11/16   "sed '/^[0-9].*$/\!d' inputfile" will print out only the lines
        of a file that start with numbers.  Supposed I want to print out
        the 1st, 6th, 10th, 16th, etc lines that begin with a number.
        How can I do that elegantly with sed or perl or whatever?
        \_ perl -ne 'print if /^\d/ && $count++ % 5 == 0'  --mconst
        \_ You can replace all the newlines with a new record separator (eg
           "FOO") and then awk '{print $1, $6, $10, $16}'.
           \- to do this either you need to understand a little bit about
              how sed works and then write a little sed program OR if
              you want a cryptic one liner, it heavily depends on the
              version of sed ... i cant think of a simple way to do
              this in "genreic sed" ... i assume your list doenst end
              at 16 ... that is trivial. i think gsed supports the +5d
              operator. --psb
        \_ perl -ne 'print if /^\d/ && <compare $. as line #>' file
           the compare could be e.g.: $. =~ /^(1|6|10|16)$/
           --dbushong
           \- if all you want to do is print out those 4 lines it is
'             trivial ... sed -n -e '1p;6p;10p;16p'  --psb
         \_ This has to work for a file that is 100000 lines long.  That's
            what I meant by etc.  I would have thought there would be an
            elegent way to basically tell it to print the first line,
            skip the next n lines, print the next line, skip the next n
            lines, etc. -op
            \_ Is n 5, 4, or 6?