i break the record for the most successive dubs ever gotten
suck my dick goyim
Name:
Anonymous2013-06-04 21:23
JACKSON 112 GET
Name:
Anonymous2013-06-04 21:24
tail dubs
Name:
Anonymous2013-06-04 21:24
consecutive number get
Name:
Anonymous2013-06-04 21:25
2n sequence get
Name:
Anonymous2013-06-04 21:25
JACKSON 53 GET
Name:
Anonymous2013-06-04 21:26
Even number get
Name:
Anonymous2013-06-04 21:26
Odd number get
Name:
Anonymous2013-06-04 21:27
27 GET
Name:
Anonymous2013-06-04 21:28
(* 3 43) GET
Name:
Anonymous2013-06-04 21:29
lE0 = 1 GET
Name:
Anonymous2013-06-04 21:30
Palindromic Prime GET
Name:
Anonymous2013-06-04 21:32
Between Two GETS GET
Name:
Anonymous2013-06-04 21:32
dubs GET
Name:
22013-06-04 21:33
Translator's Note - Keikaku means plan.
Name:
Anonymous2013-06-04 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:
Anonymous2013-06-04 22:11
she
She is free to go and suck a nigger dick.
Name:
Anonymous2013-06-04 22:27
>>136
Niggers are eradicated in the Third Great Race War of 2015.
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
>>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:
Anonymous2013-06-05 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))
>>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:
Anonymous2013-06-05 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:
Anonymous2013-06-05 2:28
/dubs/ challenge: find the natural number that has dubs 11/11 in base 11
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:
Anonymous2013-06-05 2:37
>>142
I'm sure it could be useful for cryptography.
Name:
Anonymous2013-06-05 2:39
Could we assign a dubs coefficient for how dubs rich a number is?
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>.
nikita would like dubs theory because it is all finite so far.
Name:
Anonymous2013-06-06 1:27
A twin dub is a dub number that differs from another dub number by two in any base, for example the twin dub pair in base10 (99, 100). Sometimes the term twin dub is used for a pair of twin dubs; an alternative name for this is dub twin or dub pair.
The ``twin dubs of order k in base b'', denoted ㎎(k,b), where k is a natural number {1,2,3...∞} and b the base.
㎎(k,b) = {b²*k-1,b²*k}.
Example: twin dubs of order 7 in base8.
mg(7,8) = {8²*7-1,8²*7} = {447,448}
Name:
Anonymous2013-06-06 1:29
>>162
Where I said it differs by two in any base, I meant by 1 (ONE).
>>160
The dubs coefficient doesn't need special notation. If B denotes the base set of n, then the dubs coefficient is |B| or card(B).
We also need to agree on nomenclature. Instead of ``dub number", my suggestion is ``dubb", and instead of ``twin dub [number]", it should be ``twin dubb" etc.