[seqfan] Re: from Steven Finch about A087687

[Martin Fuller]

Conjecture without proof:

a(2^k) = 2^(k+ceil(k/2))
For odd primes p, a(p^(2k-1)) = p^(3k-2) (p^k+p^(k-1)-1) and a(p^(2k)) = p^(3k-1) (p^(k+1)+p^k-1)

This is the same as A062775 for odd n.

PARI program to calculate A087687 directly:
A087687(n)=local(v=vector(n),w);for(i=1,n,v[i^2%n+1]++);w=vector(n,i,sum(j=1,n,v[j]*v[(i-j)%n+1]));sum(j=1,n,w[j]*v[(1-j)%n+1])

Martin Fuller

--- On Mon, 26/1/09, franktaw at netscape.net <franktaw at netscape.net> wrote:

> From: franktaw at netscape.net <franktaw at netscape.net>
> Subject: [seqfan] Re: from Steven Finch about A087687
> To: seqfan at list.seqfan.eu
> Cc: sfinch9 at hotmail.com
> Date: Monday, 26 January, 2009, 11:49 AM
> Note that A087687 has the "mult" keyword, which
> says that it is 
> multiplicative.
> 
> Franklin T. Adams-Watters
> 
> 
> -----Original Message-----
> From: 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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