This turing machine recognizes L = {a^nb^m | n>=1, n!=m}

This is the solution to Linz, Sect 9.1 number 4c)

The next set of quintuples are the from Example 9.7 with one removed.
These accept a^nb^n for n>=1.  F = {4}

(0ax1r
(1aa1r
(1yy1r
(1by2l
(2yy2l
(2aa2l
(2xx0r
(0yy3r
(3yy3r
!(3  4r    This accepts a^nb^n

The next set are those that must be added to cause acceptance of the
given language L.

(1  4r    accepts for m < n
(3b 4r    accepts for m > n