This substitution, s/([^^#])/$1$1/g, will do what you want. I'd expect it to be the fastest. It replaces any non-^ non-# character with itself twice.
Jeff japhy Pinyan,
P.L., P.M., P.O.D, X.S.:
Perl,
regex,
and perlhacker How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart