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[00:39] i got some neat results
[00:39] even just in mircscript
[00:39] hehe
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[01:20] i should say that i typoed too much and actually had bad results :(
[01:22] huh.
[01:22] no results
[01:25] i wonder if that is some interesting property of primes
[02:13] 3/4 is the best it'll get from 1-65536
[02:13] pretty interesting
[02:14] i'm fairly certain this method is testing all possibilities too, even though it's only picking two first and then a third later
[02:14] thinking about it, even if you picked three all at once, any two of them would have to all come to correct values on their own
[02:19] but some of you have perfect solutions for N=4
[02:19] which means that either i'm missing something
[02:19] or 'perfect' is less than the total combinations of (N-1) constants
[06:21] perfect is less than the total combinations
[06:41] so it would seem
[06:41] i wonder why, though
[06:41] it seems like there must be some basis in math for the fact that a+b and a-b being in the list seem mutually exclusive
[06:42] it's probably obvious too
[06:45] lol
[06:45] like the fact that
[06:45] were there any combinations like that, they would not be prime
[06:45] 'cause the constant separation is linear, and you won't have two primes along the same linear scale
[06:52] i worked it out in my head ;)
[06:52] a-s is the common factor
[06:52] er wait
[06:52] that's multiplication >:(
[06:53] fators, i mean
[06:53] n, n+step, and n+2step; these cannot all be prime
[06:54] these words are lies
[06:54] i don't know what i'm talking about
[06:54] it seems obvious but i can't define why
[06:55] unless step is a multiple of 3, then either n, n+step or n+2*step will be a multiple of 3
[06:56] that's it
[06:56] step cannot be a multiple of 3 then? i wonder why
[06:58] nevermind; all the steps are multiples of 3 it seems
[06:58] which is sensible
[06:59] but then, abstracting it one level fails
[06:59] i must have messed up somewhere.. trying to work this out in notepad sucks, i need a pencil and paper
[07:01] ah, i left a level out
[07:09] can't get it in my head
[07:09] too late at night perhaps
[07:09] i'll try tomorrow ;P
[07:09] I've just settled for the brute force method :-)
[07:09] hee.
[07:09] i want to know why i can't have a, b, a+b, a-b
[07:10] You can, providing b is 2
[07:10] but 2 doesn't have primes on either side of it
[07:10] i've got a list of about 4000 numbers that have primes +-105 from them
[07:11] a=5, b=2
[07:11] but i can't choose a and b from that list such that a, b, a+b, a-b are all within that list
[07:11] 3/4 is all i'll get
[07:12] Either a, b, a+b or a-b will be non-prime unless a=5 and b=2
[07:12] that's ok, none of the list are prime
[07:12] the list are numbers between pairs of primes
[07:13] there will be a multiple of 2 and a multiple of 3
[07:13] Oh okay
[07:13] for example: 156 338 494
[07:13] all possibilites are 156 338 494 832 4230 4568 4724 4906 5062 5218 5400 5556 5894
[07:14] any of those numbers +-105 is a prime
[07:14] er
[07:14] I've finished implementing my brute for program in assembly language :-)
[07:14] 7 of them are
[07:14] D:
[07:14] i can't get at 'any'
[07:14] hehe
[07:14] oh, smarty pants
[07:14] you know i have a thought about the relationship of the numbers too
[07:15] i think each number will be > the sum of all the others for best results
[07:15] Now I can get the same rubbish results 1000 times as fast :-/
[07:15] err, the sum of the previous numbers, assuming they are ordered low to high
[07:15] it's like binary
[07:15] with 4 digits you can have up to 16 unique values
[07:15] the digits are just worth something other than 2
[07:16] (and they have to have primes an odd number away ;)
[07:16] 3?
[07:16] ?
[07:16] You can either add a number, subtract a number or do neither
[07:17] that's three options, so maybe it's base 3?
[07:17] i can't fit that in my head
[07:17] :P
[07:17] i was just thinking about addition when that came to mind
[07:17] it seems like it might still hold but it's not accurate
[07:18] i don't know any numbering system where the digits represent either -place, 0, or +place
[07:18] hehe
[07:18] i wrote a script to convert things to negabinary though once!
[07:18] base -2
[07:18] shit is crazy
[07:18] normal base 3 should work
[07:18] but that's
[07:18] 0, place value, or 2* place value
[07:18] instead of using the digit 2, you'd be using -1
[07:19] at least for it to hold to my imaginary distribution of values hehe
[07:19] Imaging the values 1,3,9,27,81
[07:19] that works for the base 2 analogy... there are no overlapping sums
[07:19] but with subtraction i just don't know
[07:19] i guess actually it makes sense
[07:19] If you add, subtract or do nothing with each number, you can make every value up to 121 without any overlap
[07:20] you just want the 2nd place in the negative to be greater than the sum of all the ones before it
[07:20] Having no overlap is important. If there's any overlap in the combinations, you reduce the number of potential primes
[07:20] exactly
[07:20] the digits are going to increase more than that example though of course
[07:21] Yes
[07:31] so i guess then that each digit would be > twice the sum of the previous ones
[07:32] 'cause it could subtract all of them to reach that sum
[07:32] Yes :-)
[07:32] i don't know if it is best to start with a small odd number
[07:32] (maybe the best odd number is big)
[07:33] but it seems like a simple progression if you start with your odd number, then choose an even number higher than it and proceed from there
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[10:25] Hi Fiveop
[10:26] 3hi
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