>
> > why is f(0) in an ll test = 4
>
> It is, just accept it...
I think the Lehmer sequence is L(4,1) and f(n)=v_{2^n}, the should at least be correct
in
an indicative way - you should verify too.
>
> > also when i went to sshool (i'm 34) 7 mod 8 == 7
> > i have seen 7 mod 8 == -1
> > is -1 == 7 mod 8
> > which is more correct
>
> -1 + 8 = 7
>
> so 7 mod 8 = 7 or -1 depends on how you look at it...
>
Or also 7 = 23 mod 8, etc.
They are all literally the same - that's why you speak of an equivalence class.
Now in order to work with equivalence classes you like to chose predilectional
representatives. The most obvious ways to do this for the modulus m are:
A. Choose a representative 0<=a<m
B. Choose a representative -m/2<=a<m/2 (minimize the abolsute value of the
representatives
In your case -1 and 7 are the choices B. and A. for the same equivalence class.
Preda
>
> > cant belive i spelled school sshool
>
> cant believe you spelled believe belive.. ;)
>
> Sorry, couldn't resist...
>
> Otto
>
> _________________________________________________________________
> Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
> Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
