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an r.k.without obliterating
Of course this plait could be
obtained directly from S3 4b/8c thus :-
Here on a 1z k. 2 adjacent bights
have been cut & the alternate parts
x & 2, 1 & 3, joined without alternate
crossings, producing an i.k. of
c – 1 crossings = 7 on 1z.
Therefore among knots of 7c are
S2 7b/7c, the T.H; & this i.k. plait.
Presumably every 1z. k in S3
could be similarly made to produce
an i.k. on 1z. with 1c. less than
the number. (S3 5b/10c & S3 7b/14c can.)
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