Sum of primitive roots

franktaw at netscape.net franktaw at netscape.net
Wed Jan 9 17:22:59 CET 2008


No, I don't think you made a mistake.  The sequence is probably finite 
and complete as Robert gave it.

Note that if p^n (or 2*p^n) is not in the sequence, neither is p^{n+1} 
(or 2*p^{n+1}).  This is because the primitive roots of p^{n+1} are 
precisely those numbers congruent to a primitive root of p^n modulo 
p^n.  (Or taking a more expansive definition, they are exactly the same 
as the primitive roots of p^n.)  So given a sum for p^n, just bump one 
of the roots up by (p-1)*p^n, and you get a sum for p^{n+1}.  The same 
thing happens for 2*p^n.

So, the next term, if it exists, would have to be either p or 2*p, and 
it is unlikely that there are any more such primes.

Franklin T. Adams-Watters

-----Original Message-----
From: petsie at dordos.net

Maybe I made an error when coding this problem...
I find the start of this sequence until 162 but no more elements up to 
10000.
Is this correct?

Peter

...

On Tue, 8 Jan 2008, Robert Israel wrote:

> It's interesting to note, btw, that although this sequence contains
> 6=2*3, 18=2*3^2, 54=2*3^3 and 162=2*3^4, it does not contain
> 486=2*3^5, as 47 + 65 + 77 + 83 + 101 + 113 = 486.

> Why not allow all the positive integers that have primitive roots, so
> include 4, p^n and 2 p^n as candidates (for positive integers n and 
odd
> primes p)?
> Then I believe you get 2,3,4,6,7,9,11,14,18,22,38,54,162,...
>
> Robert Israel                                israel at math.ubc.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia            Vancouver, BC, Canada
>
> On Tue, 8 Jan 2008, franktaw at netscape.net wrote:
>
>> 2,3,7,11.
>>
>> These primes cannot be represented as a sum of their (distinct)
>> primitive roots (requiring the root to be positive).  Are there any
>> others?
>>
>> I have checked up to p = 167.  Note that it is not necessary to 
check
>> primes = 1 (mod 4), since for such primes, if k is a primitive root, 
so
>> is p - k.  It seems very unlikely that there are any others, but I
>> don't see  how to prove it.
>>
>> Assuming that this list is complete, should this be in the OEIS?
>>
>> Franklin T. Adams-Watters
>>
>>
>> 
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