[seqfan] Re: An algorithm of calculation of multiplicative order of 2 mod n
guninski at guninski.com
Wed Jul 21 17:30:20 CEST 2010
what suprises me is the algorithm doesn't use any modular stuff mod n ...
with a C program checked both versions to close to a million - no
On Wed, Jul 21, 2010 at 10:21:20AM -0400, Charles Greathouse wrote:
> It's worth noting that the algorithm (implemented, as below, as a
> script) is about 300 times slower than the built-in znorder(Mod(2,n))
> for 4-digit numbers and about 1500 times slower for 5-digit numbers.
> So the interest in the method is in finding interesting properties, as
> Georgi suggests.
> Charles Greathouse
> Case Western Reserve University
> On Wed, Jul 21, 2010 at 5:02 AM, Georgi Guninski <guninski at guninski.com> wrote:
> > your algorithm seems correct for odd numbers up to 20000.
> > looks like if you replace "+m(i-1)" with "-m(i-1)" the result is either the mult. order or half the mult. order.
> > i wonder can you work "backwards" to find n with prescribed multiplicative order of 2 (this is divisor of phi(n) ) ?
> > attached is a pari implementation.
> > On Tue, Jul 20, 2010 at 06:36:12PM +0000, Vladimir Shevelev wrote:
> >> Dear SeqFans,
> >> 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 (say, m(k)=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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