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

Charles Greathouse charles.greathouse at case.edu
Wed Jul 21 16:21:20 CEST 2010

```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
Analyst/Programmer
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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