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## Henry Bushby Transcription pages

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Revision as of Jun 18, 2014, 10:05:54 AM created by DPepper |
Revision as of Jun 18, 2014, 10:06:49 AM edited by DPepper |
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− | [The following 5 lines of text are crossed out by a bold diagonal line]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> | + | [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, 10:06:49 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, </s>an even</s> * ^ any* no. of bights indivisible

by n gives reversibles. e.g. I_{3} ^{7b}/_{14c}, I_{5} ^{2b}/_{8c}~~But try further~~.

I find __all__ the r.ks up to I_{3} ^{7b}/_{14c} are

reversible, & hence it may be safely inferred

that all I_{3} are reversible. The only knot~~s~~

in I_{3} that coincides with plait-twists [greek cap sigma] = 3~~are~~ is I_{3} ^{2b}/_{4c}. The others are different.

It is possible therefore that all in I_{5} are

reversible. And in I_{7}. Also that all II_{3} are

reversible, & so forth. No: II_{3} ^{4b}/_{8c} was not

reversible. But in I_{5}, ^{2b}_{8c}, ^{3b}_{12c}, ^{4b}_{16c}, ^{6b}_{24c}

were all reversible, and probably it may be

safely said that all I_{5} are so. Further,

probably all with n odd of I are so.