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# Scripto | Transcribe Page

## Current Page Transcription [history]

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.)