As tomte suggests, you can achieve this using deferred evaluation; the trick is to take advantage of the special variable $^N, which refers to "the most recently closed capture" in the match so far.

Using this, we get for example:

% perl -le 'print $& if shift =~ /(\d)(((??{$^N + 1})))+/' 4321234
1234
%
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This works because the first time through the eval we have seen (for example) the digit '1', so we return '2' as the next thing to match; if we match it, that match is itself captured and becomes the last thing captured, so the next time into the eval $^N is '2' and we return '3' as the new thing to match.

Note firstly that $^N was introduced in perl-5.8.0, so you won't be able to use this with earlier versions; secondly, enabling warnings gives the unexpected complaint:

(((??{$^N + 1})))+ matches null string many times before HERE mark in 
+regex m/((\d)(((??{$^N + 1})))+ << HERE )/ at -e line 1.
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Hugo

You may could experiment with (??{ code })

#!/usr/bin/perl

my $test = "4321234";

if ($test =~ m/(\d)((??{ $1 + 1}))/) {
    print "matched $1 $2 \n";
}
__END__
matched 1 2
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I guess you are right
Thanx, Alex