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## Henry Bushby Transcription pages

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Revision as of Jun 2, 2014, 6:00:12 AM edited by JLee |
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− | I<sub>2</sub> Only<br> | + | I<sub>2</sub> Only<br>(5) odd c gives c-1 knots <sup>c-1</sup>/<sub>2</sub> of (c+1)<br><sup>c-1</sup>/<sub>2</sub> of (c+2)}[bracket with line above]<u>[Oro?]</u><br>odd c give <s>same</s> ^<i>c-1 knots</i> for <u>not.</u><br>But some will be repetitions.<br> |

− | (5) odd c gives c-1 knots <sup>c-1</sup>/<sub>2</sub> of (c+1)<br><sup>c-1</sup>/<sub>2</sub> of (c+2)}[bracket with line above]<u>[Oro?]</u><br> | + | (6) Required: knots with y.c. Proceed<br>as follows:-<br><u>A.</u> Suppose y is odd.<br>(i) <u>[Oro?]</u>: - y c can be derived from<br>r.k. y-2, & r.k y-1.<br>(ii) With r.k.(y-2) join 1 to all even nos.<br>"<u>[Oro?]</u>" up to & including 1 [Sideways U] y-2. This will give<br><sup>y-3</sup>/<sub>2</sub> knots of y crossings.<br>Then with r.k (y-1) join 1 ^<i><u>[Oro?]</u></i> to all odd<br>nos. ^<i>except 5</i> up to & including 1 [Sideways U] y. This will <br>give <s><sup>y-3</sup>/<sub>2</sub> of y crossings.</s> <s><u>y-3</u></s> <sup>4-3</sup>/<sub>2</sub> knots. <br>(iii) <s> As y is odd these are no fresh knots</s> ^<i> With. r.k. y join 1 <u>not </u> to all even nos. up </i><br><s>as yet ascented in the "not" series.</s>^<i><s>by</s> to y-1 (i.e. y) excluding 2. This gives <sup>y-3</sup>/<sub>2</sub> knots.</i><br><u>B</u>. Suppose y is even.<br><s>(i)</s> <u>[Oro?]</u>:- y , c can be derived from <br>r.k y-2 & r.k y-1.<br>(i) With r.k. (y-2) join 1 to all even nos. <u>[Oro.?]</u><br>to y-2.<br><s>as in <u>A</u>.i.</s> This gives <sup>y-2</sup>/<sub>2</sub> <s>4</s> knots. |

− | odd c give <s>same</s> ^<i>c-1 knots</i> for <u>not.</u><br> | + | |

− | But some will be repetitions.<br> | + | |

− | (6) Required: knots with y.c. Proceed<br> | + | |

− | as follows:-<br> | + | |

− | <u>A.</u> Suppose y is odd.<br> | + | |

− | (i)<u>[Oro?]</u>: - y c can be derived from<br> | + | |

− | r.k. y-2, & r.k y-1.<br> | + | |

− | (ii) With r.k.(y-2) join 1 to all even nos.<br> | + | |

− | "<u>[Oro?]</u>" up to & including 1 [Sideways U] y-2. This will give<br> | + | |

− | <sup>y-3</sup>/<sub>2</sub> knots of y crossings.<br> | + | |

− | Then with r.k (y-1) join 1 ^<i><u>[Oro?]</u></i> to all odd<br> | + | |

− | nos. ^<i>except 5</i> up to & including 1 [Sideways U] y. This will <br> | + | |

− | give <s>y-3 | + | |

− | (iii) <s> As y is odd these are no fresh knots</s> ^<i> | + |

## Revision as of Jun 2, 2014, 6:34:49 AM

I_{2} Only

(5) odd c gives c-1 knots ^{c-1}/_{2} of (c+1)^{c-1}/_{2} of (c+2)}[bracket with line above]__[Oro?]__

odd c give ~~same~~ ^*c-1 knots* for __not.__

But some will be repetitions.

(6) Required: knots with y.c. Proceed

as follows:-__A.__ Suppose y is odd.

(i) __[Oro?]__: - y c can be derived from

r.k. y-2, & r.k y-1.

(ii) With r.k.(y-2) join 1 to all even nos.

"__[Oro?]__" up to & including 1 [Sideways U] y-2. This will give^{y-3}/_{2} knots of y crossings.

Then with r.k (y-1) join 1 ^* [Oro?]* to all odd

nos. ^

*except 5*up to & including 1 [Sideways U] y. This will

give

^{y-3}/

_{2}of y crossings.

__y-3__

^{4-3}/

_{2}knots.

(iii)

*With. r.k. y join 1*

__not__to all even nos. up~~by~~ to y-1 (i.e. y) excluding 2. This gives

^{y-3}/_{2}knots.__B__. Suppose y is even.

__[Oro?]__:- y , c can be derived from

r.k y-2 & r.k y-1.

(i) With r.k. (y-2) join 1 to all even nos.

__[Oro.?]__

to y-2.

__A__.i.

^{y-2}/

_{2}