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Old 2006-01-13, 13:59   #1
Greenbank
 
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Default Number of octoproths per n

All ranges checked from k=1 to 2^n (where 2^n-k would be 0).

All primes verified with PARI 2.1.7.

n<=26 op=0
n=27 op=1
n=28 op=2
n=29 op=1
n=30 op=1
n=31 op=2
n=32 op=6
n=33 op=2
n=34 op=13
n=35 op=26
n=36 op=11
n=37 op=92
n=38 op=28
n=39 op=83
n=40 op=331
n=41 op=110
n=42 op=453
n=43 op=632
n=44 op=1297
n=45 op=2129
n=46 op=5017
n=47 op=5278
n=48 op=3979
n=49 op=56905
n=50 op=18547
n=51 op=16870
n=52 op=219117
n=53 op=60620
n=54 op=230143
n=55 op=786971
n=56 op=285415

Full list of Octoproths in completed ranges (up to n=54):
Raw text: http://octoproth.greenbank.org/downl...o_complete.txt 12024KB (12MB!)
ZIP: http://octoproth.greenbank.org/downl...o_complete.zip 4922KB (4.9MB)

TODO - Download link for user discovered Octoproths (with attributions)

Last fiddled with by Greenbank on 2006-01-20 at 16:04
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Old 2006-01-13, 15:36   #2
R. Gerbicz
 
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"Robert Gerbicz"
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Very good work!

It is very interesting that we can predict the total number of octoproths for a given n!!! I've worked out it today: by modifying some very hard conjectures, first I define for every n value the "weight" of n: in (PARI):
Code:
w(n)=T=128.0;forprime(p=3,10^4,l=listcreate(8);g=Mod(2,p)^n;h=1/g;a=[g,-g,h,-h,2*g,-2*g,h/2,-h/2];\
a=lift(a);for(i=1,8,listput(l,a[i],i));l=listsort(l,1);T*=(1-length(l)/p)/(1-1/p)^8);return(T)
Then using it we can predict the total number of octoproths for a given n value by:
Code:
f(n)=floor(w(n)*2^n/(n*log(2))^8*1/16)
Try it!
For n=51 it gives that f(n)=16537 It is a very good approximation because Greenbank has calculated that the true number is 16870
ps you'll need also w() to use f()
Note that in w() the w(n) is also a prediction because it is using primes up to 10^4 ( to become faster the computation)

Last fiddled with by R. Gerbicz on 2006-01-13 at 15:39
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Old 2006-01-13, 16:00   #3
Greenbank
 
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Nice work.

f(49) = 55410

real count is 56905.
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Old 2006-01-14, 00:08   #4
fetofs
 
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Quote:
Originally Posted by R. Gerbicz
Very good work!

It is very interesting that we can predict the total number of octoproths for a given n!!! I've worked out it today: by modifying some very hard conjectures, first I define for every n value the "weight" of n: in (PARI):
Code:
w(n)=T=128.0;forprime(p=3,10^4,l=listcreate(8);g=Mod(2,p)^n;h=1/g;a=[g,-g,h,-h,2*g,-2*g,h/2,-h/2];\
a=lift(a);for(i=1,8,listput(l,a[i],i));l=listsort(l,1);T*=(1-length(l)/p)/(1-1/p)^8);return(T)
Then using it we can predict the total number of octoproths for a given n value by:
Code:
f(n)=floor(w(n)*2^n/(n*log(2))^8*1/16)
Try it!
For n=51 it gives that f(n)=16537 It is a very good approximation because Greenbank has calculated that the true number is 16870
ps you'll need also w() to use f()
Note that in w() the w(n) is also a prediction because it is using primes up to 10^4 ( to become faster the computation)
This may be stupid, but how do I enter the script on PARI (not the hard way, please)
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Old 2006-01-14, 01:10   #5
R. Gerbicz
 
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Quote:
Originally Posted by fetofs
This may be stupid, but how do I enter the script on PARI (not the hard way, please)
I know only one way at the end of line type \
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Old 2006-01-14, 12:15   #6
robert44444uk
 
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Default 55

Greenbank

Will take on n=55

Regards Robert Smith
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Old 2006-01-14, 17:44   #7
fetofs
 
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Quote:
Originally Posted by R. Gerbicz
I know only one way at the end of line type \
Typing IS the hard way!
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Old 2006-01-15, 23:08   #8
robert44444uk
 
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Default 55 almost there

Greenbank

I have done 55 up to 3e15, so a little more to do overnight. The file is enormous, how do I get it to you? Maybe you can send me a private message with your email address and I will send.

Regards

Robert Smith
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Old 2006-01-16, 08:41   #9
Greenbank
 
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2^55 = 3.6E16

So if you've done to 3e15 then you've only done 8.3%
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Old 2006-01-16, 10:01   #10
R. Gerbicz
 
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Quote:
Originally Posted by Greenbank
2^55 = 3.6E16

So if you've done to 3e15 then you've only done 8.3%
No! 2^55>3.6E16
You have to use a larger number than -1+2^n, if you want to search the full range for n.
I think Robert444444uk correctly search this interval for n=55, but he mistyped here ( just seeing previous submissions from him ), 3e15 isn't a large range for him.
And you can also use my formula to give a prediction for the total number.
Using my formula it'll be about: f(55)=772430 ( if we are using rounding instead of floor )
So the size of the file will be larger than the size of all previous files altogether ( up to n=54 )
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Old 2006-01-16, 11:09   #11
Greenbank
 
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Yeah, I meant to say 2^55 ~ 3.6E16. I was just trying to point out that 3E15 was an order of magnitude out.

2^55 = 36028797018963968
3.6E16 = 36000000000000000

robert44444uk, I'll send you an email about getting the n=55 stuff.
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