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Subject: Re: matrix indexing
Date: Mon, 27 Feb 2012 20:11:14 +0000 (UTC)
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"Roger Stafford" wrote in message <jifaca$4j5$1@newscl01ah.mathworks.com>...
> "merew" wrote in message <jif55e$jfa$1@newscl01ah.mathworks.com>...
> > for l=0:L
> > for m=-l:l
> >      value=x;
> >      D(?)=x;
> > end
>  - - - - - - - -
>  D(l^2+l+m+1) = x;
- - - - - - - -
  This would give you a different ordering.  Try it out.

 D(l+1-max(m,0),l+1+min(m,0)) = x;

Roger Stafford