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## Henry Bushby Transcription Pages

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273

(xvi.) Given C *& z = 1*; how many *regular* knots with C will

there be on 1 cord? (=z cord.)

Since (n-1) is always a ~~factor~~ *measure* of C,

C can only fall into series indicated by

1 + measures of itself. ~~??~~ (xiii.) And *(xiv)*

it will be on 1 cord only when n, (= 1 + such

a measure,) & C are prime to one another. ~~xiv by xiv~~ Therefore set out the measures

of C from 1 to C. Add 1 to each of them.

Strike out numbers which are multiples of any unaugment-

ed measure greater than unity. The remaining

nos. are the indices of the series in *each of* which

C occurs *once* on 1 cord & there are no other

such regular knots on 1 cord. By X

the highest series gives a twist knot.

Thus let C =24.

Measures of 24: 1, 2, 3, 4, 6, 8, 12, 24

Augmented by 1: 2, 3, 4, 5, 7, 9, 13, 25

Multiples of 2: 2, 4

Multiples of 3: 3, 9~~Leaving~~

Indices of series: 5,7,13, 25 of

which S25 1B[over]24C is a twist knot. (p268 follows.)