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Charles HAMEL – 2011 June Page 1 sur 36 

Release V1.3.1 

QUICK AND DIRTY EXPLORATION OF A PARTICULAR 

SERIES OF THK 

The origin of this quick and dirty trail blazing stems from a question Dominic TAYLOR posed 

to me ( Thank you Dear Dominic for the inspiration ) : 

[open quote] 

am interested in finding other sequences of TH like the 5x3/7x5/9x7/11x9 

etc sequence. Are there any others What makes it happen 

[end quote] 

As at first, quite in a hurry, a few hours from driving to Brittany and preparing the trip I 

somehow misunderstood and answered about enlargement so Dominic added : 

[open quote] 

They are just a family of 'starter' knots with cordage routes which are 

obviously related. 

5x3 goes to O2U1, 7x5 goes to O3U1, 9x7 goes to O4U1 &.c. 

What i can't see myself is any other "starter families" of sobre or casa 

THK. [end quote] 

First a remark I cannot do without making : 

sobre and casa linked by ‘or’ may let the readers think that both words and concepts can be 

equated one with the other. That is not so. 

sobre is a Spanish word used to state that the table of Half-Periods Coding is ‘economical in 

effort’ and specifies a lower number of moves for ‘UNDER’ passages of the cord while making 

the knot which crossings are predominantly ‘OVER’. 

A given THK can have two forms : either one with a maximal number of ‘OVER’ crossings or 

one with a maximal number of ‘UNDER’. 

The so called ‘sobre’ is the form with a maximum number of ‘OVER’.(or the one beginning 

with an OVER in case of equality) 

Casa (House –-House why on earth -) is a rather queer mannerism –my opinionfavoured, 

God may know why but not me, by one Tom HALL (hiding behind that pseudonym 

is one HICKEY) ; a search for originality or personal fame 

Anyway I find it useless beyond its gimmicky queerness even if it is blindly followed by too 

many without it being given much logical analyse. 

It is my considered opinion that this term that too many persons make a synonym for Turk’s 

Head Knot (THK) is better left forgotten. 

Casa could barely be acceptable as a label for a O1 U1 or U1 O1 coding as many knots 

besides THK can have it on their cordage route, but alas the so-called CASA == O1-U1 is a 

mistaken equivalence as exist some O1-U1 coding which nevertheless DO NOT comply with 

the COLUMN AND ROW coding of the THK.



This shows what I consider to be the ill thought aspect of the equivalence CASA == O1-U1 

THK . 

So sobre and casa are two different concepts in two different logical planes so one may not 

equals cabbages with kings. 

Sobre : the ratio of OVER and UNDER in a THK 

Casa : a very fuzzy concept without any interest that sometimes, according to who is using it, 

refers to the coding pattern, sometimes is taken as a synonym of THK, sometimes as both. 

Polluting junk ! 

Anyway a THK can only be O1-U1 (U1-O1) due to the Row AND Column coding ; any other 

coding than O1-U1 (U1-O1) on a THK cordage route makes it something entirely different 

from a THK. 

( Head Hunter K, Gaucho K, Fan K , Aztec K , AZTEC-FAN K, Samuel K for example are 

made on a THK cordage route but their -all different- coding are COLUMN coding but NOT 

Row AND Column coding ). 

Yet many persons ignore this FACT which has its roots in mathematical criterion and not in 

whimsical personal opinion. 

Let ‘little I’ be dogmatically clear –based on mathematics criterion- : 

THK should no be called casa knot or by an other name than THK or Regular Cylindrical 

Knots (single strand) O1-U1 and its coding should not be called other than COLUMN AND ROW 

coding which says that it can only be O1-U1 (U1-O1) 

IMPORTANT for common ground building : in this document I use 

*** the horizontal mandrel frame of reference used by Schaake. 

Knot edges or Bight edges are on the LEFT side and on the RIGHT side of the mandrel. 

*** the sobre form of the THK 

xxxxxxxxxxxxxxxxxxxx



The series proposed by Dominic TAYLOR: 

5L 3B-----7L 5B-----9L 7B….. 

The number of LEAD of knot with rank N is the number of BIGHT of knot with rank (N+1) 

SPACING = 2 STEP = 2 (Definitions are coming later on.) 

I never read or heard (books, web pages, forums, private mail, conversation) anything 

whatsoever on *this* series of THK (I prefer the term ‘series’ to ‘sequence’ ; series as in the 

mathematics TAYLOR’s series or MACLAURIN’s series where there is a ‘development’ of the terms in 

the series) and Dominic TAYLOR’s question is the first occasion that brings my attention to it. 

This is no wonder, first as I certainly may not allege to have read all that has been written 

and second considering the level of mathematical interest or rather lack thereof, appetite for 

theoretical exploration or rather the endemic absence of it and general level of intellectual 

curiosity in the world of knots tyers as can be evaluated from what can be seen in forums 

and personal web pages. 

I have known for years about two terms in the proposed series : the 7L 5B and 9L 7B THK. 

They were labelled “quick start” knots by Ron EDWARDS who is IMO a far better and 

more interesting writer than any other author on THK can ever hope to be considered to be 

(save Georg Schaake and John C.Turner who must come first in any list). 

Quick start : there is a long run of HP with only OVER crossings before the first appearance 

of an UNDER. For practical purpose the run of OVER is considered “broken” by the first 

appearance of an UNDER crossing. 

7L 5B : 21 OVER 9 UNDER ; the first 9 crossings are OVER (9/30 = 30 %) 

9L 7B : 40 OVER 16 UNDER ; the first 16 crossings are OVER ( 16/56 =28.6 %) 

Almost one third of the knot is made before making the first UNDER. 

For professional ‘braiders’ this is of interest as “production time” is an important factor in 

pricing. 

Along this particular series successively appear, knot after knot, consistent smooth and 

progressive runs of ‘O’ : O1, O1, O2 O1,O2,O3, O1, O2, O3, O4, 

O1, O2, O3, O4, O5….. 

Mind that you note there are TWO perspectives to consider in the determination of the 

highest numbered Half-Period (HP) in which the maximum of OVER is attained : 

either the last HP beginning with O max or O max-1 without any U (‘isolated’ or ‘pure’ OVER) 

The longest run of HP with pure ‘O’ without any U. 

or the last HP with O max – U1 (longest run of HP beginning with ‘O x ’ ) could even be the 

first HP with O max – U1 if you decide ‘breaking point’ is important in there : the one of 

interest for the professional ‘braider’ and corresponding to the ‘quick start’. Longest run of 

‘O’



17L 19B spacing =2 step=2 

B>L 

19L 17B spacing=2 step=2 L>B 

HP1 Free Run 1 Wrap 

HP2 Free Run 


HP4 O1 

HP5 O1 

HP6 O2 

HP7 O2 

HP8 O3 

HP9 O3 

HP10 O4 

HP11 O4 

HP12 O5 

HP13 O5 

HP14 O6 

HP15 O6 

HP16 O7 

HP17 O7 

HP18 O8 

HP19 O8 

HP20 O8 

HP21 O8 HP21 O8 

HP22 U1 – O8 

HP23 U1 – O8 

HP24 U1 – O1 – U1 – O7 

13L 17B spacing=4 step=2 B>L 




HP4 O1 

HP5 O1 

HP6 O2 

HP7 O2 

HP8 O3 

HP9 O3 

HP10 O3 

HP11 O3 HP11 O3 

HP12 U1 – O3 

HP13 U1 – O3 


HP2 O1 

HP3 O1 

HP4 O2 

HP5 O2 

HP6 O3 

HP7 O3 

HP8 O4 

HP9 O4 

HP10 O5 

HP11 O5 

HP12 O6 

HP13 O6 

HP14 O7 

HP15 O7 

HP16 O8 

HP17 O8 

HP18 O9 – U1 

HP19 O9 – U1 

HP20 O8 – U1 – O1 – U1 

17L 13B spacing=4 step=2 L>B 


HP2 O1 

HP3 O1 

HP4 O2 

HP5 O2 

HP6 O3 

HP7 O3 

HP8 O4 – U1 

HP9 O4 – U1 

HP10 O3 – U1 – O1 – U1



It is an immediate observation that : 

--- this series is of ODD LEAD ODD BIGHT THK with L>B 

--- SPACING between L & B numbers in any given knot in the series is equal to 2. 

spacing = (absolute value of L-B or | L – B | = 2) 

--- L in knot with rank N – L in knot with rank (N-1) = B in knot with rank N – B in knot 

with rank (N-1) = 2 , this will be denoted STEP. Here STEP=2 

--- As here STEP is of EVEN PARITY it follows that the ODD PARITY of LEAD and BIGHT will 

stay unchanged by the addition of “EVEN” step so we keep ODD LEAD ODD BIGHT.



NUMBER 

OF LEAD 

L – 1 : 

number of 

crossings 

in a HP 

Max number of 

crossing at one go in 

the first term of a HP 

code in the series 

form O-U 

Max number of 

crossing at one go in 

the first term of a HP 

code in the series 

form O without any U 

5 4 2 1 

7 6 4 3 

9 8 6 5 

11 10 8 7 

13 12 10 9 

15 14 12 11 

17 16 14 13 

19 18 16 15 

21 20 18 17 

23 22 20 19 

25 24 22 21 

This series can be considered to span either from O 1 to O ((L-1)/2) or from O 1 to O ((L-1/)2)-1 

*** GENERAL FORMULA FOR THIS SERIES 

3 + (2 * n ) LEAD ( 2 * n ) + 1 BIGHT IF n lowest value is 1 

n=1 3 + (2*1) LEAD 1 + ( 2 * 1 ) BIGHT == 5L 3B 

n=2 3 + (2*2) LEAD 1 + ( 2 * 2 ) BIGHT == 7L 5B 

n=3 3 + (2*3) LEAD 1 + ( 2 * 3 ) BIGHT == 9L 7B 

5 + (2 * n ) LEAD ( 2 * (n + 2) ) - 1 ) BIGHT 

better 5 + (2 * n ) LEAD ( 2 * n ) + 3) BIGHT for n from 0 

Where ‘n’ is an integer which can take value 0 to …as large as needed. 

n=0 ( 5 + ( 2 * 0 ) LEAD (2 * (0 + 2) – 1 BIGHT == 5L 3B 

n=1 ( 5 + ( 2 * 1 ) LEAD (2 * (1 + 2) – 1 BIGHT == 7L 5B 

n=2 ( 5 + ( 2 * 2 ) LEAD (2 * (2 + 2) – 1 BIGHT == 9L 7B 

*** GENERAL FORMULA FOR THE MAXIMAL NUMBER OF ‘OVER’ AT THE BEGINNING 

OF A HP 

INT ( Number of LEAD / 2 ) or (L – 1 ) / 2 

Read Integral Part of number of LEAD divided by 2 

e.g 19/2= 9.5 INT PART = 9 

INT ( Number of BIGHT /2 ) + 1 

if the HP is in the form O Z and nothing else following it or in the form O Z+1 -U1 the formula 

is not identical. 

The above formula is for O Z+1 -U1, for the other form O Z it is ONE less, 

INT ( Number of LEAD / 2 ) – 1 or ( (L-1) / 2 ) – 1 

or also 

INT ( Number of BIGHT / 2 )



*** GENERAL FORMULA FOR THE NUMBER OF THE LAST HP IN WHICH THE 

MAXIMAL NUMBER OF OVER OCCURS 

Number of BIGHT in the knot + 2 ( about the mid-way point in the number of HP (B*2) in 

the knot or more precisely in the HP numbered B+2 or Number of LEAD 

If ‘isolated’ OVER (O Z ) without any U in the HP the formula is 

Number of BIGHT or Number of LEAD – 2 

xxxxxxxxxxxxxxxxxxxx 

After giving our attention to the BL ODD LEAD ODD BIGHT 

It is just the proposed xL yB series and made it into yL xB series. 

3L 5B-----5L 7B-----7L 9B… 

The number of BIGHT of knot with rank N is the number of LEAD of knot with rank (N+1) 

spacing = 2 step = 2 

*** GENERAL FORMULA FOR THIS SERIES 

1 + ( 2 * n ) LEAD 3 + ( 2 * n) BIGHT IF n from 1 

n=1 1 + ( 2*1) LEAD 3+(2*1) BIGHT == 3L 5B 

n=2 1 + ( 2*2) LEAD 3+(2*2) BIGHT == 5L 7B 

n=3 1 + ( 2*3) LEAD 3+(2*3) BIGHT == 3L 5B 

3 + ( 2 * n ) LEAD 5 + ( 2 * n ) BIGHT for n from 0 

n=0 3 + ( 2*0) LEAD 5+(2*0) BIGHT == 3L 5B 

n=1 3 + ( 2*1) LEAD 5+(2*1) BIGHT == 5L 7B 

n=2 3 + ( 2*2) LEAD 5+(2*2) BIGHT == 7L 9B 

*** GENERAL FORMULA FOR THE MAXIMAL NUMBER OF ‘OVER’ AT THE BEGINNING 

OF A HP 

INT ( Number of LEAD / 2 ) or (Number of LEAD – 1) / 2 there is only one form : O Z 

INT ( Number of BIGHT / 2 ) - 1 

*** GENERAL FORMULA FOR THE NUMBER OF THE LAST HP IN WHICH THE 

MAXIMAL NUMBER OF OVER OCCURS 

Number of BIGHT + 2 there is only one form O Z , there is no O Z+1 -U1 only U1 - O Z 

Number of LEAD + 4



Let us RECAPITULATE those two series ( n from 0 ) 

L>B 

3 + (2 * n ) LEAD 1 + ( 2 * n ) BIGHT IF n values from 1 

5 + (2 * n ) LEAD 3 + ( 2 * n ) BIGHT for n values from 0 

L



For the sake of maintaining comparability between L>B and LB) 

INT ( Number of LEAD / 2 ) - 1 or ( (L-1) / 2 ) - 1 

or also 

INT ( Number of BIGHT / 2 ) 

On the other hand (B>L) 

INT ( Number of LEAD / 2 ) or (Number of LEAD – 1) / 2 there is only one form O Z 

or also 

INT ( Number of BIGHT / 2 ) - 1 

In each case BOTH equalities must be simultaneously verified to get O1, O2, O3, O4, O5, 

O6….. 

In each case for the two equalities to be SIMULTANEOUSLY verified there must exist a 

particular value of SPACING and of STEP 

Let us have a look at the tautological verification of the formula I made. 

INT ( Number of LEAD / 2 ) - 1 = INT ( Number of BIGHT / 2 ) for L>B 

INT ( Number of LEAD / 2 ) = INT ( Number of BIGHT / 2 ) – 1 for LB 

Let us forget the INT PART 

(Number of LEAD /2) – 1 = Number of BIGHT/2 

Let us multiply each side by 2 

Number of LEAD – 2 = Number of BIGHT 

or Number of LEAD – Number of BIGHT = 2 

INT ( Number of LEAD / 2 ) = INT ( Number of BIGHT / 2 ) – 1 for L



This is to be put side by side this with the equations for calculating the rank number of the 

Columns where crossings happen as given by Schaake. 

------ 

As a comparison think’ about ‘coincidence of cycles 

Say 22 years, 16 years, 15 yeas as the duration of a cycles 

First and second will “coincide” every 22*16 =352 

First and third 22*15=330 

Second an third 16*15=240 

The three will “coincide” every 22*16*15=7429 

For the coincidence of any two cycles you cannot infer the coincidence of the three, you 

need to use a ‘special’ operation while knowing each individual cycles. For example 240= 6 * 

40 or 12*20 or 60*4 or… how do you make your choice when just knowing the coincidence 

Knowing that 2 cycles coincide every X years you cannot know without “field observation” 

the individual duration of the two and much less the three cycles, I suspect here the same 

sort of problem. 

----- 

PIN-STEP, ∆ , ∆ * did not shed much light on the series, at least as far as I am concerned. 

----- 



Release V1.3.1



How to calculate the Column in which the crossing(s) happen is explained in The Braiding of 

Regular Knots, Schaake & Turner’s book.



To get the desired ‘smooth’ runs of ‘OVER’ along the successive HP of a knot we need a 

rare conjunction of events resulting in a ‘cadence’ of crossing placing that, just like someone 

jumping on stepping stones to cross a brook at the ford, must fall on the right sequence ! 

See next page for illustration : 11L 9B 

HP AFTER HP developpement. 

Points to keep in mind (horizontal mandrel with bight rim on the left and on the right side) are: 

**** ODD numbered HP go from LOW LEFT to UP RIGHT, read Column number from 

LEFT to RIGHT 

and 

**** EVEN numbered HP go from LOW RIGHT T to UP LEFT, read Column numbers from 

RIGHT to LEFT. 

The OVER crossing of one parity of HP is the UNDER crossing of HP of the other parity.



The first exploration was stopped here but a few days later this interesting problem was 

examined again. 




Using RKnot Builder in HP by HP mode it is easy to observe, on the 11L 9B ( L>B) : 

( pin step 5.5 ( 5 and 6) ; Delta = 5 , Delta*=4) 

HP 

number 

direction 

Column 

first 

crossing 

Column 

second 

crossing 

Column 

third 

crossing 

Column 

fourth 

crossing 

Column 

fifth 

crossing 

Column sixth 

crossing 

HP-1 L to R FREE FREE FREE FREE FREE FREE 

HP-2 R to L 9 9 is the LEFTMOST ‘O’ Column of this EVEN numbered HP 

HP-3 L to R 9 9 is the RIGHTMOST ‘O’ Column of this ODD numbered HP 

HP-4 R to L 9 7 

HP-5 L to R 9 7 

HP-6 R to L 9 7 5 

HP-7 L to R 9 7 5 

HP-8 R to L 9 7 5 3 

HP-9 L to R 9 7 5 3 

HP-10 R to L 9 7 5 3 1 10 – first UNDER 

At the beginning ONLY ODD numbered Columns are used, till there is no more ‘OVER’ left 

over then it is an EVEN numbered Column that is used and we get the first UNDER crossing 

in the knot. 

The maximum of ‘O’ available in a HP is ( L – 1 ) / 2 so above ( 11 – 1 ) / 2 = 5 

11L 9B L-1 or 10 crossings per HP with code read along the first HP in the finished knot : 

O – U – O – U – O – U – O – U – O – U (note that the first crossing on HP-1 (finished 

knot) for L>B is OVER) 

Column number 

L to R 1 2 3 4 5 6 7 8 9 10 

CODE seen by 

HP-1 on finished 

knot 

O O U U 

O U 

O O 

U U 

CODE for 

R to L Column 

O O O O O 

U 

U 

Column number 

R to L 10 9 8 7 6 5 4 3 2 1 

U 

U 

U 

Put the two tables immediately above in relation with the formula given to calculate the 

number of the Column(s) where crossing(s) appear in the considered HP. 

There are L-1 COLUMNs in this knot == 11 -1 = 10 

EVEN 2 4 6 8 10 

ODD 1 3 5 7 9 

“cadence” is ( ( L - 2 ) – ( n * 2 ) ) MODULO L where ‘n’ take values from 0 to 

m with ‘m’ such that cadence=1 at its lowest. 

(11-2) – (0*2) modulo 11 = 9 (11-2) – (1*2) modulo 11 = 7 

(11-2) – (2*2) modulo 11 = 5 (11-2) – (3*2) modulo 11 = 3 

(11-2) – (4*2) modulo 11 = 1 (11-2) – (5*2) modulo 11 = -1 =10 

(11-2) – (6*2) modulo 11 = -3 = 8 (11-2) – (7*2) modulo 11 = -5 = 6 

(11-2) – (8*2) modulo 11 = -7 = 4 (11-2) – (9*2) modulo 11 = -9 = 2



7L 5B (( pin step 3.5 (5 and 6) ; Delta = 3 , Delta*=2) . 

L-1 or 6 crossings per HP code seen by HP-1 in the finished knot O – U – O – U – O – U 

HP 

number 

direction 

Column 

first 

crossing 

Column 

second 

crossing 

Column 

third 

crossing 

Use either my Excel worksheet or, better, use RKnot BUILDER and make you own 

verifications. 

Again “ODD” PARITY for the COLUMNs where the ‘O’ crossings are made 

Same PARITY as Pin Step (lowest integer value - L/2 ) and Delta---- 

Same PARITY as L modulo B ---- this is the easiest in my opinion. (it seems that the first 

column with a ‘O’ is numbered L modulo B + 2 in this series) 

So to get a regular run O1 O1-O2 O1-O2-O3 O1-O2-O3-O4….. 

Rather it is O1,O1 O1, O1, O2, O2 O1, O1, O2, O2, O3, O3 ….. 

“cadence” is ( ( L - 2 ) – ( n * 2 ) ) MODULO L where ‘n’ take values from 0 to m with ‘m’ 

such that cadence = 1 at its lowest. 

For ( L – 2 ) – ( n * 2 ) modulo L to be ODD it is obvious that LEAD MUST be ODD 

L=14 even n=3 (14 -2 ) – ( 3 * 2 ) modulo 14 = 12 – 6 modulo14 = 6 even 

L=15 odd n=3 (15 -2 ) – ( 3 * 2 ) modulo 15 = 13 – 6 modulo14 = 7 odd 

spacing=2 so BIGHT IS of the same ODD PARITY that LEAD is. 

spacing = 2 hence ipso facto to keep ODD L ODD B step must also be equal to 2 or a 

multiple of 2 

For a given knot in the series ‘n’ is at most equal to 

INT( L / 2) -1 

or 

INT( ( L - 2) / 2 ) 

or 

INT ( ( L - 2 ) / spacing 

Colum 

fourth 

crossing 

HP-1 L to R FREE FREE FREE FREE 

HP-2 R to L 5 5 is LEFTmost ‘O’ on this HP 

HP-3 L to R 5 5 is RIGHTmost ‘O’ on this HP 

HP-4 R to L 5 3 

HP-5 L to R 5 3 

HP-6 R to L 5 3 1 6 

‘UNDER’



*** If spacing = 3 then with ODD LEAD we will get EVEN BIGHT (with EVEN LEAD we will 

have ODD BIGHT) which as we already explored does not yield the desired result. 

*** If step = 3 the ODD LEAD ODD BIGHT will become EVEN LEAD EVEN BIGHT which is 

‘an impossibility’ for a Regular Knot cordage route (GDC) . 

*** If spacing = 3 AND step = 3 

ODD LEAD ODD BIGHT would be EVEN LEAD EVEN BIGHT (impossible) 

ODD LEAD EVEN BIGHT ( or EVEN LEAD ODD BIGHT) which would become 

EVEN LEAD ODD BIGHT ( or ODD LEAD EVEN BIGHT ) and go on alternating. 


Let us see the 9L 11B LB) , first ODD numbered COLUMN used yields the 

first UNDER. 

Same PARITY as Pin Step (lowest value) and Delta * ---- --- 

Same PARITY as L modulo B ---- this is the easiest in my opinion.



9L 11B 

L-1 or 8 crossings per HP with a code as read along the first HP (finished knot) 

U – O – U – O – U – O – U – O (note that the first crossing on HP-1 (finished knot) in 

L



It seems to me that the PARITY of the lowest value of PIN STEP is the parameter telling the 

PARITY of the first COLUMN used. 


Let us go back on what we found. 

L>B as in 11L 9B 

L ) to LEFT KNOT EDGE 

We are here dealing with ODD LEAD ODD BIGHT B>L so the ‘suite’ of crossings will 

ALWAYS be U - O……………….U – O



L-1 crossing columns so 9L == 8 crossings on each HP in the finished knot 

This is the order in which the crossings will be added. 

ODD HP 

Progression is from LEFT to RIGHT 

U O U O U O U O 

1 2 3 4 5 6 7 8 

EVEN HP 

Progression is from RIGHT to LEFT 

O U O U O U O U 

8 7 6 5 4 3 2 1 

U 

1 

O 

2 

U 

3 

O 

4 

U 

5 

O 

6 

U 

7 

O 

8 

O 

8 

U 

7 

O 

6 

U 

5 

O 

4 

U 

3 

O 

2 

U 

1 

U 

1 

O 

2 

U 

3 

O 

4 

U 

5 

O 

6 

U 

7 

O 

8 

O 

8 

U 

7 

O 

6 

U 

5 

O 

4 

U 

3 

O 

2 

U 

1 

U 

1 

O 

2 

U 

3 

O 

4 

U 

5 

O 

6 

U 

7 

O 

8 

O 

8 

U 

7 

O 

6 

U 

5 

O 

4 

U 

3 

O 

2 

U 

1 

O 

8 

U 

7 

O 

6 

U 

5 

O 

4 

U 

3 

O 

2 

U 

1 

Directions of progression are different for BL. 

Whatever the direction of the progression for a given HP PARITY the ‘pace’ is every two’ 

Columns so starting from an ‘O’ it keeps the ‘O’ until the possible ‘O’ are all used up and that 

a ‘U’ is made. 

SPACING is one parameter governing the “stone stepping” on the Columns of crossings, as 

long as its PARITY is EVEN you may go from one ‘O’ Column to the other and PIN STEP 

PARITY is the parameter governing the PARITY of the COLUMNS used. 


Now for spacing = 2 let us see what the STEP value changes if it is not of EVEN PARITY. 

This (not being EVEN) is impossible as we have already seen : the ODD LEAD ODD 

BIGHT would become EVEN LEAD EVEN BIGHT and the series explode. 

15L 13B 18L 16B BOOM!! 

17L 15B 20L 18B BOOM!!



If the STEP is of EVEN PARITY then it is either 2 or a multiple of 2 (in the later case we 

are jumping to a higher level in the series like climbing a staircase every two steps, still 

having a run but with another rhythm). 

Step=2 (factorisation : 1 * 2 ) 

5L 3B (O1) 

7L 5B (O1-O2) 

9L 7B (O1-O2-O3) 

11L 9B (O1-O2-O3-O4 

13L 11B (O1-O2-O3-O4-O5) 

15L 13B (O1-O2-O3-O4-O5-O6) 

17L 15B (O1-O2-O3-O4-O5-O6-O7 

19L 17B (O1-O2-O3-O4-O5-O6-O7-O8 

21L 19B (O1-O2-O3-O4-O5-O6-O7-O8-O9) 

NINE KNOTS 

Step=4 (factorisation : 1 * 2 * 2) 

5L 3B (O1) 

9L 7B (O1-O2-O3) 

13L 11B (O1-O2-O3-O4-O5) 

17L 15B (O1-O2-O3-O4-O5-O6-O7) 

21L 19B (O1-O2-O3-O4-O5-O6-O7-O8-O9) 

FOUR KNOTS TO GET TO THE SAME ‘RUN’ (O1-O2-O3-O4-O5-O6-O7-O8-O9) 

Step=6 (factorisation : 1 * 2 * 3) 

5L 3B (O1) 

11L 9B (O1-O2-O3-O4 ) 

17L 15B (O1-O2-O3-O4-O5-O6-O7) 

23L 21B (O1-O2-O3-O4-O5-O6-O7-O8-O9-O10) 

OUT OF STEP with the other table ending with 21L 19B (factorisation contains digit ‘3’) for 

run (O1-O2-O3-O4-O5-O6-O7-O8-O9-O10) 

Step=8 (factorisation : 1 * 2 * 2 * 2) 

5L 3B (O1) 

13L 11B (O1-O2-O3-O4-O5) 

21L 19B (O1-O2-O3-O4-O5-O6-O7-O8-O9) 

THREE KNOTS TO GET TO THE SAME ‘RUN’ (O1-O2-O3-O4-O5-O6-O7-O8-O9) 

Using a STEP value that is a multiple of 2 ADD STEP/2 incremented groups of ‘O’ to the 

preceding run. (incremented group examples O2-O3 or O4-O5-O6)



So it comes out that STEP = 2 seems the ‘most complete and smooth’ run and the others 

are “extract” or “composition” that are ”as complete in the long run” but have a different 

gradient of increase in ‘O’ groups. 

The series can only FULLY persists if spacing=step=2 with ODD LEAD ODD BIGHT THK 

(B>L or B



PROVISIONAL CONCLUSION : 

The FIRST NECESSARY condition is SPACING=2 where k is an integer with values 1 to as 

large as needed to ensure that that the “stone stepping’ on the Columns of crossings will first 

offer a continuous run of OVER. 

This value of spacing entails as corollary ODD LEAD ODD BIGHT. 

To ensure that the continuous run of OVER is SMOOTH -add only one unit- AND 

PROGRESSIVE 1 1, 2 1, 2, 3 1, 2, 3, 4 the SECOND NECESSARY condition is 

STEP=2 

there is probably no other series with this COMPLETE AND PROGRESSIVE particular 

run of OVER at the beginning of the knot other than the pair :



L>B 

3 + (2 * n ) LEAD 1 + ( 2 * n ) BIGHT IF n from 1 

5 + (2 * n ) LEAD 3 + ( 2 * n ) BIGHT for n from 0 

LB 

5 + (2 * n ) LEAD 1 + ( 2 * n ) BIGHT IF n from 1 ( 7L 3B) 

7 + (2 * n ) LEAD 3 + (2 * n ) BIGHT for n from 0 (7L 3B) 

9L 5B makes a better start THK : 

7 + (2 * n ) LEAD 3 + ( 2 * n ) BIGHT IF n from 1 ( 9L 5B) 

9 + (2 * n ) LEAD 5 + (2 * n ) BIGHT for n from 0 (9L 5B) 

L



After examining the 2 cases (L>B ; L



Spacing=3 

(fact=1*3) 

Step=2 

2 5 (3 5 would lead to 6 8 ) 2 5 

4 7 5 8 

6 9 BOOM!! 8 11 

8 11 11 14 

10 13 14 17 

12 15 BOOM!! 

There is no regularity that it is possible to sustain 

such as O1, O2, O3… beyond the HP with a FREE RUN 

(2L 5B or 5L 2B) 

Spacing=3 

(fact = 1*3) 

step=3 

2L 5B FR, FR, FR, FR, FR, O1, U1-O1 5L 2B FR, O2, O2, U1-O1… 

5L 8B FR, FR, FR, O1, O1, O1, O1, O2, O2, O2-U1 8L 5B FR, O1, U1 

8L 11B FR, FR, FR, O1, U1 11L 8B FR, O1, O1, U1-O1 

11L 14B FR, FR, FR, O1, O1, O1-U1 14L 11B FR, O1, U1 


Not really satisfying ! 

with ODD LEAD EVEN BIGHT or EVEN LEAD ODD BIGHT inevitably 

*** with ODD Step the ODD becomes EVEN and the EVEN becomes ODD so preserving the 

series 

*** with EVEN Step the EVEN stays EVEN and ODD will stay ODD but if a common divisor 

appears ( one of the prime factor of the factorisation of spacing >1 ) then BOOM!! 


Spacing=4 (factor = 1*2*2) Spacing=5 (factor = 1 * 5 

3 7 3 8 

Step=2 step=3 Step=2 step=3 

3 7 3 8 3 8 

5 9 (or 9 5 ) 6 10 BOOM!! 5 10 BOOM!! 6 11 

7 11 9 13 9 14 

9 13 12 16 BOOM!! 12 17 

11 15 15 19 15 20 BOOM!! 

15 19 

17 21 

19 23 

21 25 

23 28



5L 9B FR, FR, FR, O1, O1, O1, O1, 

U1-O1 

7L 11B FR, FR, FR, O1, O1, O1, O1, 

U1-O1 

9L 13B FR, FR, FR, O1, O1, O2, O2 

O2, O2, U1-O2 

11L 15B FR, FR, FR, O1, O1, O2, O2, 

O2, O2, U1-O2 

13L 17B FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O3, O3, U1-O3 

15L 19B FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O3, O3, U1-O3 

17L 21B FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O4, O4, U1- 

O4 

19L 23B FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O4, O4, U1- 

O4 

9L 5B 

11L 

7B 

13L 

9B 

15L 

11B 

17L 

13B 

19L 

15B 

21L 

17B 

23L 

19B 

FR, O1, O1, O2-U1 

FR, O1, O1, O2-U1 

FR, O1, O1, O2, O2, O3- 

U1 

FR, O1, O1, O2, O2, O3- 

U1 

FR, O1, O1, O2, O2, O3, 

O3, O4-U1 

FR, O1, O1, O2, O2, O3, 

O3, O4-U1 

FR, O1, O1, O2, O2, O3, 

O3, O4, O4, O5-U1 

FR, O1, OFR, O1, O1, O2, O2, O3, O3, 

OO4, O4, O5-U1 

Starting with ODD LEAD ODD BIGHT and using AN EVEN STEP INESCAPABLY WE 

WILL KEEP ODD LEAD ODD BIGHT so preserving the series. 

With ODD LEAD EVEN BIGHT or EVEN LEAD ODD BIGHT whatever the parity of the STEP 

we will either get EVEN LEAD EVEN BIGHT BOOM!! or encounter such L and B that they 

have as common divisor >1 (one of the factor of the spacing) BOOM!! 




spacing=6 ( 

(factor = 1*2*3) 

2 8 and 3 9 

impossible 

5 11 5 11 

7 13 8 14 

9 15 BOOM 

11 17 

13 19 

15 21 BOOM 

17 23 

19 25 

21 27 BOOM 21 28 

BOOM 

spacing=7 (factor = 1*7) 

3 10 2 9 

step=2 step=3 step=2 step=3 

5 12 6 13 4 11 5 12 

7 14 9 16 6 13 8 15 

BOOM 

12 19 8 15 11 18 

15 22 10 17 14 21 

BOOM 

18 25 12 19 

14 21 

BOOM 

EVEN LEAD ODD BIGHT or ODD LEAD EVEN BIGHT: with an EVEN Step the 

ODD/EVEN or EVEN/ODD stay BUT IF there is a common divisor ( a factor of the spacing ) 

for L and B then BOOM! 


Spacing=8 (factors = 1*2*2*2) Spacing=9 (factors = 1*3*3*3 

Step=2 Step=3 Step=2 Step=3 

2 10!! 3 11 2 11 

2 11 

or 3 12 !! 

so 6 14 

4 13 5 14 

BOOM!! 

3 11 6 15 8 17 

BOOM!! 

5 13 11 20 

7 15 14 23 

9 17 17 26 

11 19 20 29 

13 21 23 32 

15 23 26 35 

17 25 29 38 

19 27 32 41



3L 11B FR, FR, FR, FR, FR, FR, FR, 

O1, O1 , O1, O1, O1, O1, O1, 

O1, U1-O1 

5L 13B FR, FR, FR, FR, FR, O1, O1 , 

O1, O1, O1, O1, O2, O2, O2, 

O2, O2-U1 

7L 15B FR, FR, FR, FR, O1, O1, O1, 

O1, O1-U1 

9L 17B FR, FR, FR, O1, O1, O1, O1, 

U1-O1 

11L 19B FR, FR, FR, O1, O1, O1, O1, 

U1-O1 

13L 21B FR, FR, FR, U1 

for sobre 132 ‘O’ 120 ‘U’ 

15L 23B FR, FR, FR, O1, O1, O1, O1, 

U1-O1 

17L 25B FR, FR, FR, O1, O1, O2, O2, 

O2, O2, U1-O2 

19L 27B FR, FR, FR, O1, O1, O2, O2, 

O2, O2, U1-O2 

11L 3B 

13L 5B 

15L 7B 

17L 9B 

19L 11B 

21L 13B 

23L 15B 

25L 17B 

27L 19B 

FR, O1-U1-O1 

FR, U1, O1 

for sobre 32’O’ 28 ‘U’ 

FR, O1-U1 

FR, O1, O1, O2-U1 

FR, O1, O1, O2-U1 

FR, O1, O1, O2-U1 

FR, O1, O1, O2-U1 

FR, O1, O1, O2, O2, O3-U1 

FR, O1, O1, O2, O2, O3-U1 

2L 11B FR, FR, FR, FR, FR, FR, FR, 11L 2B FR, O5, O5; U1-O1….. 

FR, FR, FR, O1, U1 

5L 14B FR, FR, FR, FR, FR, O1, O1, 14L 5B FR, O1-U1 

O1, O1, O1, U1-O1 

8L 17B FR, FR, FR, FR, FR, O1, U1 17L 8B FR, O2, O2, U1-O1….. 

11L 20B FR, FR, FR, O1, O1, O1, O1, 20L 11B FR, O1, U1 

O2, O2, O2, O2, O3, O3, O3,O3, 

O4, O4, O4, O4, O5, O5, O5-U1 

14L 23B FR, FR, FR, O1, U1 23L 14B FR, U1 

for sobre 159 ‘O’ 149 ‘U’ 

17L 26B FR, FR, O1, O1, O1, O1, O2, 26L 17B FR, O1, U1 

O2, O2-U1 


23L 32B FR, FR, FR, U1 

32L 23B FR, O1, U1 

for sobre 365 ‘O’ 339 ‘U’ 


With an EVEN Step ODD LEAD ODD BIGHT stay ODD L ODD B but ODD LEAD ODD 

BIGHT soon explodes with an ODD Step leading to EVEN L EVEN B and with an EVEN 

Step ODD LEAD EVEN BIGHT or EVEN LEAD ODD BIGHT will meet its end with a 

common divisor that is one of the factors of the spacing BOOM!! 

No use to go on as we now have the ‘rules’ 




Let us quickly have a glance at B>L ODD LEAD ODD BIGHT with different spacing and 

step values 

7L 3B----9L 5B----11L 7B----13L 9B----15L 11B----17L 13B-- 

--19L 15B----21L 17B---- 

L>B ODD LEAD ODD BIGHT spacing=4 step=2 

7L 3B 

9L 5B 

11L 7B 

13L 9B 

15L 11B 

17L 13B 

19L 15B 

21L 17B 

(FR, O1-U1) or (FR, U1-O1) both 9 over 9 under 

(FR, O1, O1, O2-U1) 

(FR, O1, O1, O2-U1) 

(FR, O1, O1, O2, O2, O3-U1) 

(FR, O1, O1, O2, O2, O3-U1) 

(FR, O1, O1, O2, O2, O3, O3, O4-U1) 

( FR, O1, O1, O2, O2, O3, O3, O4-U1) 

(FR, O1, O1, O2, O2, O3, O3, O4, O4, O5-U1) 

3L 7B----5L 9B----7L 11B---9L 13B---- 

LL ODD LEAD ODD BIGHT spacing=8 step=2 

3L 11B 

5L 13B 

7L 15B 

(FR, FR, FR, FR, FR, FR, FR, O1, O1, O1, O1, O1, O1, O1, O1, U1-O1 

(FR, FR, FR, FR, FR, FR,O1, O1, O1, O1, O1, O1, O2, O2, O2, O2, O2-U1 

(FR, FR, FR, FR, FR, O1, O1, O1, O1, O1-U1) 

11L 3B…13L 5B…15L 7B … 


11L 3B (FR, O1-U1….) 

13L 5B (FR, U1-O1) ; 

15L 7B (FR, O1-U1) 




2L 11B…5L 14B…8L 17B…11L 20B----14L 23B 

B>L ODD LEAD ODD BIGHT spacing=9 step=3 

2L 11B (FR, FR, FR, FR, FR, FR, FR, FR, FR, FR, FR, O1, U1) 

5L 14B (FR, FR, FR, FR, FR, O1, O1, O1, O1, O1, O1, U1-O1) 

8L 17B (FR, FR, FR, FR, FR, O1, U1) 

11L (FR, FR, FR, O1, O1, O1, O1, O2, O2, O2, O2, O3, O3, O3, O3 , O4, O4, O4, O4, 

20B 

14L 

23B 

O5, O5, O5-U1) 

(FR, FR, FR, O1, U1) 

11L 2B…14L 5B…17L 8B…20L 11B----23L 14B 


11L 2B (FR, O5, O5, U1-O1….) 

14L 5B (FR, O1-U1) 

17L 8B (FR, O2, O2, U1-O1….) 

20L 11B (FR, O1, U1) 

23L 14B ( FR, U1 for sobre 159 ‘O’ 149 ‘U’) 


PROVISIONAL CONCLUSION 

Reader, please understand that all this is not a formal mathematical demonstration but an 

experimental statement that will remain standing on even keel till someone can bring at 

least one series that does not comply with the conditions stated and that will nevertheless 

yields the same smooth and progressive run of OVER O1 O1, O2 O1, O2, O3 O1, 

O2, O3, O4 O1, O2, O3, O4,……….,O(n-2), O(n-1), O(n) 

Nevertheless there exist other type of regular runs that do not have the progressivity of the 

princeps series (see also ADDENDA) 


…/…



ADDENDA 

A WARNING FLAG FOR THOSE WANTING TO EXPLORE FARTHER 

(sorry about some duplication of some of the tables in the preceding pages. 

There is a way ( coming back to ‘cycles coincidence’ ) that one could imagine as successful. 

But first let us “batter down some already open doors” (meaning that all this is so evident for 

someone having made a few THK that it does not seem necessary to say it, “ it goes without 

saying but it goes so much better with saying it” ! 

Number of FREE RUN will ALWAYS be ODD 

The first HP is ALWAYS a free run ( 1 is ODD ) 

For n = 1 to B HP( 2*n) and HP ( (2*n)+1) have the same coding so if HP-2 is a FR then 

HP-3 MUST BE a FREE RUN, hence 3 (odd ) FREE RUN 

HP( 2*n) and HP ( (2*n)+1) have the same coding (illustration n=3 HP (2*3) HP6 ) 

EVEN numbered HP = right to left ODD numbered HP = left to right 

Let us try and amend the SPACING = 2 to SPACING = k * 2 where k may have a value 

from 1 to any larger integer value 

k=1 spacing = 1 * 2 = 2 

k=2 spacing = 2 * 2 = 4 

k=3 spacing = 3 * 2 = 6 

k=4 spacing = 4 * 2 = 8 

----------------------------------------------------- 

47L 25B SPACING = 22 is the larger spacing before disappearance of O1 with the 

47L 23B spacing=24 

45L 23B has this O1 but 43L 21B does not have ‘pure O’ ( it seems that as long as 2*B > L 

there are ‘O’)



The larger the SPACING the smaller will be the length of the run. 

--------------------------------------------------------- 

Take Spacing = 6 step=2 

let us try 31L 25B (first term in this hypothetical series is 13L 7B ) 

ok 

Sure 

In fact it is a dead idea as the series is not sustainable, it will go BOOM!! on GDC rule. 

13/7 15/9 (boom!) 17/11 19/13 21/15 boom! 23/17 25/19 27/21 boom! 29/23 

31/25 33/27 boom! 35/29 37/31 39/33 boom! 41/35 43/37 45/39 boom! 47/41 

49/43 51/45 boom! 

Reason why 

6 factorise into prime number as 6 = 1 * 2 * 3 

When one term of the factorisation is not EVEN this leads to BOOM!! 

Prime numbers : 0, 1, 2 , 3 , 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 563, 59, 61, 67, 

71, 73, 79, 83, 89, 97 

NO USE GOING ON = THE ONLY PRIME NUMBER OF EVEN PARITY IS 2 

It follows that ANY number that is not a POWER of 2 (2, 4, 8, 16, 32 for example) is 

DOOMED 

2 factorise in 1 * 2 

4 factorise in 1 * 2 * 2 

6 factorise in 1 * 2 * 3 explodes on GCD=3 

8 factorise in 1 * 2 * 2 * 2 


12 factorise in 1 * 2 * 2 * 3 explodes on GCD =3 

14 factorise in 1 * 2 * 7 explodes on GCD =7 

16 factorise in 1 * 2 * 2 * 2 * 2 


20 factorise in 1 * 2 * 2 * 5 explodes on GCD=5 


Better amend SPACING = k * 2 to SPACING = 2 k 

So let us take the best bet : 

ODD LEAD ODD BIGHT THK L>B AND L>B with 

SPACING being a power of 2 and 

---------------------------------------- 

STEP being multiple of 2



(3 and 7) (3 and 11) being pairs of PRIME will stay pairs of PRIME with the addition of equal 

EVEN number on both. 

SPACING = 4 step= 2 SPACING = 4 step= 2 

3L 7B FR, FR, FR, FR, FR O1, O1, O1, 

O1, O1-U1 

5 9 FR, FR, FR, O1, O1, O1, O1, 

U1-O1 

7 11 FR, FR, FR, O1, O1, O1, O1, 

U1-O1 

9 13 FR, FR, FR, O1, O1, O2, O2, 

O2, O2, U1-O2 

11 15 FR, FR, FR, O1, O1, O2, O2, 

O2, O2, U1-O2 

13 17 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O3, O3, U1-O3 

7L 3B FR, O1-U1, O1-U1 

or (FR, U1-O1, U1-O1) both 9 over 

9 under 

This start THK is not suitable 

9 5 FR, O1, O1, O2-U1 

11 7 FR, O1, O1, O2-U1 

13 9 FR, O1, O1, O2, O2, O3-U1 

15 11 FR, O1, O1, O2, O2, O3-U1 

17 13 FR, O1, O1, O2, O2, O3, O3, O4- 

U1 

15 19 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O3, O3, U1-O3 

17 21 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O4, O4, U1-O4 

21 23 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O5, O5, O6, 

O6, O7, O7, O8, O8, O9, O9, 

O10, O10, O10, O10, U1-O10 

19 15 FR, O1, O1, O2, O2, O3, O3, O4- 

U1 

21 17 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5-U1 

23 19 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5-U1 

Goes out of step 25 21 FR, O1, O1, O2, O2, O3, O3, O4, 

23 25 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O5, O5, O6, 

O6, O7, O7, O8, O8, O9, O9, 

O10, O10, O11, O11, O11, O11 

U1-O11 

25 27 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O5, O5, O6, 

O6, O7, O7, O8, O8, O9, O9, 

O10, O10, O11, O11, O12, O12, 

O12, O12, U1-O12 

O4, O5, O5, O6-U1 

27 23 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5, O5, O6-U1 

29 25 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5, O5, O6, O6,O7-U1 


3L 7B FR, FR, FR, FR, FR O1, O1, O1, 7B 3L FR, O1-U1 

O1, O1-U1 

7 11 FR, FR, FR, O1, O1, O1, O1, 11 7 FR, O1, O1, O2-U1 

U1-O1 

11 15 FR, FR, FR, O1, O1, O2, O2, 15 11 FR, O1, O1, O2, O2, O3-U1 

O2, O2, U1-O2 

Just the above but every two 

15L 19B 21L 23B 

Just the above but every two 19L 

15B 23L 21B 


5L 9B FR, FR, FR, O1, O1, O1, O1,U1- 9L 5B FR, O1, O1, O2-U1 

O1 

13 17 FR, FR, FR, O1, O1, O2, O2, 17 13 FR, O1, O1, O2, O2, O3, O3, O4-



O3, O3, O3, O3, U1-O3 U1 

21 25 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O5, O5, O5, 

O5, U1-O5 

29 33 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4,O5, O5, O6, O6, 

O7, O7, O7, O7, U1-O7 

37 41 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O5, O5, O6, 

O6, O7, O7, O8, O8, O9, O9, 

O9, O9, U1-O9 

25 21 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5 O5, O6-U1 

33 29 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5 O5, O6, O6, O7, O7,O8- 

U1 

41 37 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5 O5, O6, O6, O7, O7, O8, 

O8, O9, O9,O10-U1 


3L 11B FR, FR, FR, FR, FR, FR, FR, 11L 3B FR, O1-U1-O1 

O1, O1, O1, O1, O1, O1, O1, 

O1, U1-O1 

5 13 FR, FR, FR, FR, FR, O1, O1, 13 5 FR, U1-O1 

O1, O1, O1, O1, O2, O2, O2, 

O2, O2-U1 

7 15 FR, FR, FR, FR, FR, O1, O1, 15 7 FR, O1-U1 

O1, O1, O1-U1 

9 17 FR, FR, O1, O1, O1, O1, U1-O1 17 9 FR, O1, O1, O2-U1 

11 19 FR, FR, FR, FR, O1, O1, O1, 19 11 FR, O1, O1, O2-U1 

O1, U1-O1 

13 21 FR, FR, FR, U1 21 13 FR, O1, O1, O2-U1 

SPACING = 8 step=4 SPACING = 8 step= 4 



SPACING = 8 step= 6 SPACING = 8 step=6 

3L 11B FR, FR, FR, FR, FR, FR, FR, 

O1, O1, O1, O1, O1, O1, O1, 

O1, U1-O1 

9L 17 FR, FR, FR, O1, O1, O1, O1, 

U1-O1 

15 23 FR, FR, FR, O1, O1, O1, O1, 

U1-O1 

21 29 FR, FR, FR, O1, O1, O2, O2, 

O2, O2, U1-O2 

27 35 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O3, O3, U1-O3 

33 41 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O4, O4, U1-O4 

39 47 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O4, O4, U1-O4 

45 53 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O5, O5, O5, 

O5, U1-O5 

51 59 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O5, O5, O6, 

O6, O6, O6, U1-O6 

57 65 FR, FR, FR, O1, O1, O2, O2, 

O3, O3, O4, O4, O5, O5, O6, 

11L 3B 

FR, O1-U1-O1 

17 9 FR, O1, O1, O2-U1 

23 15 FR, O1, O1, O2-U1 

29 21 FR, O1, O1, O2, O2, O3-U1 

35 27 FR, O1, O1, O2, O2, O3, O3, O4- 

U1 

41 33 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5-U1 

47 39 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5-U1 

53 45 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5, O5, O6-U1 

59 51 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5, O5, O6, O6, O7-U1 

65 57 FR, O1, O1, O2, O2, O3, O3, O4, 

O4, O5, O5, O6, O6, O7, O7, O8-

Charles HAMEL – 2011 June Page 36 sur 36 


O6, O7, O7, O7, O7, U1-O7 U1 

Only SPACING = 4 STEP = 2 (with a particular start THK that is not the lowest in the 

series) offer hope of a long series with smooth and progressive run. 

The trail is hopeless IMO for STEP that is NOT EQUAL to 2 AND L>B 

Let us try ( knowing that it is doomed as 6 = 1*2*3) 

SPACING = 6 (not a power of 2 ) STEP = 2 

SPACING = 6 STEP = 2 SPACING = 6 STEP = 2 

5 11 FR, FR, FR, FR, FR, O1, O1, 11 5 

O1, O1, O1-U1 

7 13 FR, FR, FR, O1, O1, O1, U1- 13 7 

O1 

9 15 BOOM! factor of spacing 

GDC=3 

15 9 BOOM! factor of spacing

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