& corresponds to the matched string for the given pattern. Parenthesized portions of a regular expression can be matched with \#:

    $ echo this is digit 7 in a number | sed 's/digit \([0-9]\)/\1/'
    this is 7 in a number

The preceding command replaces digit 7 with 7. The substring matched is 7. \(pattern\) matches the substring. The pattern is enclosed in () and is escaped with backslashes. For the first substring match, the corresponding notation is \1, for the second, it is \2, and so on.

    $ echo seven EIGHT | sed 's/\([a-z]\+\) \([A-Z]\+\)/\2 \1/'
    EIGHT seven

([a-z]\+\) matches the first word and \([A-Z]\+\) matches the second word; \1 and \2 are used for referencing them. This type of referencing is called back referencing. In the replacement part, their order is changed as \2 \1, and hence, it appears in the reverse order.