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Henry Bushby Transcription pages

VM533B94v4-p018.jpg

Revision as of Jun 18, 2014, 10:06:49 AM
edited by DPepper
Revision as of Jun 18, 2014, 10:08:27 AM
edited by DPepper
Line 1: Line 1:
[The following 5 lines of text are crossed out by a bold diagonal line]<br>It looks as if with r. ks. When n<br>is even, an odd no. of bights indivisible<br>by n gives reversibles = none; and when<br>n is odd, </s>an even</s> <i> ^ any</i> no. of bights indivisible<br>by n gives reversibles. e.g. I<sub>3</sub> <sup>7b</sup>/<sub>14c</sub>, I<sub>5</sub> <sup>2b</sup>/<sub>8c</sub><br><s>But try further</s>.<br>I find <u>all</u> the r.ks up to I<sub>3</sub> <sup>7b</sup>/<sub>14c</sub> are<br>reversible, & hence it may be safely inferred<br>that all I<sub>3</sub> are reversible. The only knot<s>s</s><br>in I<sub>3</sub> that coincides with plait-twists [greek cap sigma] = 3<br><s>are</s> is I<sub>3</sub> <sup>2b</sup>/<sub>4c</sub>. The others are different.<br>It is possible therefore that all in I<sub>5</sub> are<br>reversible. And in I<sub>7</sub>. Also that all II<sub>3</sub> are<br>reversible, & so forth. No: II<sub>3</sub> <sup>4b</sup>/<sub>8c</sub> was not<br>reversible. But in I<sub>5</sub>, <sup>2b</sup><sub>8c</sub>, <sup>3b</sup><sub>12c</sub>, <sup>4b</sup><sub>16c</sub>, <sup>6b</sup><sub>24c</sub><br>were all reversible, and probably it may be<br>safely said that all I<sub>5</sub> are so. Further,<br>probably all with n odd of I are so.<br>
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[The following 5 lines of text are crossed out by a bold diagonal line]<br>It looks as if with r. ks. When n<br>is even, an odd no. of bights indivisible<br>by n gives reversibles = none; and when<br>n is odd, <s>an even</s> <i> ^ any</i> no. of bights indivisible<br>by n gives reversibles. e.g. I<sub>3</sub> <sup>7b</sup>/<sub>14c</sub>, I<sub>5</sub> <sup>2b</sup>/<sub>8c</sub><br><s>But try further</s>.<br>I find <u>all</u> the r.ks up to I<sub>3</sub> <sup>7b</sup>/<sub>14c</sub> are<br>reversible, & hence it may be safely inferred<br>that all I<sub>3</sub> are reversible. The only knot<s>s</s><br>in I<sub>3</sub> that coincides with plait-twists [greek cap sigma] = 3<br><s>are</s> is I<sub>3</sub> <sup>2b</sup>/<sub>4c</sub>. The others are different.<br>It is possible therefore that all in I<sub>5</sub> are<br>reversible. And in I<sub>7</sub>. Also that all II<sub>3</sub> are<br>reversible, & so forth. No: II<sub>3</sub> <sup>4b</sup>/<sub>8c</sub> was not<br>reversible. But in I<sub>5</sub>, <sup>2b</sup><sub>8c</sub>, <sup>3b</sup><sub>12c</sub>, <sup>4b</sup><sub>16c</sub>, <sup>6b</sup><sub>24c</sub><br>were all reversible, and probably it may be<br>safely said that all I<sub>5</sub> are so. Further,<br>probably all with n odd of I are so.<br>

Revision as of Jun 18, 2014, 10:08:27 AM

[The following 5 lines of text are crossed out by a bold diagonal line]
It looks as if with r. ks. When n
is even, an odd no. of bights indivisible
by n gives reversibles = none; and when
n is odd, an even ^ any no. of bights indivisible
by n gives reversibles. e.g. I3 7b/14c, I5 2b/8c
But try further.
I find all the r.ks up to I3 7b/14c are
reversible, & hence it may be safely inferred
that all I3 are reversible. The only knots
in I3 that coincides with plait-twists [greek cap sigma] = 3
are is I3 2b/4c. The others are different.
It is possible therefore that all in I5 are
reversible. And in I7. Also that all II3 are
reversible, & so forth. No: II3 4b/8c was not
reversible. But in I5, 2b8c, 3b12c, 4b16c, 6b24c
were all reversible, and probably it may be
safely said that all I5 are so. Further,
probably all with n odd of I are so.