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

[franktaw at netscape.net]

There was some confusion here (at least on my part) between quadratic 
fields and quadratic forms.

I do think the following should be done:

1) Either adjust the definition for A002143, or add a comment, making 
it explicit that these are class numbers for binary quadratic forms.

2) Include a link from A002143 to some suitable discussion of class 
numbers for binary quadratic forms.  Possibly the link Victor provides 
(below) would be appropriate.

3) Add a page to Wikipedia discussing class numbers of binary quadratic 
forms.  Include a reference to it from the disambiguation page for 
"class number", as well as from the quadratic forms article.

4) Include discussion of class number for groups (as found on 
Wikipedia) and for binary quadratic forms in Mathworld.

Victor, perhaps you could take care of at least items 1 and 2?

Franklin T. Adams-Watters

-----Original Message-----
From: victor miller <victorsmiller at gmail.com>

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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>>
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>>
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