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

[Max Alekseyev]

For prime p==3 (mod 4),

A165186(p) = p*h(-p) + p*(p-1)/2

where h(-p) is the class number (listed in A002143)

For example, h(-19)=1 and
A165186(19) = 19*1 + 19*18/2 = 190.

btw, A165186 has wrong offset - it should be 1, not 0.

Regards,
Max

On Sun, Nov 22, 2009 at 4:25 AM,  <franktaw at netscape.net> wrote:
> 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.
>
>
>
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