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IF TRIPS

Name: Anonymous 2013-06-04 15:05

I WILL KILL YOU

Name: Anonymous 2013-06-04 21:23

JACKSON 112 GET

Name: Anonymous 2013-06-04 21:24

tail dubs

Name: Anonymous 2013-06-04 21:24

consecutive number get

Name: Anonymous 2013-06-04 21:25

2n sequence get

Name: Anonymous 2013-06-04 21:25

JACKSON 53 GET

Name: Anonymous 2013-06-04 21:26

Even number get

Name: Anonymous 2013-06-04 21:26

Odd number get

Name: Anonymous 2013-06-04 21:27

27 GET

Name: Anonymous 2013-06-04 21:28

(* 3 43) GET

Name: Anonymous 2013-06-04 21:29

lE0 = 1 GET

Name: Anonymous 2013-06-04 21:30

Palindromic Prime GET

Name: Anonymous 2013-06-04 21:32

Between Two GETS GET

Name: Anonymous 2013-06-04 21:32

dubs GET

Name: 2 2013-06-04 21:33

Translator's Note - Keikaku means plan.

Name: Anonymous 2013-06-04 21:36

Imagine a historian in the year 3000, looking through the old internet archives to try to learn of our culture. Then she stumbles upon this thread. What on earth could she possibly imagine that might explain what has transpired in this thread in the last six hours.

Name: Anonymous 2013-06-04 22:11

she
She is free to go and suck a nigger dick.

Name: Anonymous 2013-06-04 22:27

>>136
Niggers are eradicated in the Third Great Race War of 2015.

Name: Anonymous 2013-06-04 22:28

>>137
And so were womyn.

Name: Anonymous 2013-06-04 22:45

>>137
Jews were the only race left. Truly the best war the world ever had.

Name: Anonymous 2013-06-04 22:47

>>111
Still waiting, OP.

Name: Anonymous 2013-06-05 1:25

A natural number i has dubs in basen if the last two digits are the same. The dubs predicate in scheme: (lambda (i n) (= (remainder i n) (remainder (floor (/ i n)) n)))
For every natural number i, where i > 2, there exists ``trivial dubs'' xxx..11 in basei-1
A natural number i is a ``dubless number'' if it only has trivial dubs. In other words, it doesn't have dubs in any other base than basei-1

Bateman conjecture:
The set of dubless numbers is infinite

The first 100 dubless numbers:
{5 6 29 41 61 113 317 509 569 761 797 1213 1229 2617 5297 6221 8017 12457 14621 16889 21773 23117 23357 25657 30313 30713 30809 32717 33721 38329 46073 52817 59113 61729 67961 69809 72973 73121 78797 96797 98017 102497 103613 118913 120209 140317 153757 169241 182489 204797 210097 224669 225529 226397 241229 246557 249521 256721 259517 266009 297469 313129 317921 338153 352753 385129 401329 420221 425309 429361 437113 465809 486329 488617 490169 499129 511961 550541 568441 570173 588173 591089 597353 602717 615313 661217 706973 709117 734221 761357 815897 818173 821489 841213 875129 886429 891929 929953 934057 942521}

Name: Anonymous 2013-06-05 1:39

>>141
I really wish we could expand dubs theory, and maybe even co-publish a book on it, but everything about it collapses into trivialities. We need to think of some harder dubs problems.

Name: Anonymous 2013-06-05 2:01

For fun:
(define (bases i)
    (define (base-iter l i n)
      (cond ((> i (- n 2)) l)
            (else (if (= (remainder n i) (remainder (floor (/ n i)) i)) (base-iter (cons i l) (+ i 1) n) (base-iter l (+ i 1) n)))))
    (base-iter '() 2 i))

(define (dubs n)
  (display (foldr string-append "" (list "[b]" (number->string n) "[/b] has")))
  (let ((base-list (bases n)))
    (cond ((empty? base-list) (display " no dubs."))
          (else (begin (display " dubs ") (print-dubs base-list n))))))

(define (print-dubs l n)
  (cond ((empty? l) (void))
        (else (let ((punctuation (cond ((empty? (cdr l)) ".")
                                       ((empty? (cddr l)) " and ")
                                       (else ", ")))
                    (dub-digit (number->string (remainder n (car l)))))
                (display (foldr string-append "" (list dub-digit "|" dub-digit " in base[sub]" (number->string (car l)) "[/sub]" punctuation))))
              (print-dubs (cdr l) n))))


>>141
141 has dubs 3|3 in base46 and 6|6 in base9.
>>142
142 has dubs 2|2 in base70.
>>143
143 has dubs 11|11 in base12, 5|5 in base6, 3|3 in base5, 3|3 in base4, 2|2 in base3 and 1|1 in base2.

Name: Anonymous 2013-06-05 2:27

>>143
Neat. I bet we can make some claims about the dubs in base n sequences. Also, check em base 11 dubs.

Name: Anonymous 2013-06-05 2:28

/dubs/ challenge: find the natural number that has dubs 11/11 in base 11

Name: Anonymous 2013-06-05 2:29

>>145
Sadly it would be 10|10 only.

Name: Anonymous 2013-06-05 2:34

Imena base 10. *commits soduko*

Name: Anonymous 2013-06-05 2:37

144 is a very dub rich number.
144 has dubs 2|2 in base71, 3|3 in base47, 4|4 in base35, 6|6 in base23, 8|8 in base17, 9|9 in base15, 0|0 in base12, 4|4 in base10, 0|0 in base6, 0|0 in base4, 0|0 in base3 and 0|0 in base2.

Name: Anonymous 2013-06-05 2:37

>>142
I'm sure it could be useful for cryptography.

Name: Anonymous 2013-06-05 2:39

Could we assign a dubs coefficient for how dubs rich a number is?

Name: Anonymous 2013-06-05 3:05

>>144
Fun, but trivial, property:

If a number i has dubs 0|0 in basen then i has dubs 0|0 in basem, where m is a factor of n

Name: Anonymous 2013-06-05 3:08

>>150
yes.

Let ℡(n) denote the dubs coefficient of n, defined as the size of the set, { b | n has dubs in base b }.

Name: Anonymous 2013-06-05 3:11

>>149

It would be so funny if we actually produced a strong encryption algorithm based on dubs. Even funnier if it gained adoption.

Name: Anonymous 2013-06-05 3:39

k-dubs coefficient of n (k-℡(n)): denotes the number of k|k dubs of n.

Name: Anonymous 2013-06-05 3:54

>>150-154
check em

Name: Anonymous 2013-06-05 6:36

>>142
dubs in complex base systems

Name: Anonymous 2013-06-05 17:01

In any case, it gives a whole new meaning to checking 'em.

Name: Anonymous 2013-06-05 17:09

>>149
OR PROBABLY JUST STEGANOGRAPHY

Name: Anonymous 2013-06-05 17:50

We need to draft the DubsCoin protocol.

Name: Anonymous 2013-06-06 0:19

The dubs chain of an natural number n is a finite length sequence:

<s1, s2, s3, ..., sm>

where n has dubs si|s[sub]i/[sub] in base bi, with bi < bj when i < j. That is, the dubs chain is the finite list of dubs obtained by n, sorted in increasing order by the base. The ``base set'' of a natural number n is set of all bases in which n has dubs.

The ``dubs coefficient of n'', denoted ℡(n), is defined as the size of it's base set.



For example, 144 has dubs 2|2 in base71, 3|3 in base47, 4|4 in base35, 6|6 in base23, 8|8 in base17, 9|9 in base15, 0|0 in base12, 4|4 in base10, 0|0 in base6, 0|0 in base4, 0|0 in base3 and 0|0 in base2.

144 has dubs chain <0, 0, 0, 0, 4, 0, 9, 8, 6, 4, 3, 2>.
The base set of 144 is {2, 3, 4, 6, 10, 12, 15, 17, 23, 35, 47, 71},
and ℡(144) = 12.


A natural number n is a ``perfect dubs of order k'' if n has a dubs chain of <1, 2, 3, ..., k>.

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