[seqfan] Re: A088144 and A076409

Richard Guy rkg at cpsc.ucalgary.ca
Wed Feb 23 18:58:05 CET 2011


Dear JM Bergot,
                I'm copying my reply to
the seqfan network, as there are one or
two things they may like to add to OEIS.

                The sum of the primitive roots
of  p  (A088144) is a multiple of  p  whenever
p  is of shape  4k+1.  This is because the
primitive roots come in pairs,  r  and  p-r,
and their number is  phi(phi(p)) = phi(p-1),
and the total is  p * phi(p-1) / 2,  though I
didn't see that this was mentioned in  A088144.
The sum is also a multiple of  p  for  p = 19;
are there other examples?  Searching for the
sequence  5, 13, 17, 19, 29, ... didn't give
any hits.  Perhaps it would be of interest to
list the subsequence

2 = 1*3 - 1,  8 = 1*7 + 1,  23 = 2*11 + 1, 
57 = 3*19,  139 = 6*23 + 1,  123 = 4*31 - 1,
257 = 6*43 - 1,  612 = 13*47 + 1, 886 = 15*59 + 1,
669 = 10*67 - 1, 1064 = 15*71 - 1, 1105 = 14*79 - 1,
1744 = 21*83 + 1, ...

and of the coefficients

1, 1, 2, 3, 6, 4, 6, 13, 15, 10, 15, 10, 14, 21, ...

and even the ``errors''

-1, 1, 1, 0, 1, -1, -1, 1, 1, -1, -1, -1, 1, ...
(are there any more zeroes? or any terms other
than 1 & -1 ?)

       In answer to your second question, the
sum of the primitive roots is equal to the sum
of the quadratic residues (A076409) for  p = 5
and  p = 17  and perhaps (only?) for the Fermat
primes.  Here it IS mentioned that the sum is
k(4k+1)  when  p= 4k+1  and  A076410  confirms
that the sum is always a multiple of  p, though
it would be good to mention this at A076409.

       Best,   R.

On Mon, 21 Feb 2011, JM Bergot wrote:

> Do you think it is possible for:
> (1)The sum of the primitive roots of some 
>      prime p is divisible by p.
>  
> (2)The sum of the quadratic residues for
>      some prime p is the same as the sum
>      of its primitive roots, say for some p
>     of the form 2*q+1=p for prime q.


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