[seqfan] Re: Generally known (to those who know of such things)
victor miller
victorsmiller at gmail.com
Mon Nov 23 02:08:07 CET 2009
As Neil said they are class numbers of binary quadratic forms. Here's
a quick description of them :
http://math.berkeley.edu/~nsnyder/tutorial/lecture7.pdf
Victor
On Sun, Nov 22, 2009 at 6:38 PM, <franktaw at netscape.net> wrote:
> Let's hear it for reuse of mathematical nomenclature.
>
> Wikipedia lists two meanings for "class number"; neither matches this
> one. Mathworld only knows about one of them.
>
> It isn't at all obvious to me why this should be called a "class
> number"; what are the classes being numbered?
>
> I know, we can't fix all the problems with mathematical notation and
> nomenclature. But somehow this one is somehow particularly annoying to
> me.
>
> Franklin T. Adams-Watters
>
> -----Original Message-----
> From: Max Alekseyev <maxale at gmail.com>
>
> 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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>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
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