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## Henry Bushby Transcription Pages

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271

prime to one another, z = 1. ~~Therefore~~ In the formula

when n, & ^{a}⁄_{n}

C is always (n-1)^{2}c. ~~Divide this~~

by (n-1) to bring it to the form ^{c}⁄_{n-1>}

Therefore if n & ^{(n-1)2 c}⁄_{(n-1)} are prime

to one another, __z__ = 1. Therefore if

n & (n-1) are so, __z__ = 1. But these

must be so (except when n = z). Hence

z = 1. And when __n__ = z, __z__ also equals

1 (ii.230.)

The reduced knot therefore is on one cord also.

(xi) It is probable that all the other

knots in any series K_{n} are ir-

reducible.

(xii) Except that in K the crossings

are governed by ix[alpha]. supr. (p266) &

that there is (one) reducible knot

in every series, K & S are ana-

logous in every way.

(xiii) S_{1} & K_{1}, S_{2} & K_{2} are identical.