As we underlined in Section 1.11, we can distinguish generating
noninvertible knots and noninvertible knots that are members of
already derived families. From every generating noninvertible
pretzel knot its corresponding family is obtained by the following
substitutions, respecting symmetry conditions:
For n=10
crossings there is only one noninvertible pretzel
knot: 2 2,3,2 1.
For n=11
there are four of them:
For the first knot 4 1,2 1,3, its generating knot is 2 1,2 1,3 with n=8 crossings that satisfies the necessary condition for noninvertibility of pretzel knots: its type [0,0,1] consists only from 0s and 1s. It contains the tangles (2p) 1 and (2q) 1, so according to the symmetry condition it generates the family of noninvertible pretzel knots (2p) 1,(2q) 1,(2r+1) for p ¹ q. Hence, the first noninvertible knot in this family is 4 1,2 1,3.
For n=11, the four noninvertible knots are divided into two
subsets:
The first subset contains the knot 4 1,2 1,3 that belongs to the family of noninvertible knots (2p) 1,(2q) 1,(2r+1) with the additional symmetry condition p ¹ q, and the knots from the other subset generate families of noninvertible knots without additional requirements on parameters.
