[seqfan] Re: An algorithm of calculation of multiplicative order of 2 mod n

[Vladimir Shevelev]

Of course, one can calculate ord_n(a) by this algorithm ((n,a)=1).
Examples: 
to calculate ord_10(3) we have
10+1   0  11
10+11 1    7
10+7   0  17
10+17 3    1
 Thus ord_10(3)=1+3=4;
to calculate ord_11(4) we have
11+1    1    3
11+14  0   25
11+25  1    9
11+9    1   5
11+5    2   1
 Thus ord_11(4)=1+1+1+2=5.

Regards,
Vladimir




----- Original Message -----
From: Max Alekseyev <maxale at gmail.com>
Date: Wednesday, July 21, 2010 23:05
Subject: [seqfan] Re: An algorithm of calculation of multiplicative order of 2 mod n
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> The algorithm iteratively performs divisions by 2 modulo n.
> It starts with n+1 == 1 (mod n) and ends with 1. The total 
> number of
> divisions k is then nothing else but the multiplicative order of 
> 2 mod
> n.
> 
> Max
> 
> On Tue, Jul 20, 2010 at 2:18 PM, Vladimir Shevelev 
> <shevelev at bgu.ac.il> wrote:
> > Dear Seq Fan,
> >
> > I  would like to present an easy (till now-conjectural!) 
> algorithm of calculation of multiplicative order of 2 mod n (n 
> arbitrary positive odd number). It is based on trivial 
> calculation of A007814(x) in base 2.
> >
> > Step1. l(1)=A007814(n+1),  m(1)=(n+1)/2^l(1);
> > Step i(i>=2). l(i)=A007814(n+m(i-1)),  m(i)=(n+m(i-1))/2^l(i);
> > The process ends when m=1.
> >
> > Now we have: A002326(n)=l(1)+...+l(k).
> >
> > Example (here A007814=a(n)).
> >
> > a(17+1)=1, (17+1)/2=9;
> > a(17+9)=1, (17+9)/2=13;
> > a(17+13)=1, (17+13)/2=15;
> > a(17+15)=5, (17+15)/(2^5)=1.
> >
> > Thus A002327(17)=1+1+1+5=8.
> >
> > Regards,
> > Vladimir
> >
> >  Shevelev Vladimir‎
> >
> > _______________________________________________
> >
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> >
> 
> 
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> 

 Shevelev Vladimir‎