[seqfan] Re: from Steven Finch about A087687

Torleiv.Klove at ii.uib.no Torleiv.Klove at ii.uib.no
Mon Jan 26 07:52:51 CET 2009


To show that a(n) is multiplicative is simple number theory.
If gcd(n,m)=1, then any solution of x^2+y^2+z^2 = 0 mod (n)
and any solution (mod m) can combined to a solution (mod nm) using the
chinese reminder theorem, and any solution (mod nm) gives solutions  
(mod n) and (mod m). Hence a(nm)=a(n)a(m).


Quoting Olivier Gerard <olivier.gerard at gmail.com>:

> To seqfan members: if you reply to this message,
> please be sure to put Steven Finch in copy, as he
> is not currently subscribed to the list.
> Olivier
> From: Steven Finch <sfinch9 at hotmail.com>
> Date: Sun, 25 Jan 2009 20:55:11 -0500
> Subject: A087687
> Hello!
> I believe that a(n), as defined by A087687, is a multiplicative function.
> It seems clear that a(p)=p^2 for any prime p, a(4)=8, a(8)=32 and
> a(p^2)=p^2*(p^2+p-1),    a(p^3)=p^4*(p^2+p-1)
> for any odd prime.  Formulas for a(p^k) for k>=3 seem to be hard to
> find.  Does anyone know how to do this?  Thank you!
> Steve Finch
> http://algo.inria.fr/bsolve/
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