It's a good guess, because there is a the fact that $1 points into the original string, and can change when the original string changes. But it's not the source of the confusion here.

You can see that by changing the original code to use m/(\d+)/g instead of s/(\d+)//, thus not modifying the original string at all:

sub R {
   printf qq{before: \$_ is '$_'};
   printf qq{  \$1 is %s \n}, defined($1) ? qq{'$1'} : 'undefined';
   m/(\d+)/g ? $1 + R() : 0;
   printf qq{after: \$_ is '$_'};
   printf qq{ \$1 is %s \n}, defined($1) ? qq{'$1'} : 'undefined';
}
$_ = 'a81b2d34c1';
R();
__END__
before: $_ is 'a81b2d34c1'  $1 is undefined 
before: $_ is 'a81b2d34c1'  $1 is '81' 
before: $_ is 'a81b2d34c1'  $1 is '2' 
before: $_ is 'a81b2d34c1'  $1 is '34' 
before: $_ is 'a81b2d34c1'  $1 is '1' 
after: $_ is 'a81b2d34c1' $1 is '1' 
after: $_ is 'a81b2d34c1' $1 is '1' 
after: $_ is 'a81b2d34c1' $1 is '1' 
after: $_ is 'a81b2d34c1' $1 is '1' 
after: $_ is 'a81b2d34c1' $1 is '1'
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Let me repeat what I wrote earlier, but hopefully a bit clearer this time: There is only one variable $1. The first call to m/()/ or s/()// creates the dynamic variable $1, and all subsequent calls modify the existing variable $1. Since there is no mechanism for resetting $1 to a previous value, you can see the last value of $1 in all stack frames that have access to it.