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.

Page 2 sur 36

Charles HAMEL – 2011 June

Release V1.3.1 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

Page 3 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

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 Omax or Omax-1 without The longest run of HP with pure ‘O’ without any U.

any U (‘isolated’

or ‘pure’ OVER)

or the last HP with Omax – U1 (longest run of HP beginning with ‘Ox’ ) could even be the first HP with Omax – 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’

Page 4 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

17L 19B spacing =2 step=2

B>L

HP1 Free Run 1 Wrap HP2 Free Run HP3 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 HP1 Free Run 1 Wrap HP2 Free Run HP3 Free Run

HP4 O1 HP5 O1 HP6 O2 HP7 O2 HP8 O3 HP9 O3 HP10 O3

HP11 O3 HP11 O3 HP12 U1 – O3 HP13 U1 – O3

19L 17B spacing=2 step=2 L>B HP1 Free Run 1 Wrap

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 HP1 Free Run 1 Wrap

HP2 O1 HP3 O1 HP4 O2 HP5 O2 HP6 O3

HP7 O3 HP8 O4 – U1 HP9 O4 – U1 HP10 O3 – U1 – O1 – U1

Charles HAMEL – 2011 June

Page 5 sur 36

Release V1.3.1

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.

Page 6 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

NUMBER OF LEAD

5 7 9 11 13 15 17 19 21 23 25

L–1: number of crossings in a HP 4 6 8 10 12 14 16 18 20 22 24

Max number of crossing at one go in the first term of a HP code in the series form O-U 2 4 6 8 10 12 14 16 18 20 22

Max number of crossing at one go in the first term of a HP code in the series form O without any U 1 3 5 7 9 11 13 15 17 19 21

This series can be considered to span either from O1 to O((L-1)/2) or from O1 to

O((L-1/)2)-1

*** GENERAL FORMULA FOR THIS SERIES 3 + (2 * n ) LEAD n=1

( 2 * n ) + 1 BIGHT IF n lowest value is 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 OZ and nothing else following it or in the form OZ+1-U1 the formula is not identical. The above formula is for OZ+1-U1, for the other form OZ it is ONE less, INT ( Number of LEAD / 2 ) – 1 or ( (L-1) / 2 ) – 1 or also

INT ( Number of BIGHT / 2 )

Page 7 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

*** 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 (OZ) 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 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 n=0

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

5 + ( 2 * n ) BIGHT for n from 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 : OZ 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 OZ , there is no OZ+1-U1 only U1 - OZ Number of LEAD + 4

Page 8 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

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


B) 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 OZ 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 LB) : ( pin step 5.5 ( 5 and 6) ; Delta = 5 , Delta*=4) HP number

direction

HP-1 HP-2 HP-3 HP-4 HP-5 HP-6 HP-7 HP-8 HP-9 HP-10

L to R R to L L to R R to L L to R R to L L to R R to L L to R R to L

Column Column Column Column Column Column sixth first second third fourth fifth crossing crossing crossing crossing crossing crossing FREE FREE FREE FREE FREE FREE 9 is the LEFTMOST ‘O’ Column of this EVEN numbered HP 9 9 is the RIGHTMOST ‘O’ Column of this ODD numbered HP 9 9 7 9 7 9 7 5 9 7 5 9 7 5 3 9 7 5 3 10 – first UNDER 9 7 5 3 1

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 CODE seen by HP-1 on finished knot CODE for R to L Column

1

O U

Column number R to L

2 3

10

4

O

5

6

O

7

8

O

9

10

O

U

U

U

U

U

O

O

O

O

O

U 9 8

U 7

6

U 5

4

U 3

2

1

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) – (7*2) modulo 11 = -5 = 6 (11-2) – (6*2) modulo 11 = -3 = 8 (11-2) – (8*2) modulo 11 = -7 = 4 (11-2) – (9*2) modulo 11 = -9 = 2

Page 16 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

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 HP

direction

number

HP-1 HP-2 HP-3 HP-4 HP-5 HP-6

L to R R to L L to R R to L L to R R to L

Column first crossing FREE 5 5 5 5 5

O–U–O–U–O–U

Column Column Colum second third fourth crossing crossing crossing FREE FREE FREE 5 is LEFTmost ‘O’ on this HP 5 is RIGHTmost ‘O’ on this HP 3 3 3 1 6 ‘UNDER’

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

Page 17 sur 36

Charles HAMEL – 2011 June

Release V1.3.1 *** 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.

xxxxxxxxxxxxxxxxxxxx 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.

----

Page 18 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

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 LL so the ‘suite’ of crossings will ALWAYS be U - O……………….U – O

Page 20 sur 36

Charles HAMEL – 2011 June

Release V1.3.1 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 B L and B L.

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.

xxxxxxxxxxxxxxxxxxxx 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!!

Charles HAMEL – 2011 June

Page 21 sur 36

Release V1.3.1 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 ) (O1) (O1-O2) (O1-O2-O3) (O1-O2-O3-O4 (O1-O2-O3-O4-O5) (O1-O2-O3-O4-O5-O6) (O1-O2-O3-O4-O5-O6-O7 (O1-O2-O3-O4-O5-O6-O7-O8 (O1-O2-O3-O4-O5-O6-O7-O8-O9) NINE KNOTS

5L 3B 7L 5B 9L 7B 11L 9B 13L 11B 15L 13B 17L 15B 19L 17B 21L 19B

Step=4 (factorisation : 1 * 2 * 2) (O1) (O1-O2-O3) (O1-O2-O3-O4-O5) (O1-O2-O3-O4-O5-O6-O7) (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)

5L 3B 9L 7B 13L 11B 17L 15B 21L 19B

Step=6 (factorisation : 1 * 2 * 3) (O1) (O1-O2-O3-O4 ) (O1-O2-O3-O4-O5-O6-O7) (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)

5L 3B 11L 9B 17L 15B 23L 21B

Step=8 (factorisation : 1 * 2 * 2 * 2) (O1) (O1-O2-O3-O4-O5) (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)

5L 3B 13L 11B 21L 19B

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)

Charles HAMEL – 2011 June

Page 22 sur 36

Release V1.3.1 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 BB 3 + (2 * n ) LEAD 5 + (2 * n ) LEAD

LB 5 + (2 * n ) LEAD 7 + (2 * n ) LEAD

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

LB ; L1 ) then BOOM!!

xxxxxxxxxxxxxxxxxxxx

Spacing=4 3 Step=2 3 7 5 9 (or 9 5 ) 7 11 9 13 11 15 15 19 17 21 19 23 21 25 23 28

Spacing=5 (factor = 1 * 5 3 8

(factor = 1*2*2) 7 step=3 6 10 BOOM!! 9 13 12 16 BOOM!! 15 19

3 5

Step=2 8 10 BOOM!!

step=3 3 8 6 11 9 14 12 17 15 20 BOOM!!

Page 27 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

5L 9B 7L 11B 9L 13B 11L 15B 13L 17B 15L 19B 17L 21B

19L 23B

FR, FR, FR, O1, O1, O1, O1, U1-O1 FR, FR, FR, O1, O1, O1, O1, U1-O1 FR, FR, FR, O1, O1, O2, O2 O2, O2, U1-O2 FR, FR, FR, O1, O1, O2, O2, O2, O2, U1-O2 FR, FR, FR, O1, O1, O2, O2, O3, O3, O3, O3, U1-O3 FR, FR, FR, O1, O1, O2, O2, O3, O3, O3, O3, U1-O3 FR, FR, FR, O1, O1, O2, O2, O3, O3, O4, O4, O4, O4, U1O4 FR, FR, FR, O1, O1, O2, O2, O3, O3, O4, O4, O4, O4, U1O4

9L 5B

FR, O1, O1, O2-U1

11L 7B 13L 9B 15L 11B 17L 13B 19L 15B 21L 17B

FR, O1, O1, O2-U1

23L 19B

FR, O1, O1, O2, O2, O3U1 FR, O1, O1, O2, O2, O3U1 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!!

xxxxxxxxxxxxxxxxxxxx

Page 28 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

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

spacing=7 (factor = 1*7) 3 step=2 5 12 7 14 BOOM

10

2

9

step=3 6 13 9 16

step=2 4 11 6 13

step=3 5 12 8 15

12 19 15 22

8 15 10 17

11 18 14 21 BOOM

18 25 21 28 BOOM

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!

xxxxxxxxxxxxxxxxxxxx

Spacing=8 (factors = 1*2*2*2) Step=2 2 10!!

Step=3 3 11

so

6 14 BOOM!!

3

11

5 7 9 11 13 15 17 19

13 15 17 19 21 23 25 27

Spacing=9 (factors = 1*3*3*3 Step=2 2 11 or 3 12 !! 4 13 6 15 BOOM!!

Step=3 2 11 5

14

8

17

11 14 17 20 23 26 29 32

20 23 26 29 32 35 38 41

Page 29 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

3L 11B

5L 13B

7L 15B 9L 17B 11L 19B 13L 21B 15L 23B 17L 25B 19L 27B

2L 11B 5L 14B 8L 17B 11L 20B

14L 23B 17L 26B 20L 29B 23L 32B 26L 35B

FR, FR, FR, FR, FR, FR, FR, O1, O1 , O1, O1, O1, O1, O1, O1, U1-O1 FR, FR, FR, FR, FR, O1, O1 , O1, O1, O1, O1, O2, O2, O2, O2, O2-U1 FR, FR, FR, FR, O1, O1, O1, O1, O1-U1 FR, FR, FR, O1, O1, O1, O1, U1-O1 FR, FR, FR, O1, O1, O1, O1, U1-O1 FR, FR, FR, U1 for sobre 132 ‘O’ 120 ‘U’ FR, FR, FR, O1, O1, O1, O1, U1-O1 FR, FR, FR, O1, O1, O2, O2, O2, O2, U1-O2 FR, FR, FR, O1, O1, O2, O2, O2, O2, U1-O2

11L 3B

FR, O1-U1-O1

13L 5B

FR, U1, O1 for sobre 32’O’ 28 ‘U’

15L 7B

FR, O1-U1

17L 9B

FR, O1, O1, O2-U1

19L 11B

FR, O1, O1, O2-U1

21L 13B

FR, O1, O1, O2-U1

23L 15B

FR, O1, O1, O2-U1

25L 17B

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

27L 19B

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

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

11L 2B

FR, O5, O5; U1-O1…..

14L 5B

FR, O1-U1

17L 8B 20L 11B

FR, O2, O2, U1-O1….. FR, O1, U1

23L 14B

FR, FR, O1, O1, O1, O1, O2, O2, O2-U1 FR, FR, FR, O1, U1 FR, FR, FR, U1 for sobre 365 ‘O’ 339 ‘U’ FR, FR, FR, O1, U1

26L 17B

FR, U1 for sobre 159 ‘O’ 149 ‘U’ FR, O1, U1

29L 20B 32L 23B

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

35L 26B

FR, O1, O1, U1-O1

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’

xxxxxxxxxxxxxxxxxxxx

Page 30 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

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 … L>B ODD LEAD ODD BIGHT spacing=8 step=2 11L 3B 13L 5B 15L 7B

(FR, O1-U1….) (FR, U1-O1) ; (FR, O1-U1)

xxxxxxxxxxxxxxxxxxxx

Page 31 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

2L 11B…5L 14B…8L 17B…11L 20B----14L 23B B>L ODD LEAD ODD BIGHT spacing=9 step=3 2L 11B 5L 14B 8L 17B 11L 20B 14L 23B

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

11L 2B…14L 5B…17L 8B…20L 11B----23L 14B L>B ODD LEAD ODD BIGHT spacing=9 step=3 11L 2B 14L 5B 17L 8B 20L 11B 23L 14B

(FR, O5, O5, U1-O1….) (FR, O1-U1) (FR, O2, O2, U1-O1….) (FR, O1, U1) ( FR, U1 for sobre 159 ‘O’ 149 ‘U’)

xxxxxxxxxxxxxxxxxxxx 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)

xxxxxxxxxxxxxxxxxxxx …/…

Page 32 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

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

Let us try and amend the SPACING = 2 to SPACING 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

= k * 2 where k may have a value

----------------------------------------------------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’)

Page 33 sur 36

Charles HAMEL – 2011 June

Release V1.3.1 The larger the SPACING the smaller will be the length of the run. --------------------------------------------------------Take Spacing = 6 let us try 31L 25B

step=2 (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 10 factorise in 1 * 2 * 5 explodes on GCD=5 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 18 factorise in 1 * 2 * 9 explodes on GCD=9 20 factorise in 1 * 2 * 2 * 5 explodes on GCD=5 22 factorise in 1 * 2 * 11 explodes on GCD=11 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 ----------------------------------------

Page 34 sur 36

Charles HAMEL – 2011 June

Release V1.3.1 (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

7L 3B

5

9

9

7

11

9

13

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

11 15 13 17

15 19 17 21 21 23

5

FR, O1-U1, O1-U1 or (FR, U1-O1, U1-O1) both 9 over 9 under This start THK is not suitable 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, O4U1

19 15

FR, O1, O1, O2, O2, O3, O3, O4U1 FR, O1, O1, O2, O2, O3, O3, O4, O4, O5-U1 FR, O1, O1, O2, O2, O3, O3, O4, O4, O5-U1

21 17 23 19

Goes out of step

25 21

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 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 SPACING = 4 step= 4

27 23

FR, FR, FR, FR, FR O1, O1, O1, O1, O1-U1 FR, FR, FR, O1, O1, O1, O1, U1-O1 FR, FR, FR, O1, O1, O2, O2, O2, O2, U1-O2 Just the above but every two 15L 19B 21L 23B SPACING = 4 step= 8

7B 3L

FR, O1-U1

11 7

FR, O1, O1, O2-U1

15 11

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

FR, FR, FR, O1, O1, O1, O1,U1O1 FR, FR, FR, O1, O1, O2, O2,

9L 5B

FR, O1, O1, O2-U1

17 13

FR, O1, O1, O2, O2, O3, O3, O4-

25 27

3L 7B 7

11

11 15

5L 9B 13 17

29 25

FR, O1, O1, O2, O2, O3, O3, O4, O4, O5, O5, O6-U1 FR, O1, O1, O2, O2, O3, O3, O4, O4, O5, O5, O6-U1

FR, O1, O1, O2, O2, O3, O3, O4, O4, O5, O5, O6, O6,O7-U1

SPACING = 4

step= 4

Just the above but every two 19L 15B 23L 21B SPACING = 4 step= 8

Page 35 sur 36

Charles HAMEL – 2011 June

Release V1.3.1

21 25

29 33

37 41

O3, O3, O3, O3, U1-O3 FR, FR, FR, O1, O1, O2, O2, O3, O3, O4, O4, O5, O5, O5, O5, U1-O5 FR, FR, FR, O1, O1, O2, O2, O3, O3, O4, O4,O5, O5, O6, O6, O7, O7, O7, O7, U1-O7 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 SPACING = 8

3L 11B

5

13

7

15

9 17 11 19 13 21

3L 11B

9L 17 15 23 21 29 27 35 33 41 39 47 45 53

51 59

57 65

step= 2

25 21

33 29

41 37

U1 FR, O1, O1, O2, O2, O3, O3, O4, O4, O5 O5, O6-U1 FR, O1, O1, O2, O2, O3, O3, O4, O4, O5 O5, O6, O6, O7, O7,O8U1 FR, O1, O1, O2, O2, O3, O3, O4, O4, O5 O5, O6, O6, O7, O7, O8, O8, O9, O9,O10-U1

SPACING = 8

step= 2

FR, FR, FR, FR, FR, FR, FR, O1, O1, O1, O1, O1, O1, O1, O1, U1-O1 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 FR, FR, O1, O1, O1, O1, U1-O1 FR, FR, FR, FR, O1, O1, O1, O1, U1-O1 FR, FR, FR, U1 SPACING = 8 step=4

11L 3B

FR, O1-U1-O1

13

5

FR, U1-O1

15

7

FR, O1-U1

17 19

9 11

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

Just the above but every two SPACING = 8 step= 6

Just the above but every two SPACING = 8 step=6

FR, FR, FR, FR, FR, FR, FR, O1, O1, O1, O1, O1, O1, O1, O1, U1-O1 FR, FR, FR, O1, O1, O1, O1, U1-O1 FR, FR, FR, O1, O1, O1, O1, U1-O1 FR, FR, FR, O1, O1, O2, O2, O2, O2, U1-O2 FR, FR, FR, O1, O1, O2, O2, O3, O3, O3, O3, U1-O3 FR, FR, FR, O1, O1, O2, O2, O3, O3, O4, O4, O4, O4, U1-O4 FR, FR, FR, O1, O1, O2, O2, O3, O3, O4, O4, O4, O4, U1-O4 FR, FR, FR, O1, O1, O2, O2, O3, O3, O4, O4, O5, O5, O5, O5, U1-O5 FR, FR, FR, O1, O1, O2, O2, O3, O3, O4, O4, O5, O5, O6, O6, O6, O6, U1-O6 FR, FR, FR, O1, O1, O2, O2, O3, O3, O4, O4, O5, O5, O6,

FR, O1, O1, O2-U1 21 13 SPACING = 8 step= 4

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, O4U1 FR, O1, O1, O2, O2, O3, O3, O4, O4, O5-U1 FR, O1, O1, O2, O2, O3, O3, O4, O4, O5-U1 FR, O1, O1, O2, O2, O3, O3, O4, O4, O5, O5, O6-U1

41 33 47 39 53 45

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-

Page 36 sur 36

Charles HAMEL – 2011 June

Release V1.3.1 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 5 11 FR, FR, FR, FR, FR, O1, O1, O1, O1, O1-U1 7 13 FR, FR, FR, O1, O1, O1, U1O1 9 15 BOOM! factor of spacing GDC=3

SPACING = 6 STEP = 2 11 5 13 7 15 9

BOOM! factor of spacing