## About the Item

__Title__

## About the Item

__Transcription__

71

Take S_{4} 6b/18c z = 2

Here it is clear by inspection that

there are 2 exclusions, 4 & 8.

B/z - 1 = 6/2 - 1 = 3 - 1 = 2 [checkmark]

M_{2n} = M x 8 = 8, [16 exceeds 2B]

(M_{2n} = 8, 16, 24, 32, 40 &c.

(M'B = 6, 12, 18, 24, 30, &c.

The rule won't work thus.

M_{2n} - 2B = 8 - 12 [??]

16 - 12 = 4 [checkmark]

24 - 12 (~~exceeds~~

equals 2B].

The rule stands without the 3rd~~[??]~~exception, though that seems

to apply when z = 1. But even

then it won't cover the 4 in S_{6} 8b/40c

Test the rule as given on p 67

with S_{4} 9b/27c, z = 1

M_{2n} = M x 8 = __8__, __16__, [24 exceeds 2B]

M_{2n} - 2B = __6__, __24__, [22 -------

M_{2n} -2 = __6__, __14__, [22 ------

B/2 - 1= 8. Exclusions are __6__, __8__, __14__, __16__, ~~22~~

The rule holds but the numbers

overlap.

Take S_{4} 11b/33c

n = 4, B = 11, z = 1, 2B = 12

B/2 - 1 = 11/1 - 1 = 10.

M_{2n} = M_{8} = __8__, __16__, [24, 32, 40, 48,

M_{2n} - 2B = __2__, ~~12~~ __10__, 18 [26 ------

M_{2n} - 2 = __6__, __14__, [22

[therefore] exclusions are__2__, __6__, __8__, 10, 14, __16__ & 18.

4 12 20

In this case they turn out to be all

the even nos, & the rule breaks down.

I believe in every case of 1z,

1 added to an even no. gives a ring.

For there are 4 parts cut, and the

parts are even & odd alternately from

arc to arc. Therefore an odd cut part

will be even in the first bight it

emerges in, odd in the 2nd, even in

[All in pencil]

Let M = any even multiplier & O = an odd one.

Exclusions are O_{2n}

& M_{2n} + 1 [bracketed with] or do. - 2b

which make rings.

Values not to exceed 2B, ~~amend~~ group limited

by B/z - 1

Exclusions are

2 x 2n (even)

(2 x 2n) + 1

3 x 2n

(4 x 2n) + 1

5 x 2n

(6 x 2n) + 1

79

in S_{5} So in S_{6}.

Therefore it would seem that

in any S, 1 must not be joined to

M_{2n}, but this must be reduced

where necessary by subtracting 2B.

i.e. values of M_{2n} are to be taken

up to 4B - 2, & 2B subtracted from

those above 2B.

Take S_{5} 12b/48c z = 1

n = 5, b = 12, 2B = 24, B/z - 1 = 11

M_{2n} are 10, 20, 30, 40, 50,__24 24 24__

6, 16, 26

giving 6, 10, 16, 20. But this

omits 8 & 18 (p 67. B.)

In point of fact 1 joined to any

even no. in S_{5} 12b/48c gives rings.

Where z = 1, it seems obvious that

as 1 passes into every bight & emerges

as an even no, previous to emerging

as __x__, if it is joined to any even no.

77

The part ~~[?]~~ 1 emerges in the nth bight

from ~~itself~~ its own, in the 2nth, in

the 3nth, the 4nth &c.

Does it emerge as an odd or as

an even no?

In S_{2} ^{4b}/_{4c} 1 emerges as 4(=2n) in

the nth bight from its own

In S_{2} ~~4~~ ^{5b}/_{5c}< as 4(=2n) in the nth bight

& as 8(=2x2n) in the 2nth bight.

Throughout S_{2} 1 emerges always

as an even no.

Therefore in S_{2} the exclusions are~~[Vertical line]~~M2n. & these only, & the ^{B}/_{Z}-1 rule

limits their no.

In S_{3} 1 emerges in the nth bight

from its own, in the 2nth, 3nth &

so on. It emerges as an even no (Ex-

cept in S_{3} ^{3b}/_{6c} where there can be

no prohibitions)

In S_{4} it emerges as an even no. So

79

in S_{5}. So in S_{6}.

Therefore it would seem that

in any S, 1 must not be joined to

M_{2n}, but this must be reduced

where necessary by subtracting 2B.

i.e. values of M_{2n} are to be taken

up to 4B - 2, & 2B subtracted from

those above 2B.

Take S_{5} ^{12b}/_{48c} z=1

n=5, b=12, 2B=24, ^{B}/_{z}-1=11

M_{2n} are 10, 20, 30, 40, 50,

24 24 24

[horizontal line]

6, 16, 26

giving 6, 10, 16, 20. But this

omits 8 & 18 (p67. B.)

In point of fact 1 joined to any

even no. in S_{5} ^{12b}/_{48c} gives rings.

Where z=1, it seems obvious that

as 1 passes into every bight & emerges

as an even no, previous to emerging

as __x__, if it is joined to any even no.

81.

other than __x__ it must form a ring.

But this is not the case ~~[in?]~~

when z is greater than 1.

Can this be proved theoretically?

1 emerges in the nth, 2nth, 3nth

bight &c. & where these exceed [?]B in

these bights less B. (The nos. of its

parts are 2n, 4n, 6n, 8n &c. less

2B if necessary.

Nos. are M_{2n} & M_{2n - 2B}.

up to 2B.

Where n=2 & B=3 we have

4, 8, 12, 16, 20, 24, 28

6 6 6 6

[horizontal line]

4 2, 6

Here we have 2,4,6 - all the even

nos up to 2B.

Where 2=2 & B=5

M2n=4, 8, 12, 16, 20, 24

10 10 10

[horizontal line]

2 6 10

and here we have them all.

These are both in S_{z}.

83

Take S_{3}. ^{4b}/_{8c} 2B=8

M2n=6, 12, 18, 24, 30, 36

8, 8

[horizontal line]

6, 4, [10

Here we have 4 & 6 only. Yet 2

gives a ring.~~It app~~

Divide ~~[?]~~ by n. If there is no

remainder the exclusions are 2n,~~3n,~~

4n, *8n* &c. up to 2B. If there is a

remainder it represents a certain

no of bights less than __n__ to the left

of x. ~~Through~~ Under these the line

will pass & emerge in the (nth *from x* - this

remainder) bight, then in the (2nth

- this remainder) bight & so on. There-

fore to the even no. in that bight it

must not be joined. i.e. to 2 (n-r)

2(2n-r) &c. or M2(n-r.) up to

2B. And if there is still a re-

85

mainder it will continue to 2(n-r^{1}),

2(2n-r^{1}) &c. until there is no remainder.

Take S_{6} ^{12b}/_{12c}. Divide 12 by 6 =2: no

remainder. Exclusions = 12, ~~18~~ 24 (=2B.)

Take S_{6} ^{10b}/_{50c}. Divide 10 by 6 =1+4

2B=12. Exclusions are 2n(=12)

It may be said at once that

the first exclusion is 2n, the 2nd

4n, the 3rd 6n, the 4th 8n, but

where ~~these nos. [?]~~ one of these nos.

equals x, there the thing ends, i.e.

if 2B is divisible by 2n (i.e. B by

n). But if it is not then the next

exclusion after the one lower than

2B is the difference from 2B + balance

of 2n, i.e. it is a [sum?] equal to 2n.

Take S_{6} ^{10b}/_{50c}. First exclusions are

12,[24 &c]; but ~~20/6=3+2 & 2+4=6~~ *20-12=8 & 2z-8=4*

87

(=2n-r, where nr is remainder). This

2n-r=4. Next exclusions are 4,~~10~~, 16, [28 &c). But 16 from 20 = 4

& 2n-4=12-4=8. Next exclusions

are 8, 20 Now 20=2B, & that ends

it. Therefore total exclusions are

12, 4, 16, 8 or 4, 8, 12, 16.

This is correct.

Therefore the Rule of Rings may

be stated thus: Where z=1, an odd

no must not be addes to an even no,

nor an even no to an odd; where z

exceeds 1, if B is divisible by n, then

2n, 4n, 6n &c. up to 2B are excluded;

where B is indivisible by n, then (1)

2n, 4n, 6n &c up to 2B are excluded

& the ~~remainder ~~ __r__*difference, d,*.

of these nos is next below 2B

from 2n, & this no. +2n, 4n &c. are

excluded & so on till a number is reached

which equals 2B.

[Blank Page]

89

Query: ~~Ist~~ Is it a series of which

the difference is always 2 x the G.C.M.

of B & 2? Thus in S_{6}^{10b}/_{50c} the G.C.M of

6 & 10 is 2, & 2x2=4.

In S_{6} ^{12b}/_{60c}, the GCM is 6 & the double

is 12.

In S_{6} ^{9b}/_{45c}, the G.C.M. is 3 & the double

is 6.

In S_{6}^{11b}/_{55c} the GCM is 1 & the double

is 2.

In S_{5} ^{10b}/_{40c} the GCM is 5 & the double

is 10

In S_{5} ^{11b}/_{44c} the GCM is 1 & the double is

2

In S_{5} ^{5b}/_{20c} the GCM is 5 & double is 10.

In S_{4} ^{6b}/_{18c} the GCM is 2 & the double is 4

Yes

? Is it true that

y joined to y-1 + Mn,z in any Series n,

gives rings? No -

[Caption: S_{3} ^{4b}/_{8b}]

In S_{2} ^{7b}/_{7c} there are 16 parts 1 + 1 [?:suit]

2=1

M2z = M1 = 2, 4, 6, 8, 10, 12, 14

leaving ~~1~~ 3 5 7 9 11 13

= 6 other cases, & length. This

would give 7 h, but Tait says there are 8

In S_{2} ^{5b}/_{5c} there are 10 pts.

M2z= 2, 4, 6, 8, 10

leaving 3 5 7 9 {Deleted: 4], so 4 other[? rings]

but Tait says there are only 4 knots

Test the knots so formed.

91

The Rule of Rings is this in all

cases:*^[cannot read pencil notation]* 1 must not be joined to

M2 (GCM of B&n)=

M2 (GCM of ^{C}/_{n-1}~~+~~*^&*n)

or y must not be joined to y-1

+ M2(GCM of B&n)=

y-1+M2(GCM of ^{C}/_{n-1}+n)

But by XIV. (ii.267) the GCM of

B & n = z.

Therefore the Rule of rings is *^? [* 1 must

^{1}

_{m}?] S

_{2}^only 2 bights being cut

not be joined to M2Z, or, generally,

y must not be joined to y-1+M2Z.

If it is so joined Z becomes Z+1

Problems: Rule for reducing 2 to 1

Rule for further exclusions

*^or additions*so that C may

remain the same

In S

_{3}

^{4b}/

_{8c}join 1 to 4 (=M2Z ie. 2x2x1)

no rings result - Yes they do, & so they

do when joined to 6

S_{2} ^{6b}/_{6c}

Multiples of 4 barred

6c in 1, 2, 3, 6,

5 4 7

4c

3c in 1, 3

2, 4

5c knots [running?]

5c 1, 3-__+ 7__

2,0 6,

S_{2} {^{1}/_{1}}[parenthesis extends to next line]

S_{6} {^{1}/_{1}}[continuation of parenthesis from line above] = 4

93

If 1 must not be joined to M2z

[deleted: &z = 1], 1 must not be joined to an even

no.

It may be joined to an odd no. *^Where z = 1*

The number of odd nos to which it

may be joined therefore are (^{2b}/_{2} - 1) where

b = no of bights, or (^{x}/_{2} - 1), x being

the top left hand number as on p1.

T S_{2} ^{5b}/_{5c} x = 10 ^{x}/_{2} - 1 = 4

but there are only 2 knots (? for each 5c

form in each [seven?]?)

S_{2} ^{6b}/_{6c} x = 12 ^{x}/_{2} - 1 = 5

but there are only 4 knots [But here

Z=2]

S_{2} ^{7b}/_{7c} x = 14 ^{x}/_{2}-1= = 6

& there are 8 knots

[blank page]

95.__Experiments in Twisting.__

z cords are arranged ~~at equal distances~~

on the [deleted: rad] circumference of a ~~circle~~

circular tube at equal distances from

one another, thus: -

[Caption: (1)

2 cords

(2)

3 cords

&c -

(3)]

If the tube is now

simply bent so that the

ends are joined, as

is being done in fig 3,

the ends of each

cord will meet, and*^separate unlinked* circles of cord will result. But if*^the end of* the tube is twisted through ^{360[degrees]}/_{z} (where

z = no of cords) this will no longer

be the case, & other results will fol-

low which is proposed to investi-

gate.^{360[degrees]}/_{z} is called "[deleted: one] *^an* arc", whatever the value

[blank page]

97

of __z__, since __z__ being given, the pre-

cise value of the arc can always be

determined if it is of any importance.

For illustration if the tube

is here bent

& one end

twisted through one arc 1 will join 3,

3 join 2, & 2 join 1 & z will be reduced

from 3 to 1. Twisted through two arcs

1 will join 2, 3 join 1, 2 join 3, & __z__

again is reduced from 3 to 1. But

if it is twisted through 3 arcs (=

z arcs [deleted: )] = 360[degrees]) 1 joins 1, 2 joins 2, 3 joins

3, & z remains 3.

It is of no importance that, twisted

the other way, 1 would join 2, 2 join 3 &c

Since the reverse will be formed & in any

case the same pairs join.

K_{3} ^{2a}/_{4c}

2 arcs 3 cords. Result.

4 Crossings, not alternate,

& [Deleted: not] __reducible__ to the

overhand S_{2} ^{3b}/_{3c}

Crossings over 2 & under 2.

Note: A strip of paper given 2 half twists &

cut along 2 lines (forming 3 bands) gives

[Deleted: S_{3} ^{3b}/_{[3 corrected to 6]c} not S_{2}.] [Deleted: S_{3} ^{3b}/_{6c}]*^ 3 linked rings* not S_{2} ^{3b}/_{3c}. This is

because in twisting flat paper the central^{1}/_{3} part longitudinally must always join itself.

(But in any S_{n} where * n* is even the expected

result ought to occur? or might sometimes)

99.__2 cords__.

[Revolved?] through __Result__.

arcs

1 S2 1b/1c [tickmark]; tested .& with paper strip

2 S2 2b/2c [tickmark];

3 S2 3b/3c [tickmark]paper strip[tickmark][ bracketed from 1 to 6: everything

regular.

]
4 S2 4b/4c [tickmark];

5 S2 5b/5c [tickmark]; tested with paper strip [tickmark];

6

__3 cords.__

arcs. result. z

1 S3 1b/2c [tickmark]; circle 1

+2 S2 3b/3c [3 tickmarks];__overhand__ 1

3 ~~S3 3b/6c [tickmark]~~ 3 circles 3

4 anomalous. p 101. z=1.

5

6

K3 3a/6c

3 arcs 3 cords: result.

[Caption: non-alternate.

crossings over 2 under 2.]

[Caption: =]

[Caption: K3 6a/8c]

best form

4a

K3 -- 101

8c

__4 arcs, 3 cords__ : result .

not a Carrick.

This knot is undoubtedly

the result, having been

tried many times, and

appears to be an ir-

regular irreducible knot

of 8 crossings. There are ~~2 pairs of~~ non-

alternative crossings.

5a

K3 -- 103

10c

5 arcs 3 cords.~~results.~~

[Caption: Crossings not alternate.]

6a

K3 --- 105

12c

6 arcs 3 cords. Results 2 = 3

[Caption: non-alternate.]

7a

K3 --- 107

14c

7~~6~~ arcs 3 cords-

[Caption: Drawn from rule.]

8a

K3 --- 109

16c

8 arcs 3 cords

[Caption: rule.]

(Empty page)

K3 __9a__ 111

18c

[Caption: Drawn from rule]

(Empty page)

K3 __10a__ 113

20c

10 arcs 3 cords

(Empty page)

K3 __11a__ 115

22c

11 arcs 3cords

(Empty page)

K3 __12a__ 117

24c

12 arcs 3 cords

(Empty page)

K4 __1a__ 119

3c

1 arc 4 cords

a twist.

(empty page)

121

K_{4} ^{2a}/_{6c}

2 arcs 4 cords

[Caption: Drawn from rule.]

[Editor: this page contains manuscript in pencil which has been overwritten in ink. The text below reflects the manuscript written in ink. The manuscript in pencil is illegible in places, and has not been transcribed. The manuscript in pencil appears to be a fragmentary earlier version of the manuscript in ink. The content of this page needs to be reviewed.]

[Editor: First four lines have a curly brace to their right with the text "Tested" to the right of the curly brace.]

Tested.

K_{2} ^{1a}/_{c} = K_{1} ^{2a}/_{c}

K_{3} ^{2a}/_{4c} = K_{2} ^{3a}/_{3c}

K_{4} ^{3a}/_{9c} = K_{3} ^{4a}/_{8c}

K_{5}^{4a}/_{16c} = K_{4} ^{5a}/_{15c}

|| Does K_{n} ^{(n-1)a}/_{c} = K_{(n-1)} ^{na}/_{c}? Yes.

If so K_{5} ^{4a}/_{16c} = K_{4} ^{5a}/_{15c} Yes. [checkmark]

(Note that the crossings are 1 less & the arcs

1 more.) & that the reducible knots have^{2a}/_{2zc}

{ Does K_{n} ^{(n-2)a}/_{c} = K_{(n-2)} ^{(n+1)a}/_{c}? No.

If so K_{4} ^{2a}/_{c} = K_{2} ^{5a}/_{c}. No.

[Editor: mixed pencil and ink manuscript again. I have tried to sort out the mixed sentence and figure that starts with "Reduces to..." in the way that seems most faithful to the manuscript and also makes mathematical sense, as the mathematical expressions and the figures are two representations of the same concept. The content of this page needs to be reviewed.]

123

K_{4} ^{3a}/_{9c}

3 arcs 4 cords

[Caption: Drawn from rule.]

Reduces to

K_{4}^{3a}/_{9c} reduces to = [FIGURE]

by experiment

= K_{3} ^{4a}/_{8c.}

Full formula:^{4a}/_{8c}

K_{n} ^{(n-1)a}/_{(n-1)2c} = K_{(n-1)} ^{na}/_{[(n-1)2-1]c}

This series may be expressed, instead of

e.g. S_{3} ^{4b}/_{8c} as K_{3} ^{4a}/_{8c} = 3 cords, re-

volved through 4 arcs, give a knot of

the form S_{3} ^{4b}/_{8c} with 8 crossings in alter-

nate sequences of (3-1) = 2 (q. v. p 101[alpha]) z

in the result being the same in both series.

125

4 arcs 4 cords

[Caption : b w 9 2]

[Editor, I'm unsure about the "b" and "w"]

Crossings not alternate but over 3 &

under 3.__Rule of Twists__.

If in the Plait series we read

arc for bight, & no. of original cords

for series number then for [Editor: illegible text crossed out] __n__ cords

& x arcs we shall get a form

similar to S_{n} ^{xb}/_{c}, __c__ being the same

BUT, instead of being alternate they

are in alternate sequences of (n-1) over

& (n-1) under. Z is also the same in

the result.

K_{4} ^{5a}/_{c}

[Caption: rule.]

Does this = K_{5} ^{4a}/_{c}? Yes [checkmark]

127

[Caption: 5 arcs 4 cords]

[Caption: fill in this]

[blank page]

129

6 arcs 4 cords

7 arcs 4 cords

[blank page]

131

8 arcs 4 cords

9 arcs 4 cords

[Un-numbered]

[Page left blank]

133

10 arcs 4 cords

11 arcs 4 cords

12 arcs 4 cords.

[Un-numbered]

[Page left blank]

135

K_{5} 1a/c

K_{5} 2a/c

[Un-numbered]

K_{5} 4a/c

(rule)

Does this = K_{4} 5a/c ? Yes [Tick/check mark].

137

K_{5} 3a/c

K_{5} 4a/c

[Un-numbered]

[Page left blank]

139__ General Principles. __

(1) A r.k. is a knot expressed in

one of the figures of which ^ * the * charac-

ter is shown in vol ii. __ ad fin. __

(2) Every complication can be expressed

as a r.k. but in this form it

may or may not show its least

no. of crossings.

(3) Every group (G) of ^ * r. * knots ~~ such ~~ ^ * e.g. *~~ as ~~ Plaits (P) ~~ or ~~ ^ * and * Torsions (K) has

alternate equal sequences of crossings

over & under.

(4) The sequence defines the group

& must be a (& may be any) measure

of (n - 1) from n - (n - 1) [=1] to

n - 1. Thus the seq. in P is 1 or

n - (n - 1); in K it is (n - 1). These

two groups represent the two ex-

tremes. Between them lie :-

[Un-numbered}

Groups will be in Series with n = 1 + [M ?]

141

n - (n - 2) = 2

n - (n - 3) = 3

& so on up to n - 1.

Therefore any ~~ factor ~~ measure of

(n - 1) may be a sequence number,~~ but ~~ and form a Group, but only

those ~~ figures will be rated in~~ values of n will

which the given

supply figures, in which the sequence

no. selected is a ~~ factor ~~ measure

of (n - 1). Thus the P group with

seq. of 1 will embrace every figure; so

will the K group with seq. (n - 1) for

both these seq. nos. are measures of

(n - 1). But a group "II" with seq.

2 will only embrace the series

in which n is odd (since then

(n - 1) is divisible by 2. Group III

will embrace only ~~ S4, ~~ S4, S7, S10 &

so on; Group IV will be in S5, S9

& so on. And in some cases the

groups ^ *or series of different groups * will coincide. Thus ~~ S2, being ~~

[Un-numbered]

{1 = n - (n - 1)

{n/n n - 1/n - 1

{ n - 1 = n - (n + 1) = n2 - n/n = n - n/n

{= n2 - 2n + 1/ n - 1 = (n - 1)2/n - 1~~ n ~~ Seq. = n - (n +/- 1) covers K & P

n~~ 3 odd n - 1 = II ~~ 3 - 1 = II

4 - 2 = III

5 odd n - 5 - 3 = IV

6 - 4 = II ( [ ? ] )

3 - 1 = II

5 - 3 = II

4 - 3 = I

143

"I"(~~=P~~) appears in S[subscript]2 ~~as~~ with

a sequence n-(n-1)=P[subscript]1, or ~~as~~

with a seq. (n-1)=K[subscript]1 for both

are equal to 1. In S[subscript]3 (factors

of n-1=1,2,)we have "I"(P) &~~2~~"II" coinciding with K.(n-1). and

no others. Thus

Series no (n) (n-1) Factors Sequences

1 0 0 0

2 1=(n-1) 1 __I(P,K.)__

[curly bracket]3 2 1 ~~?~~ I(P)~~4~~ 2=(n-1) __II(K.)__

[curly bracket]4 3 1 I(P)

3=(n-1) __III(K.)__

[curly bracket]5 4 1 I(P)

2 [Circled:II]

4 __IV (K.)__

[curly bracket]6 5 1 I(P)

5 __V(K.)__

[curly bracket]7 6 ~~2~~1 II(P)

~~3~~2 [Circled:II]

~~6~~3 [Circled:III]

6 __VI(K.)__

*?shld reduce to [vertical line]K _{4} ^{9B}/_{27C}-4 then to less*

Parallel Case wld be III

_{10}

^{3B}/

_{27C}reducing to 1K

_{4}

^{ 9B}/

_{27C}& then

*Parallel Knot III*

_{10}---to less

So I

_{2}

^{1B}/

_{1C}reducing to I

_{2}

^{1B}/

_{1C}is also parallel.

*I*[Vertical:

_{2}^{1b}/_{1c}parallel*Cutting line +-+--+-+*

--+--+-+]

--+--+-+

II

_{5}

^{2b}/

_{8c}(before ?rings)

reduces to

1K

_{3}

^{4b}/

_{8c}(after ?rings)

reduces to

6c

which is "A" in Vol ii pp 100-1.

145

n (n-1) Measures Sequences

8 7 1,~~7~~ I (P)

7 VII (K).)

9 8 1 I (P)

2 [Circled:II]

4 [Circled:IV]

8 VIII (K.)

10 9 1 I (P)

3 [Circled:III]

9 IX (K)

and so on.

It seems probable that in

every [?] group except I there is

a law of reduction for knots before

rings. But observe P always = I,

but K=I,II,III,IV ... in succession~~& the~~ sequences vary. The law of__reduction for K is stated in vol vi.__

It was found that II_{5} ^{2b}/_{8c} reduced

to I.K._{3} ^{4b}/_{8c}, & that this reduced

to informal knot of 6c, as shown

on opposite page.

[Blank]

147

It is unsafe to ~~def~~ deduce a

law from this ^*[?G]K* because :K

may be any number, but here

we deal with II only, though K

also is II; hence what is uni-

versal for K ought to be universal

for all groups.

[On the analogy of the K

formula we should have~~II~~_{5} ^{2}/

II_{n} ^{xb}/_{x(n-1)c}=I.K._{x+1}^{(n-1)b}/_{x(x-1)c}

Observe, here the reduction is to an

1.K, & even that is not ultimate.

For x(n-1)c we may write simply

yC as usual.

Although it is convenient to

have P=Group I, & equally convenient to

have K=Groups I,II &c. whenever

equal to (n-1) yet it seems probable

that one must choose between the

- Fallacy. Because a rule applies to
__Torsions__

in every Numeral Group (I,II &c.) it does

not therefore apply to every__[??]__individual

of each group. vi.268.

We have

IV_{5}^{2b}/_{8c}reducing to I_{2}^{5a}/_{5b}

II_{5}^{2b}/_{8c}[Horizontal line] 1K_{3}^{4b}/_{8c}

149

systems & decide that I shall be

single & complete as a Group; II a

complete group & so on. In that

case K would enter into every

group instead of every group entering

into K.

Since K enters into every group

any rule universal for all Ks

is presumably applicable to each

& every of I,II, III &c. But it

is found that the Law of Reduction

needs modification to embrace the

case of ~~K~~ II_{5} ^{2b}/_{8c} reducing to an

1K_{3} ^{4b}/_{8c}. According to the K. law it

should reduce to S_{2} ^{5[?]}/_{5c}, which

it certainly does not, being an ir-

[Vertical Line ? for emphasis?]reducible i.k.of [?6c] ultimately. Of

course the Torsion S_{5}^{2b}/_{8c} is IV_{5}^{2b}/_{8c}

& the other is II_{5}^{2b}/_{8c}

N^{3}/_{2}^{1}/_{3} [?] 1=1 s.2

Combinations of __n__ things __r__ at a time are

[Fomula: __|n__ / __|r__ __|(n-r)__] [In modern mathematical notation this would be

n!/r!(n-r)! where n! = nx(n-1)x(n-2).... e.g. 5! = 5x4x3x2x1 = 120]

[Bracket spanning 3 lines:] Permutations __n__ things __r__ at a time are

n(n-1)(n-2) . . . (n-r+1)

[No of Permutations = n!/(n-r)!]

Permutations of __n__ things all together are [n?] [n!]

Possibilities of __y__ crossings 2^y. [2 to the power y, 2^{y}]

__Notes on Knots. Vol 4.__

__HNGB.__

Ap. 1902.

1

__Irregular Knots,
with alternate crossings.__

__Preliminary Investigations as to the con-
nection (if any) between these and the
Regular Knots.__

—

For purposes of clear reference in any

diagram of a regular knot (2.230 ff) the

*^parts of the* bights, if cut, are supposed to be numbered

from the top by the right from 1 to x.

Thus in the case of S_{2}5b/5c, the numbers

would be as follows:--

[Diagram] | The same plan
can be adopted |

bight can be indicated. In forming i.ks

from r.ks. since more than 1 bt. must

always be cut, the top bt. 2-1 is as-

2.__Reversibility__.

1z 2z 3z 4z

2c (1B knot) I2 2b/2c [checkmark]

3c None

4c I3 2b/4c [checkmark] [pencil hook? knot] ~~I2 4b/4c [checkmark)~~

5c

6c "A" (6c twist.) [checkmark] [pencil 14 2b/6c I failed?]

7c

8c I5 2b/8c [checkmark]

9c

10c

11c

12c Twist Knot. [checkmark]

3

[?] to be one of the cut bts. in every case.__Provisional Rules.__

Let __o__ = an odd no & __E__, an even no. in

the above system. Let [symbol] = joined

(1) __o__ [symbol] __o__ +1 or __e__[symbol]__e__ -1} makes a twist & annuls 1c.

(2) __o__ [symbol] __o__ -1 __e__ [symbol] __e,__ +1} restores original bight.

(3) __o__ [symbol] __o__ + 2x? -1 __e__ [symbol] __e__ [symbol __e__ -2x? +1 } rings (x? = S width.)

(4) __o__ [symbol] __o__+2 __e__ [symbol] __e__ +2} ~~knots~~encloses?/loose end involosing?

/ new C & then cutting of 2 adjacent B outside enclosure.

(5) __o__ [symbol] [symbol?]__o__ +3 either encloses / uncut B

or 2 l. ends involosing? 2 new C.

[library number 39891]

To reverse a twist knot: Kew round

[Oro?] 180 [degrees]; & put bight across from right

to left or [v.v.?] ~~The only twist knots~~

I could reverse were I3 2b/4c = Figure

of 8), "A" & the 12c knot, and ? is this

last [egnel?] to I5 3b/12c ?

It seems probable that twists knots of MGc

are reversible.

I could reverse no T.H. except I3 2b/4c

nor the Weaver, nor I2 6b/6c<be>nor I4 2b/6c

Probably I7 2b/12c is reversible, in which case

a series, [riff?] 4c [wld?] be indicated, x=2

& is always odd.

See I. p. 152-3. Every other twist knot of this

series 3, 6, 9, 12 [etc?] (viz 6, 12, 18 [etc?] and of

the series 4, 7, 10, 13 (viz 4, 10 [etc?] are reversible.

? of the 2z series 5, 8, 11, 14. Probably 8 & 14 are

reversible. Anyhow 15 2b/8c is. Is this a twist knot.

The 2z 8c twist is reversible.

5.

(6) __e__ [symbol] __e__ + 3 encloses 2 l. ends involving

a twist (and [onff?]) or 2 new C.

(7) __o__ [symbol] __o__ +4 } 1 loose end involving 1 new C__e__ [symbol] __e__ + 4 } & an uncut B, or 3 l. ends

of [?.] 2 may make a twist

counteracting the 1 new C

made by the 3rd, or 3 ~~[?]~~[?]

involving 3 new C.

(8) __o__ [symbol] + 5 encloses 2B.

(9) __e__ [symbol] + 5 " 1 B, 2 l.ends.

(10) __o__ [symbol] + 6 " 2B, 1 l.end

(11) __e__ [symbol] + 6 " 2 B, 1 l.end

(12) __o__ {symbol] to any other __o__=__o__' encloses (o'-1)B

(13) __e__ [symbol] __higher__ __o__=o'-2) B & 2 l.ends.

(14) __o__ [symbol] to any higher __e__ = __e'__ encloses E'-2/2 B4 & 1 l.end

6

[diagram] [diagram] 5c [diagram] 8c [diagram] 7c

{[diagram] 5c [diagram] 4c [diagram] 3c

[diagram] 1 [diagram] 6c

[diagram]

[diagram] 2c

7.

(15) __e__ [symbol] to any other higher __e__=__e__' encloses also

e'-2/2 B + 1 l.end.

(16) There is always a loose end when __o__ is

joined to a higher __o__'.

(17) Always 2 l.ends (which may be joined

subject to __o__ rules) when __e__ [symbol] __e__'.

(18) For every single enclosed l.end a twist

must be supplied by cutting 2 B adja-

cent to one another.

Note: A join, for purposes of Description

is supposed to be made by the right (as

the hands of a clock.) But in effect

it makes no difference how it is made.

Crossings must be kept alternate.

.

8.

__Plait or Twist Knots as on p152 [ft?].
of vol 1.__ Their [double underline] reversibiltiy.

The law of reversibilty for these knots

is as follows:-

Let [sigma] = any odd no. of strands >1

Then [sigma]-1 is the difference of an

Arithmetical series of which it is also

the first [terin?], and in this series every

[terin?] represents a reversible Plait (Twist)

Knot with crossing equal to the [terin?].

But except in the case of [sigma]=3 the first

[terum] duplicates a knot in

__n__^

*(not the)*preceding

series, and therefore, when all the series

are being considered from 3([sigma]) onwards,

it is best, except with [sigma]=3, to make

2([sigma]-1) the 1st [terin?].

Thus: Required reversible knots: [sigma]=5.

[sigma] = 5, [sigma] - 1 = 4 = 1 st terin (since the series

alone is being considered.

Are = 4c, 8c, 12c, 16c, 20c . . . . . knots.

(p g follows)

9.

Required: Reversibles of ~~12c~~. 16c.

[sigma] diffces are 2, 4, 6, 8 etc.

16 is divisible by 2, 4, 8 only.

Values of [sigma] are 3, 5, 9. _ 3 knots.

Therefore, except in the case of 2c, of

which there is 1 knot, given y (the

coefficient of c) there are as many

knots as there are measures equal

to 2 or multiples of 2 ^* less than y* which will

divide y, & values of [sigma] = these +1.

Example : Required, reversibles of 24.

24 is divisible by 2 _ [sigma] = 3

[symbol] 2 x 2 _ [sigma] = 5

[symbol] 3 x 2 _ [sigma] = 7

[symbol] 4 x 2 _ [sigma] = 9

[symbol] 6 x 2 _ [sigma] = 13[bracket] 5 = AND.

These knots are not necessarily on /z.

__Determination of z.__

In [sigma] = 3, z = 1 except in the case of

2c, 2 + 6 (or a multiple of 6)c, in which

case z = 2. Every 3rd knot is on 2z.

(p 10 follows) 1st = 1 ([sigma]-1)c

10.

[sigma] = 5. (4c z=1) [tick]

~~ 8c z = 3~~ 8c z=3 [tick]

~~12c z = 1~~ 12c z=1 [tick]

~~10c z = ~~ 16c z=1 [tick]

~~20c z = 1 ~~ 20c z=1 [tick]

~~34c z = ~~ 24c z=1 [tick]

~~30c z = ~~ 28c z=3 [tick]

~~42c z = ~~ 32c

Every 5th knot is on 3z. 1st 3z = 2([sigma]-1)c

[sigma] = 7. (6c z = 1)[tick]

12c z = 1 [tick]

18c z = 4

24c z =

30c z =

36c z =

42c z =

48c z =

54c z =

60c z = 4

Every 7th is on 4z. 1st 4z = 3 ([sigma]-1)c

11.

Examina__tion__ of Sz 5b/5c.

[diagram][Caption: 1 2 3 4 5 6 7 8 9 10] Let B 10.1 be cut:

Join 1, 2. A twist is

formed [annabling?] a

crossing thus:-

[diagram] [Caption: 10 [2/1?] 3 4 5 6 7 8 9] If 10 & 3 are [symbol] no

new C is made to com-

pensate.^but I2 6/u results If one of

them is taken under & over the new B

2 new C result. Let 3 be taken under & over to 10.

[diagram] [Caption: 3 4 5 6 7 10] There are now 6C on

2 cords (S2 6b/6c) which

can be reduced only

by cutting 2 adjacent

B. Cut 4, 5, & 6, 7. &

join 5, 6, to make a twist.

[diagram] [Caption: 7 6 5 4] = [diagram] [Caption: 7 4] There are now 5C again. Join 4, 7.

Result: Original Twist.

[vertical line in left margin]The law for determining z is this:-

In any series [symbol], (where [symbol]=no.

of strands), the reversible having

([symbol]-1)2/2 c, and those having this

no. of c ~~multiplied by [symbol] or any~~+ [symbol] ([symbol-1])c or + any multiple of

[symbol] ([symbol]-1) c ~~multiple of [symbol]~~ are on [symbol +1/2 z

and the others are on 1z.

[end of vertical line in left margin]Thus: Required reversibles on 1z, 8c.

8 is divisible by 2, 4

Values of [symbol]= 3, 5

[symbol]=3. (3-1)2/2 c = 2 c~~[symbol] x 2 = 6~~

[symbol] ([symbol]-1)c = 3 (3-1)c = 6c.

2c + 6c = 8c. But this then is not on

1z, and is no good.

[symbol] = 5 (5-1)2/2 = 8c. This then is no good.

Ans: No ^[symbol] 8c twist reversible on 1z.

13

Suppose however 6,7 & 8,9 had been cut

& 7 [symbol akin to > indicating ‘join to’or ‘joined to’ or ‘joining’] 8, 6 [>] 9. Then also the re-

sult is the same, viz Original Knot.

Next begin by [>]1 & 2 as before, &

join 10,4 & 3,5, keeping the C alter-

nate where the new C is made, thus:-

This knot results which

is a new i.k. of 5C,

viz the Timber Hitch,

which may be represented

It is found by experiment that this

same T.H. can be obtained by the

following processes:-

1 [>] 2, 3 [>] 5, 10 [>] 4

1 [>] 3, 2 [>] 8, 10 [>] 9

1 [>]5, 3 [>] 4, 10 [>] 2

1 [>]7, 6 [>] 8, 10 [>] 9

1 [>] 9, 7 [>] 8, 10 [>] 6,

1[>] 8, 4 [>] 6, 5[>] 7, 10 [>] 9.

__Reversibility of Links__

[Caption: Revolve top of 1 down * ^ forwds*, 2 steady.]

{Caption: Revolve top of 1 down *forwds*, 2 steady, revolve top of

3 up.]

And generally when thus ~~regularly~~ * ^ placed at*~~linked~~ first, revolve alternates.

If the links are thus:-

[Caption: simply revolve bottom

of 2 up.]

In all * ^ these* cases the links can ~~only~~ be

connected only in one way

15.

Hence there are only 4 nos, to which

1 cannot be joined, viz itself 1; 10, re-

storing original B; 4, necessitating

a ring & [symbol for ‘therefore’] 2 cords; & 6, do. In

the case of 4 this is obvious, & it is

easy to demonstrate in the case of

6. For join 1,6.

Then if 3,4 are joined there is a ring.

If none of 2, 3, 4, & 5 are joined to one

another there will be 4 new crossings

& it is only possible to make 2 twists.

Therefore 2 of them must be joined.

It is immaterial which 2 are joined.

Join 2,3. If 4 & 5 are not now joined

there must be 2 new C &

[symbol for therefore] 2 twists. But if 2

twists are made by joining

7 8, 9 10, nothing is left to

__As to Iz Reversiblesof the series 2, 8, 14 &c.__

20,24, 28 appear in other series, but with

regard to 2 it is certain it cannot be

on Iz except as an unguarded twist. With

regard to 14c

as a twist plait knot on any but 2z. With

regard to 8c it cannot appear as do on any

but 2z or 3z, but it does appear as I

_{5}

^{2b}/

_{8c}.

Now of what series does this knot form a part, and can it include 14c ?

Values of x, with y = 14, are

We have

*^ (z=1)*I

_{3}

^{7b}/

_{14c}, I

_{8}

^{2b}/

_{14c}(z = 2)

(z = 1) II

_{3}

^{7b}/

_{14c}(VII

_{8}

^{2b}/

_{14c}reducible)

There are therefore 3 knots.

Note that I

_{2}

^{2b}/

_{2c}, I

_{3}

^{2b}/

_{4c}I

_{5}

^{2b}/

_{8c}are

reversibles. I don’t feel sure that I

_{4}

^{2b}/

_{6c}(2z) is

not so. But the only two regulars on 1z are

I

_{3}

^{7b}/

_{14c}& II

_{3}

^{7b}/

_{14c}

Examination shows conclusively that I

_{3}

^{7b}/

_{14c}

is reversible.

17

join on to 4 & 5. Therefore join 4, 5.

Now if 10 is joined

to 9, 8 must be

joined to 7 & 2 C

are annulled. [sign for therefore]

10 must be joined

to 7 or 8. It can-

not be joined to 7 or a ring will be

formed. [sign for therefore] join 10 to 8 & 7 to 9. But

2 new C. will result. Therefore 1 cannot

be joined to 6 Q.E.D.

It appears from the foregoing that

provided a twist & a l. end are

formed, & the l. end joined to the

remaining end, the T.H. will always

be formed from S_{2} ^{5b}/_{5c}.__Tait__ says there are only 2 1 cord

knots of 5c, & no doubt S_{2} ^{5b}/_{5c} and 5C T.H. are those knots.

[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. 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.

19.

It is possible to represent S_{2} ^{5b}/_{5c}

in long form thus:-

[Caption: Now join1, 2;

4,10; 3,5

The result

is obviously

the 5C. T.H.]

It might be thought that:-

obtained by [symbol resembling>] 1 8, 9 10, 4 6, 5 7

(turned upside down) was

a new k, but it is

clear that by untwisting

the bottom twist, a new

twist is added to the top, & the T.H. results.

Evidently 2 * ^ or more* unguarded twists at each

end are unstable & can be added on

to the reverse end from either until the

guard is reached. By a guard is meant

a cord passing through a bight above

the twist or below it, thus:-

[Blank except for the following lines in pencil]

And “A_{1}” ii. 100 – 1.

I [wld?] I think [srihoh?] Reef & Granny

21.__Examination of I. K. of 6C.__

There is no r. k. on one cord of 6 C ex-

cept the Twist, because the factors of

6 are 1, 2, 3 & 6 giving __n__ as 2, 3,4 & 7,

all of which except 7, the twist, have

measures in common with 6. (the same

applies to knots of 1 & 2 C.)

It is known that a ~~[i?]~~ k. of

T. H. form corresponds to every no. of C.

of 3 & upwards (that of 3 being a~~n~~.

r. k. S_{2} ^{3b}/_{3c} & that of 4 being S_{3} ^{2b}/_{4c})

Hence there is a T. H. of 6 C. Further

the Granny, the Reef & the Weaver

(or Bowline, or Crossed Running Knot)

are other knots of 6 C. on one cord. __Tait__

says there are only 4 such knots.

Problem: To derive these 4 ks. From

r. ks. of 6 C.

From S_{2} ^{6b}/_{6c} the Reef is obtained

by joining 2,4; 1,3; 5,6; 12,7. In the

form of 2 dissimilar overhands on

and endless cord. From do. the Granny

No Text to transcrive on this page

23

on 1 cord is obtained by joining 1.b;

12.7.

from S4 [2b/6c] the 6.c T.H arises by ~~a~~

untwisting 1, 2, 3, 4; 1 & 2 being joined

and 3 & 4 joined through their bight.

from S3 [3b/6c] the Weaver G. can be

made by joining 1,4; 3, 5; 2, 6; of

which 3,4 are untwisted while 2,6

give the new C.

The T.H. is made thus

Compare p19.H for the

1st form of the T.H. here

shows.

But can S2 [6b/6c] supply the T.H. & the

Weaver as well as the Rief & Granny?

Page 24 - Only one Undocumented figure in center of page.

25

It is possible to represent S2 [6b/6c] in

long form thus: --

Proceed on analogous

lines with the plan

on p 19 & join 1,2.

There are now 5c & a new one must

be made. This can be done by

joining 3,5; 12,4 as before, when

the T.H. on 1 cord with 6C. results.

In both cases 1,2; 3,5; & 2,4 were

joined to produce the T.H.

[Therefore] 3 out of 4 possible [__ks??].

have been obtained from S2 [6b/6c]: can

the Weaver be got in some similar

way?

It can be done as follows: Join 1,2;

3,4; 4, 7; 6, 10; 11,12. The stages are as

PAGE 26 - No Text to transcribe.

27

follows: --

In the above process 8,9 is the only

bight not cut. The operation is [________?] com-

plicated that it is going rather far to

say that the one knot is or can be

derived from the other. On the other

hand the T. H. is fairly derived both

on pp 19 & 25. It would be curious

if in the case of any knot in [__________???]

whether on 1 cord or more, a single

cord T.H. with the [_______??] no. of C as

the knot could always be produced by

joining 1,2; 3,5; 4 & [alpha??]. Try this. Proved

in case of Sz [5b/5c], Sz [6b/6c].

[under first fraction "z=1", under second fraction "z=2"]

Page 28 - Penciled note in lower right corner - "[2?] K regular knot."

No other text on page.

29

Join 1,2, 3,5; 2,4; in [J"S?"]2

There is still no 5 till we get to S2 [3b/3c]

& this is really a T.H. already. But

if the process is applied, result is the

reflection of the knot. (z=1)

Sz [4b/4c] (z=2) Result, Figure 8 on 1z

(a T.H.)

Sz [5b/5c], 5c T.H. (z=1) p 19.

Sz [6b/6c] (z=2): 6c T.H. (z=1)

Sz [7b/7c] (z=1): 7c T.H. (z=1)

Sz [8b/8c] (z=2): 8c T.H. (z=1)

Therefore this general rule may be

laid down: An irregular knot, the

Timber Hirch of the same no. of C. as

any knot in Sz with more than 2c, can

be obtained *on 1 cord,* [___________?] other methods by

joining 1,2; x,4; 3,5, and in the case

of Sz [3b/3c] it is merely another form of

the r.k. and in the case of Sz [4b/4c] it is

Page 30 - No text to transcribe.

31

another form of Sz [2b/4c], an r.k. so

that in these two cases the r.k. & i.k.

overlap. In other words there is no

i.k. of 3 or 4c on 1z, but these

two knots can be displayed as r.k. or

T.Hs.

Query: Does the T.H. come within the

definition of a r.k.?

No: I think not, except of course

Sz [3b/3c], and S3 [2b/4c] re-arranged.

Twice the no. of knots in any S is

infinite, & Sz contains knots of all

possible C, [therefore] all the T.Hs. can be

derived from the S2 r.ks., & this disposes

of one class of i.ks., the rule above

laid down applying to all of them.

Query: Can the rule on p29 be

stated in more general terms on the lines

of p17.m? This can be said that

in any Sz knot if ~~a twist~~ adjacent parts

lying in two bights are joined, & 2 other parts

Page 32- No text to transcribe.

33

joined over a l. end, & the l. end

joined to the remaining free part, that

the rule holds good?

It has been shown so in S2 __5b__ 5c -

Take S2 __6b__ 6c where Z = 2.

Join 10,9; 4,6; if 5

is joined to 11, there still

remain 7 & 8 to be dealt

with. But they must not

be joined or a twist will

result.

Can it be said that if

the twist is made, & the parts on each

side of it joined, & the parts on

each side of these again joined ~~to~~ through the new bt

make a bight

a T.H. will result? No. There

remain uncut bights.

The rule however may be stated

more generally thus:- Any knot in S2

with 3 or more C. can be converted into

an i.k. in the form of a T.H. * ^on 1 cord* by cutting

Nothing to be transcribed on this page (34)

35

any 3 adjacent bights, joining 2 adjacent

parts, each in a different bight (thus

making a twist) joining the part farthest

away from the twist * ^ [sorthing?] the half circumference* to the part nearest

to ~~it~~ *^ the twist* on the same side of it, and

joining the remaining *^ outside* parts, *^ on either side of the uncut bight or bights.* keeping

the crossings alternate.

(It will be noticed on p13 that

all the 3 bight methods deal with adjacent bights.)

With regard to the formation of the

granny (pp21-3) it will be noticed that

every alternate 2z knot in S2 has op-

posite bights on different cords. Such knots

are __2b__ 2c, __6b__ 6c, __10b__10c &c. while the knots__3b__ 3c__4b__ 4c, __8b__ 8c, __12b__ 12c, have them on the same.

Hence it may be laid down as a general

rule that: An i.k. ~~ of the granny type~~

may be formed with the same no. of C. on

1z from every knot in S2 (except __2b__ 2c) with

Nothing to be transcribed on this page (36)

37

an even no. of C. not divisible by 4 by * ^cutting*

joining the nearest sides of * ^ two* opposite

bights. (In the case of __2b__ 2c an i. twist

results.) Also in those even C. knots

divisible by 4, an i.k. on (2+1) = 3z

by the same process, and on (2+N)z~~for every pair~~ when N pairs of opposite

bights are joined in such knots.

With regard to the formation of the

reef knot, it may also be laid down

that “ An [i?].k. may be formed with

the same no. of C. on 1z. from every

knot in S2 (except __2b__ 2c) with an even

no.of C. not divisible by 4 by ~~joining~~

cutting 4 adjacent bights, joining 2

adjacent parts of adjacent bights, joining

an odd to an odd & and even to an even

part (in the same __1__ 2 circ. as the twist)

& joining the ~~remaining~~* ^2 outside* parts on either side

of the uncut bight or bights.

For example, take S2

__10b__10c :-

Nothing to be transcribed on this page (38)

39

so with regard to the weaver, it may

be said, From any S2 knot with even

C. indivisible by 4, an i.k. may be

formed by cutting 5 adjacent bights, say

x – 1 to 8 – 9, & joining 1,2; 3,4; 5,7; 6,9;

& x,8;

It appears probable that for every

additional bight cut, an additional i.k.

could be formed in the case of S2 knots

with even C. indivisible by 4, in which

case it might be said that the no.

of irregular knots capable of being

formed from such knots is B-2, giving

4 for S2 __6b__ 6c, 8 for __10b__ 10c, 12 for __14b__ 14c & so on.

Nothing to be transcribed on this page (40)

41

2 B must be cut to get the first

i.k., & one B must be left ap-

parently uncut.

Tait says there are 8 knots with

7c. Two of these are the r.k. S2 __7b__ 7c

& the 7c. T.H. What are the other 5 on

1z? And why is it that only 1 i.k.

can be formed from S2 __5b__ 5c & 7 i.ks from__7b__ 7c ?

According to Tait’s series the knots

are 5c 2k

6c 4k

7c 8k

The series 2,4,8 may be continued 16,

32, 64, & so on, ~~or it may be the series~~

but it seems improbable that this should

be so. If however it is so it smashes

the B-2 theory above.

42

[Caption: What is this]

Cf V. 203.

If the ring in the knot opposite were

a metal ring & the cord cut at the

twist, this knot would form rather a

good mooring knot of the Larks Head

order, thus:-~~B~~ Either end ~~will~~ *^ may*

[run?] *be strained* & I believe

this to be a new~~knot~~ *^ hitch*. If __a__ were

taken through the

left hand bight it would form an ordinary

running knot. If through the right hand

bight, a figure of 8. If instead of __a__ a

bight were put down, it might serve as a

signal hitch &c. since on pulling __a__ out

& straining __b__, the hitch runs out.

43

In * ^ any knot of* S2 the effect of a twist & join

over it (2 adjacent B cut) is to

reproduce the knot preceding it

in the series.

Try effect on S3 __4b__ 8c

Here an i.k.of 7c on 2z is produced.

a circle interlaced with ~~an [ounce?]~~. By cutting the twisted bight

bend

& the circle bight & joining the

following plait is obtained on 1z, 7c.

This cannot be joined up & made into

Nothing to be transcribed on this page (44)

45

an r.k.without obliterating

2c.

Of course this plait could be

obtained directly from S_{3} ^{4b}/_{8c} thus :-

Here on a 1z k. 2 adjacent bights

have been cut & the alternate parts

x & 2, 1 & 3, joined without alternate

crossings, producing an i.k. of

c – 1 crossings = 7 on 1z.

Therefore among knots of 7c are

S_{2 } ^{7b}/_{7c}, the T.H; & this i.k. plait.

Presumably every 1z. k in S_{3}

could be similarly made to produce

an i.k. on 1z. with 1c. less than

the number. (S_{3} ^{5b}/_{10c} & S_{3} ^{7b}/_{14c} can.)

Nothing to be transcribed on this page (46)

47__Question of rings.__ See p91.

A ring is formed in the following cases:-

S_{2}^{2b}/_{2c} no 1 & no 4 (exists as such)^{3b}/_{3c} 1,4.^{4b}/_{4c} 1,4; - 1,8 (exists)^{5b}/_{5c} 1,4; 1,8^{6b}/_{6c} 1,4; 1,8; - 1-12 (exists)^{7b}/_{7c} 1,4; 1,8; 1,12; & so on from

which it appears that any part with numeral __x__* ^y* joined to any other

part with numeral __x__* ^y* + (M2 X n-1)

gives a ring.

S_{3}^{3b}/_{3c} 1 + ~~12~~ *^ 6* (exists)^{4b}/_{8c} 1 + 6^{5b}/_{10c} 1, 6;

Nothing __in ink__ to be transcribed on this page (48) only some rough pencil jottings of formulae

49

(S_{3} cot^{d})^{6b}/_{6c} 1,6; - 1,12 (exists)^{7b}/_{14c} 1,6; 1,12;

^{8b}/_{16c} 1,6; 1,12;^{9b}/_{18c} 1,6; 1,12; - 1,18 (exists)^{10b}/_{20c} 1,6; 1,12; 1,18. & so on, from which

it appears that any part with a

numeral __x__*^ y* joined to any other part

with numeral __x__*^ y* - (M2 X ~~S~~ n) – 1 gives a

ring.

May the formula be stated generally

thus:- Rings are formed when __x__ is

joined to (M2 X n) – 1? (M2 being any

multiple of 2)

Take S_{4} ^{6b}/_{18c} . Here 1 + 4, 12 + 5 give

rings. According to the rule, 1 + (M2 X 4)

gives any rings, i.e. 1 + 8, 1 + 16 &c. Here 1 + 8 does

give a ring. 1 + 16 does not exist. The rule

holds good, but is not comprehensive enough.

No text __in ink__ to be transcribed on this page (50) only some rough pencil jottings of formulae under pencilled sketches

51

1 + 4 points to S_{2}, & it is to be

noted that S_{4} ^{6b}/_{18c} is on 2z -

In S_{4} ^{9b}/_{27c}, 1 + 4 also gives a ring

though this is on 1z.

It looks as though the rule

applying to S_{2} applied also to S_{4}

and that the rule ought to be stated

thus:-

Let P be any prime factor of __n__ -~~Then~~greater than unity. * ^ (It may equal n.)* Then rings

are formed by joining x to

(M2 X P) – 1.

Test this with S

_{6}

^{11b}/

_{55c}.

__A priori__

since P may be 2 & 3, one would

expect that

+ 1 + (1 X 2 X 2) – 1 = 1 + 3

1 + (1 X 2 X

__3__) – 1 = 1 + 5 no

1 + (2 X 2 X 2) - 1 = 1 + 5 no

[Caption: Rule breaks down] 1 + (2 X 2 X

__3__) - 1 = 1 + 11 no

+ 1 + (4 X 2 X 2

*^ 16*) – 1 = 1 + 15

1 + (4 X 2 X 3) – 1 = 1 + 23 non existent

1 + (6 X 2 X 2) – 1 =1 + 23 [ditto marks] [ditto marks]

1 + 12 ring = 1 +

No text to be transcribed on this page (52)

53~~In S~~_{6} ^{11c}/_{22c} the only rings that can

be formed is by adding 1 & * ^ 1 + 6*, 2, 2

& 13 &c. i.e. x to x + 2x.

Has it anything to do with the

number of times ~~k~~ 2 * ^ or K2* will go into C?

Thus there are here 22c, & the added

number is (^{22}/_{2} + 1) = 12.

But this would not suit S_{2} where

the C are often odd.

In S_{6} ^{9b}/_{45} it is obvious that

1 + 6 gives a ring.

It appears that in joining S_{6}

1 + 6, 1 + 12, * ^ 1+18,* & so on give rings, i.e.

1 + Mn.

The same may be said of S_{2}

when z = 1.

In all the series the part

numbered 1 regains the circumference

at 2n & 4n & 6n counting round from

the right & counting 1 for each part

gone by. The question is how to state

No text to be transcribed on this page (54)

55

this fact as a rule so as to em-

brace the cases where it emerges

e.g. a 2nd time in between its first

re-appearances.

It depends on the no. of bights.

of bights all is well

Take [I?]

_{4}

^{6b}/

_{18c}12 parts

Rings are formed at

1+8(2z)

1+16(4z)=1+, 4n-12

Take S

_{6}

^{11b}/

_{55c}

Rings are formed at

1+12(2n)=[space]1+12

1+24(4n)=1+4n-22=1+2

1+36(6n)=1

_{1}+6n-22=1+1

*4*

1+48(8n)=1+8n-22=1+26

non existent.

Therefore the rule may be stated

thus: Rings are formed when any

part numbered

__^__~~x~~

*is joined to*

__y____^__~~x~~

*y-*

-1, or, where M2 exceeds 2B, to

Usually __y__=1 to begin with, in which

case this practical rule is: Dont

join 1 to (M2×n) or to (M2×n)-2B.

& no M of 2 need be considered which

gives a value for the second part

higher than 2B, because such parts

do not exist.

57

*y*+ (M2×n)-(1+2B).

But in the case of [S?]

_{z}with 1

_{z}, a

ring is formed by joining 1+2, and

not where z=2. This apparent

exception is met by applying the

2nd part of the rule.

B=

^{C}/

_{n-1}, ∴ 2B=

^{2C}/

_{n-1}

Hence the Rule of Rings may be

stated thus:-

[Blue vertical line crossed out down to end of paragraph] A ring is formed when any

part numbered

__^__~~x~~

*y*is joined to

any other part the number of

which =

__^__~~x~~

*+(M2×n)-1, or to*

__y__any other part the number of which

=

__^__~~x~~

*+(M2×n)-(1+*

__y__^{2C}/

_{n-1}), and

for

^{2C}/

_{n-1}may be substituted 2B.

Therefore in forming i.ks. with 1z

it is necessary to set out all the

parts existing comprehended in this double

formula as prohibited joins.

Since x=2B=^{2C}/_{n-1}, more simply

thus: Don't join 1 as forbidden on

p 56 nor (~~when the joined no. is even~~)

must it equal ~~x+(M2×n)-2 or~~ (M2×n)-2;

x+~~nor when odd, to~~.

x+(M2×n) or M2×n

But x is always even: ∴ these last 2 ex-

pressions can never be odd. x+(M2×n ^*-2* can

not exist. But (M2×n)-2=M2 ? No. no.

59

With regard to this rule it is

to be noted that 2 parts cannot be

joined without 2 other parts being

joined. Therefore if only 4 parts are

to be joined (the minimum number) i.e.

if only 2 bights are cut, the joining

of the complements of these parts

would necessitate the rule being

broken in the 2nd join. In such

a case therefore, where only 2

bights are cut & y=1, to the

working rule on p.56 must be

added this:-

The no. to which 1 is joined must

not come within the formula on p.

56, nor must it, when

__even__, be such

that it equals

x+(M2×n)-(2+

^{2C}/

_{n-1});

__odd__,

to x+(M2×n) or to x+(M2×n)-

^{2C}/

_{n-1}.

[vertical line]

broken in the case of x, where x

=the part numbered 2B.

61

The complete working rule for

joining 1 where 2B are cut (which

may or may not be adjacent is):-

Don't join 1 to

M2×n

(M2×n)-2B

(M2×n)-2

M2 being any multiple of 2, which it

is no good to consider when any of

these 3 expressions represent a number

equal to or higher than 2B(=

^{2C}/

_{n-1}).

If such joins are made, rings will

result.

M2×n may be written "M2n"

when the 3 prohibitions are

2Mn or M2n

2(Mn-B) or M2n-2B

2(Mn-1) or M2n-2

Unfortunately the rule breaks down

since S

_{6}

^{12b}/

_{60c}shows that there is

only 1 exclusion. But it may perhaps

hold when z=1 to begin with.

63

When z exceeds 1

rule may be stated thus:-

The no. of ring-giving joins is

^{B}/

_{Z}-1 (By ring-giving is meant

that the no. of Z is increased.)

Cases where Z exceeds unity

S

_{2}

^{4b}/

_{4c}1 exclusion

^{4b}/

_{2}-1=1

viz 2n

S

_{2}

^{6b}/

_{6c}2 ex.

^{6b}/

_{2}-1=2, ^

*4*2n, ^

*8*4n

& so on.

S

_{3}

^{3b}/

_{6c}0 ex

^{3}/

_{3}-1=0

S

_{3}

^{6b}/

_{12c}1 ex

^{6}/

_{3}-1=1, ^

*6*2n

S

_{3}

^{9b}/

_{18c}2 ex

^{9}/

_{3}-1=2 ^

*6*2n, ^

*12*4n

& so on -

S

_{4}

^{6b}/

_{18c}2 ex

^{6}/

_{4 2}-1=2 ^

*4n-2B*no 4, ^

*2n*no 8

65

__Rule of Rings__

*p91*

In any [

ways in which

be cut and have their cords rejoined

so as to increase the existing value

of Z by forming an additional Z

(or "ring")

*cannot exceed*

^{B}/

_{Z}-1,

(or

^{C}/

_{Z(n-1)}-1),

A ring is formed when any part

in such a case numbered

__y__is joined

to any other part the number of

which is y+M2n -1 or y+M2n-(1+2B)

(= y+M2n-(1+

^{2C}/

_{n-1})) or y+M2n-3

(as in this case the complementary part

of y's bight would

*come within*the

first part of the rule.), provided always

that the number of cases is limited

to

^{B}/

_{Z}-1, and that the values of

the part-numerals obtained by this

formula are the first (

^{B}/

_{Z}-1) one or

more in order of arithmetical greatness.

67

Where 1 = y, exclusions under

this rule are:-

M_{2n}

M_{2n} - 2B

M_{2n} - 2

where M is any multiple: but subject

to the rule that there are no

more such values than B/2 - 1, and

further that these are taken in order

of arithmetical magnitude; and

of course, since x = 2B, no value

can equal or exceed 2B.

Test this rule with S5 12b/48c, z = 1.

B/z - 1 = 12/1 - 1 = 11.

M_{2x}= Mx10 = ~~10, 20, exceeds ~~__10__, __20__, [30 exceeds 2B]

M_{2n} - 2B = ~~M2~~ M x 10 - 24

= __6__, __16__, [26 exceeds 2B]

M_{2n} = 2 = __8__, __18__, [28 exceeds 2B]

[therefore] the prohibitions are__6__, __8__, __10__, __16__, __18__, __20__

69

This test is found to confirm the

rule.

Next take S_{6} 8b/40c. z = 2

B/z - 1 = 8/2 - 1 = 3

It is obvious that

there are 3 cases = 4, 8, 12.

M_{2n} = M_{12} = 12, [24 exceeds 2B]

M_{2n} - 2B = 8, [20 exceeds 2B]

M_{2n} - 2 = 10, [22 exceeds 2B]

Of these 12 & 8 are right; 10 is wrong.

Therefore strike out the (M_{2n} - 2)

part of he rule & state the exclusions thus

M_{2n}

M_{2n} - M'B

where M & M' stand for any multi-

pliers which may be the same or

different.

The rule thus stated meets the

cases of both S_{5} 12b/48c & S_{6} 8b/40c. Test

it further.

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