] -}

dubs4_c' 5 = True
dubs4_c' 10 = True
dubs4_c' 15 = True
dubs4_c' n = dubs2_c' n ||
or [[2, 2] `subset` (factorize (n - 5)),
[2, 2] `subset` (factorize (n - 10)),
[2, 2] `subset` (factorize (n - 15))]

-- base 5

dubs5_c' 6 = True
dubs5_c' 12 = True
dubs5_c' 18 = True
dubs5_c' 24 = True
dubs5_c' n = or [[5, 5] `subset` (factorize n),
[5, 5] `subset` (factorize (n - 6)),
[5, 5] `subset` (factorize (n - 12)),
[5, 5] `subset` (factorize (n - 18)),
[5, 5] `subset` (factorize (n - 24))]

-- arbitrary base

-- for some digit d and base b, contract [d, d] b = d * (b + 1)

atomic b n 0 = False -- 0 is not dubs
atomic b n k = n - (k * (b + 1)) == 0 || atomic b n (k - 1)

-- an alternate dubs decision procedure
dubs' b n = atomic b n (b - 1) || dubs'' b n (b - 1) where
dubs'' b n 0 = ds `subset` (factorize n)
dubs'' b n k = ds `subset` (factorize (n - (k * (b + 1)))) || dubs'' b n (k - 1)
ds = (factorize b) `union` (factorize b)

-- base 10

zeros n = floor (fromIntegral ((length (filter (\p -> p == 2 || p == 5) (factorize n))) `div` 2))

Name: Anonymous 2016-03-23 17:41

>>22
Oops, I forgot to mention another case - if the number has a non-zero trailing digit, but no digits in front of it. These dubs I dub the ``elementary'' dubs, because they are constant and decidable for a base, and don't seem to decompose into anything - they are the numbers k * (b + 1). Also, I'm pretty sure the zeros procedure at the end is wrong. Also, check 'em.

Name: Anonymous 2016-03-23 18:26

From here we can conjecture that all dubs in a base b are of the form b^2 * n + (k * (b + 1)). For elementary/atomic dubs, n = 0 and k =/= 0, for trailing zeros, n =/= 0 and k = 0, and for trailing non-zeros, n =/= 0 and k =/= 0. I'll think about how I might prove this, but it sounds tricky.

Name: Anonymous 2016-03-23 20:13

>>21
if for all 1 < k < n - 1

You don't need to check until n-1, only until you find the base b in which n is represented as ducks (22), since the next bases the only dubs you can get are 11 which are in base n-1

for n = 10 you would stop at 4.
2 [1, 0, 1, 0]
3 [1, 0, 1]
4 [2, 2]
5 [2, 0]
6 [1, 4]
7 [1, 3]
8 [1, 2]
9 [1, 1]

that is 2b + 2 = n -> b = (n/2)-1, so you would check for all 1 < k < floor((n/2)-1)

Name: Anonymous 2016-03-23 20:24

>>22
*factorise

Name: Anonymous 2016-03-24 1:06

If x is dubsable by n, one of either (x + n + 1) or (x + 1) (if x + 1 are divisible by n) are also dubsable by n.

Maybe we could somehow use this fact to write a Dubs Sieve of Eratosthenes to efficiently calculate dubsless numbers.

Name: Anonymous 2016-03-24 1:12

>>27
What do you mean by dubsable?

Name: Anonymous 2016-03-24 16:47

Which one of you is responsible?
https://reddit.com/r/unexpectedfactorial

Name: Anonymous 2016-03-24 17:00

>>29
Are you trying to trigger someone?

Name: Anonymous 2016-03-24 17:50

>>30
Yea, I'm a trigga nigga.

Name: Anonymous 2016-03-24 18:18

>>30
Every time i hear this word i imagine a fat tumblr feminist trying to talk me down...
This is how you look to me.

Name: Anonymous 2016-03-24 18:29

>>32
Check 'em

Name: Anonymous 2016-03-24 18:44

>>32
That's how I imagine someone that uses Reddit.

Name: Anonymous 2016-03-25 22:23

>>32
Every time I hear a real war vet getting their PTSD triggered, my hate for Tumblristas blaspheming that notion skyrockets.

Name: Anonymous 2016-08-02 21:32

NEW DUBS THEORY FUNCTIONS:

- The least common dubs of two integers a and b, usually denoted by LCDubs(a, b), is the smallest positive integer that is dubs both in base a and base b.

- The greatest common dubs base of two integers a and b, usually denoted by GCDubsB(a, b), is the largest positive integer in which base both a and b are dubs.

Name: Anonymous 2017-02-02 18:55

>>1
Please add the sequence to OEIS

Name: Anonymous 2017-02-04 1:22

>>13
2017, March: Donald Trump pledges to build the biggest ever particle accelerator, to test the predictions of the Dubs Theory.

We have less than two months to make this happen!

Name: Anonymous 2017-02-04 23:18

is 6 the only non prime dubsless number?

I didn't spend a lot of time writing and reviewing this, so there might be some errors or strange parts:

DEFINITIONS:

nⒹ Means the number n has dubs in some non-trivial base (any base other than 1 or n-1).

nⒹb Means the number n has dubs in base b.

nⒹd Means the number n has dubs ending in digit d in some base.

nⒹdb Means the number n has dubs in base b, and the dubs digits are d.

Concrete example: 35Ⓓ56 Means 35 in base 6 has dubs ending in digit 5.

THEOREMS:

The square dubs theorem: ∀x∈ℕ(x20x) or every squared natural number n2 represented in base n has dubs ending in digit 0. The proof of this simple theorem is left to the reader.

The first d-digit dubs theorem: (d+1)2-1 is the smallest number n which verifies nⒹd, the base that verifies that is d+1. Also n is a multiple of d. For example for digit 1, the first number that has dubs ending in 1, is (1+1)2-1=3 (11 in base 2). The proof of this simple theorem is also left to the reader.

The little composite dubs theorem: if we add integer multiples of d (d*i) to (d+1)2-1 (last theorem), the resulting number n = (d+1)2-1+d*i still verifies nⒹd, concretely nⒹdd+1+i. Also n continues to be a multiple of d. In other words, every multiple of d equal or greater than (d+1)2-1 has dubs in some base. Then, if a composite number c has a prime factor p for which c>=(p+1)2-1, c must have dubs.

The composite dubs theorem: If there were any other dubsless composite numbers c other than 6 they should be in some interval between (pi+1)2-1 and (pi+1+1)2-1 (both dubs, because last theorem), where pi and pi+1 are two consecutive primes. The lower bound means that if c has a prime factor p less or equal than pi, then c must has dubs (last theorem), and the upper bound can be rewritten as pi+1*(pi+1+2), which is the product of two consecutive odd numbers (in some cases both may be prime). With those restrictions, there are 3 choices for the prime factors of c:

- greater than pi+12: pi+1*(pi+1+2) this only works when (pi+1+2) is also a prime number and is the only choice because 2 is the minimum distance between two consecutive primes (excluding 2 and 3, we treat the interval defined by these two primes as a special case). But because last theorem shows us that the rewritten form of pi+1*(pi+1+2) has dubs, this choice doesn't give dubsless numbers.

- equal to pi+12: because the square dubs theorem this choice neither gives dubsless numbers.

- less than pi+12: then some prime factor p has to be less or equal than pi, so because of the lower bound, this choice also gives dubs numbers.

This proves that no other composite number than 6 can be a dubsless number. And 6 is dubsless because the distance between the primes 2 and 3 is only 1, which doesn't happen for any other pair of primes.

Name: Anonymous 2017-02-05 17:40

Can к(n) be computed in sub-linear running time?

Name: Anonymous 2017-02-05 18:19

>>39
Candidate for the fundamental theorem of dubs: ``Every composite number other than 6 has dubs in some base.''

Name: Anonymous 2017-02-07 9:50

>>39
You have a bright future ahead of you, o wise dubsman.

Name: Anonymous 2017-02-07 10:45

>>39
TED talk pls

Name: Anonymous 2017-02-07 19:16

check em

Name: Anonymous 2017-02-08 18:35

Name: Anonymous 2017-02-08 21:26

In 2016, Lord Dubs sponsored an amendment to the Immigration Act 2016 to offer unaccompanied refugee children safe passage to Britain amidst the European migrant crisis.
Shalom!

Name: Anonymous 2017-02-08 21:38

check those dubs before you wreck those dubs

Name: Anonymous 2017-02-26 13:53

age

Name: Anonymous 2017-02-26 14:12

>>46
So Lord Dubs is a cuck?

Name: Anonymous 2017-02-26 17:04

Dubs factorization and public key crypto based on dubs.

Name: Anonymous 2017-02-26 22:11

>>49
Of course. To be human is to be cuck.

Name: Anonymous 2017-02-27 1:40

>>39
so by extending this a prime number is a number that doesn't have 2-digit dubs in any trivial base with the exception of 6. This way we can redefine primes based on dubs !

Name: Anonymous 2017-02-27 2:00

>>39
>>52
and also if a number has 2-digit dubs of digit d in some base, it also has dubs of the digits of the prime factorization of d in some bases. For example the number 66 in base 10 is also 33 and 22 in some other bases.

Name: Anonymous 2017-02-27 2:09

>>52
then dubsless numbers can be considered prime numbers with more restrictions (excluding 6):

2-digit dubsless: prime number
3-digit dubsless: subset of prime numbers (``more pure primes''?)
4-digit dubsless: subset of 3-digit dubsless
.
.
.
dubsless: maximum restrictions

Name: Anonymous 2017-02-27 2:23

Name: Anonymous 2017-02-27 3:14

>>55

Name: Anonymous 2017-06-05 0:50

bamp. Where are all the dubs enthusiasts?

Name: Anonymous 2017-06-05 0:59

What programming language is this?

Name: Anonymous 2017-06-05 17:26

dubs programming language

Name: Anonymous 2018-01-12 18:18

bamp

Name: Anonymous 2018-01-13 9:39

Back to 4chan, please.

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