If the case is that simple (your string could start by /^BC/ or by /^.*BC/), you can use the replacement operator s/// to get the right answer.
#!/usr/bin/perl -w
use strict;

my $source = "BCBCBCBCBCCCB";
my $group = "BC"; # the repeating string

# capturing the left part, if any,
# and removing it from the source
my $left = substr($source,0, index($source,$group));
substr($source,0,index($source,$group)) ="";

# counting the occurrence of the repeating group
my $exp = $source =~ s/$group//g;

#print the "formula"
print "$left ($group)^{$exp}$source\n";

[download]
Output:
(BC)^{5}CCB
Changing source to "CCBCBCBCBCBCCCB";
You get the output:
CC (BC)^{5}CCB

In reply to Re: Converting BCBCBCBCBCCCB to (BC)^{5}CCB by gmax
in thread Converting BCBCBCBCBCCCB to (BC)^{5}CCB by fishfork

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