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 FFR4EVA_00 10-24-2011 07:27 AM

i'm seriously advocating that the quickest method would be starting at the set for Co(100) and drawing each Co(n+1) from Co(n)

 FFR4EVA_00 10-24-2011 11:26 AM

it's predictable though
basically it comes down to an iterative process
i think the hardest part would be how to arrange the arrays for optimal speed
i was thinking that you would basically set it up like this:
[a_2, a_3, a_5, a_7, ..., a_p]
p is the largest prime below n, and the a_i are not necessarily distinct, so that when you actually wanted to find Co(n) you'd remove duplicates and add 1

 FFR4EVA_00 10-24-2011 02:29 PM

 FFR4EVA_00 10-24-2011 02:50 PM

like i said: the set doesn't have to be distinct in this arrangement
i could easily just leave empty spaces but hey, programming doesn't really like those
and 16 only has one factor i have no idea what you're talking about
what's going on with 21 is this: if a < b < c, then ac+b > ab+c
i just threw in the snippet about testing 20 because i figured a program would try it

 LongGone 10-24-2011 04:26 PM

 FFR4EVA_00 10-24-2011 04:58 PM

longgone is correct on 1 and 2
here's what you're confused about if i am right:

Quote:
 [a_2, a_3, a_5, a_7, ..., a_p] p is the largest prime below n, and the a_i are not necessarily distinct, so that when you actually wanted to find Co(n) you'd remove duplicates and add 1
Quote:
 20 gets a bit more complex 20 drags in 16 and 15, but 15 is also a multiple of 3 it quickly becomes apparent that the only thing one could do with the 3 is use 9 20+9 < 16+15, don't insert 20 : [16, 15, 15, 7, 11, 13, 17, 19] 21+5 > 15+7, insert [16, 21, 5, 21, 11, 13, 17, 19] 20 < 16+5, don't attempt

 Knut.Angstrom 10-24-2011 05:18 PM

Quote:
 Originally Posted by Reincarnate (Post 3556057) A step-by-step runthrough of the Co(100) process:
Your sum 1355 is not correct since you use 99 instead of 88 giving 1356

 Knut.Angstrom 10-24-2011 05:24 PM

1+64+81+25+49+11+13+17+19+23+29+31+37+41+43+47+53+59+61+67+71+73+79+83+89+97+
(95-25-19)+(91-49-13)+(88-64-11) =1356

 fido123 10-24-2011 05:25 PM

Worked on this for a little bit one night. Did 1, 2, and 6 all in C:

Question 1:

Question 2:

Question 6:

Easy stuff so far but I'm not really all that knowledgeable about math at all so I'm not sure how far I'll get with this lol.

 FFR4EVA_00 10-24-2011 05:28 PM

Quote:
 Originally Posted by Reincarnate (Post 3556421) Okay but what about needing to retrieve something lost? For instance, the solution to Co(100) includes a 17, which gets removed back in Co(51)

Quote:
 Originally Posted by Reincarnate (Post 3556421) Also some numbers will have more than 1 unique prime factor
i heard whispers that this is true but i am trying to rigorously disprove it

 FFR4EVA_00 10-24-2011 05:56 PM

yeah, i never said what those were, i'm sorry
a_p is just the number that is divisible by p in whatever set we're looking at (since we're supposed to be looking at coprime sets there is exactly one a_p for every p but not all a_p are unique)
the reason why i'm duplicating the terms is that i think it will be easier for a computer to traverse

 LongGone 10-24-2011 06:18 PM

Quote:
 Originally Posted by Reincarnate (Post 3556421) Also some numbers will have more than 1 unique prime factors (e.g. 30 -> unique primes 2, 3, and 5)

 Reincarnate 10-24-2011 06:22 PM

Quote:
 Originally Posted by LongGone (Post 3556451)
I think this is a reasonable assumption to make

 FFR4EVA_00 10-24-2011 07:04 PM

@Reincarnate- that is correct, yes
now i'm messing around with a brute-force solution of sorts though, NOTHING like what i was doing before

 FFR4EVA_00 10-24-2011 08:42 PM

 Reincarnate 10-24-2011 09:19 PM

congrats FFR4, nicely done

 stargroup100 10-25-2011 01:44 AM

http://img268.imageshack.us/img268/5532/eulero.jpg

I got 0 eulerian points for this. mad

 iironiic 10-25-2011 09:53 AM

Got #71 with pencil and paper xD

EDIT: Nice FFREva!

 FFR4EVA_00 10-25-2011 10:41 AM

http://projecteuler.net/eulerians
"For the twelve most recent problems the difference, d, in the length of time to solve the problem (in minutes) between each member and the slowest in the table is calculated and log2(max(d,2)) points are awarded."
so every time someone solves the problem, the points of everyone who already solved it go up
in other words cosmovibe has 8 points now

 Reincarnate 10-25-2011 11:58 AM