> for $d in distinct-values($seq) return $d[count($seq[. eq $d]) ge $i]
> 
> equivalent?  I think it is, and probably a lot more 
> efficient, although it is longer.
> 

They are both O(n^2).

What about:

<xsl:for-each-group select="$seq" group-by=".">
  <xsl:sequence select="current-group()[$i]"/.
</xsl:for-each-group>

which also scores quite well on brevity, I think - in syntax tree form, it
has 6 nodes which is the same as Dimitre's expression; and I think it rates
higher on both efficiency and clarity.

Michael Kay
http://www.saxonica.com/