[QUOTE=CRGreathouse;252199]Can you prove that it behaves the way you expect mod 7? Mod 3? Mod 4?[/QUOTE]

[CODE]for(n=1,100,for(x=1,#MeVec,print1(M(x,n)","));print("::"n))[/CODE]

the code I showed to prime95 was :

[CODE](11:01)>for(n=1,#mersenne1,print((2^mersenne1[n]-1)%n))
0
1
1
3
1
1
1
7
1
1
6
7
5
1
7
15
1
1
12
7
10
1
15
7
1
1
1
3
26
1
1
31
7
7
31
31
31
31
1
7[/CODE]

[QUOTE=CRGreathouse;252199]Can you prove that it behaves the way you expect mod 7? Mod 3? Mod 4?[/QUOTE]

what the valuation code if so no not really, I can't even find a use of valuation because I have no idea what it does.

it can't act modulo( checked with second argument as 3) so I'm stumped, oh never mind I think I realize which valuation it's talking of after going through :

[url]http://en.wikipedia.org/wiki/Valuation_(mathematics[/url])

[QUOTE=CRGreathouse;252199]Can you prove that it behaves the way you expect mod 7? Mod 3? Mod 4?[/QUOTE]

[CODE]0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,::3
3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,::4
3,0,3,1,1,3,1,1,1,3,3,1,3,1,1,1,1,1,3,1,3,3,3,3,3,1,1,3,1,1,1,3,3,1,3,3,3,3,1,::7
3,7,31,43,43,31,43,43,43,31,31,43,31,43,43,43,43,43,31,43,31,31,31,31,31,43,43,31,43,43,43,31,31,43,31,31,31,31,43,::84[/CODE]

84 is a multiple of 3,4 and 7 so it's a matter of proving 31 and 43 align with the modulo 3,4 and 7 need according to this. 31%3 == 1 that aligns 43%3== 1 that also aligns, 31%4==3 that aligns and 43%4==3 so that also aligns, 31%7 =3 that aligns, 43%7 = 1 this also can align with the modulo so it is in fact plausible that this is true. Plausible is not a proof I know that. The proof needs to be more rigorous, if it's true maybe we can eliminate more checks when combined with 24x+7. though we know what x are mersenne numbers(it doesn't look to help though).

[CODE]minimumprime(x) = if((x-2)%2==1,print(6*((x-2)-floor((x-2)*.5))-1),print(6*floor((.5*(x-2)))+1))[/CODE]

just thought I'd put this up.

[QUOTE=science_man_88;252271][CODE]minimumprime(x) = if((x-2)%2==1,print(6*((x-2)-floor((x-2)*.5))-1),print(6*floor((.5*(x-2)))+1))[/CODE]

just thought I'd put this up.[/QUOTE]

and one thing I just learned by playing alt + 1(numpad) gets you back to the start of text and yes I get it's the control character start of header. something tells me if i every make long scripts that act as one line in PARI I'll be able to use it.

[QUOTE=science_man_88;252271][CODE]minimumprime(x) = if((x-2)%2==1,print(6*((x-2)-floor((x-2)*.5))-1),print(6*floor((.5*(x-2)))+1))[/CODE]

just thought I'd put this up.[/QUOTE]

Don't print, just return! That way you can actually use the value instead of merely looking at it.

[QUOTE=CRGreathouse;252305]Don't print, just return! That way you can actually use the value instead of merely looking at it.[/QUOTE]

I changed it to returning it the hard part now is I was trying to make a script that could use that value to get the real answer because by x = 1000 it's off by about 5000:

[CODE]v=vector(13,n,minimumprime(n));v[1]=2;v[2]=3;c=[];a=0;for(x=1,#v,for(y=1,x-1,if(v[x]%v[y]==0,a=a+1));if(!a>1,c=concat(c,[v[x]])))[/CODE] is my best attempt so far, but anyways I'm kinda repeating the same task I know.

I don't know what the function is supposed to return, nor by what measure it's off by 5000.

[QUOTE=CRGreathouse;252307]I don't know what the function is supposed to return, nor by what measure it's off by 5000.[/QUOTE]

[CODE](21:44)>prime(1000)
%38 = 7919
(22:00)>minimumprime(1000)
%39 = 2995[/CODE]

[QUOTE=science_man_88;252310][CODE](21:44)>prime(1000)
%38 = 7919
(22:00)>minimumprime(1000)
%39 = 2995[/CODE][/QUOTE]

Are you trying to have it return the n-th prime? Why not use prime() in that case? :confused:

[QUOTE=CRGreathouse;252313]Are you trying to have it return the n-th prime? Why not use prime() in that case? :confused:[/QUOTE]

because I want to try and make one that works without that so it's not limited by primelimit.