[seqfan] Re: Generally known (to those who know of such things)

[franktaw at netscape.net]

Yes, now it makes sense.

But, sum(k=1..p, mod(k(p-k), p)) = sum(k=1..p, mod( -k^2, p)).  But the 
squares are symmetrically arranged modulo primes == 1 (mod 4); so for 
such primes, A165186(p) = p*(p-1)/2; and of course this is increasing.

For primes == 3 (mod 4), the squares are antisymmetric, so there is no 
such simple expression for A165186(p).  Thus there is no reason why it 
has to always be increasing; and, in fact, it is not.

Franklin T. Adams-Watters

-----Original Message-----
From: wouter meeussen <wouter.meeussen at pandora.be>

my Apologies,
cutting (and pasting) corners,
should have known better!

If anything, I should have said:
A165186 (p1)=Sum(k=1..p1;  mod( k (p1-k) , p1)  ) seems to be strictly
increasing in function of p1, while
A165186 (p3)=Sum(k=1..p3;  mod( k (p3-k) , p3)  ) is not (in function of
p3).
(* why? *)
with again p1 and p3 the primes congruent to 1 resp. 3 mod 4.

in other words,
A165186 (p1)  seems to be strictly increasing in function of p1, while
A165186 (p3) is not (in function of p3).

with, of course,
A165186(n) = Sum(k=1..n;  mod(k*(n-k) ,n)  )
(a sequence originating from my earlier mail here
Sent: Sunday, September 06, 2009 4:08 PM
Subject: [seqfan] Trivium.)

and Sum(k=1..n;  mod(k*(n-k) ,n)  )  =  Sum(k=1..n-1;  n-mod(k^2 ,n)  )

Mma code:
----------------------------------------------------
Table[p=Prime[n];
If[Mod[p,4]===1,Sum[Mod[k(p-k),p],{k,p}],z],{n,1000}]/.z->Sequence[];
Flatten[Position[Rest[%]-Drop[%,-1],_?Negative]]=={}
----------------------------------------------------

Wouter.