PLEASE NOTE: Even though seeming almost infinite in variations and kinds, Golden Harmonic Ratios and Fibonnaci Numbers do have limits. Their bewildering array is consolidated by knowing of the existence of a set of inter harmonic permutive ratios by which any powers of the Golden Section number are directly calculated by the addition and subtraction of certain pieces of other powers of the Golden Harmonic. From this, direct bridge between Fibonnaci and Lucas numbers (oppositive geometric sequence to Fibonnaci) is correctly and instantly shown. This interface between the two sequences (Fibonnaci and Lusas) is shown in Golden.txt

GOLDEN RODS This document has ascii characters. It reads fine at Dos or in any compatable editor but not in Windows. In Windows the ascii characters are substituted by Ansi surrogates such as capital letters and strange looking symbols. In a Windows Browser you cannot read this document's equations, since the strange codes surround and change the equational content's appearance too drastically. The best view is in any Dos environment such as an editor. 旼컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴커 � ANCIENT BOOTY REVEALS A TREASURE CHEST FOR THE FUTURE � 읕컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴켸 The Golden Rectangle is an example of an ABSOLUTE image. So is 5 sided geometry. The mathematical understructure is duplicated in thermodynamic expressions in nature through numerous dissimilar regions of molecular behavior including crystal lattices, golden spirals, the branching of certain leaves in vines and plants, and sunflower seed spirals, etc. In lesser practices, spinoffs of the GOLDEN ratio (in Pentagrams for instance) are used for magic and casting of spells, which is nonsense in the higher Realities of existence. This document explores where the GOLDEN RATIOS have absolute and abstract foundations which are clear of malpractices such as stardom entertainment, religious rituals, self serving magic, and the power trips of witches. It is impossible to tie diabollica ambitions to GOLDEN HARMONIC information which is genuinely Cosmic. Cosmic GOLDEN HARMONICS are sourced in intricacies of a six-sided Cube and Sphere. So let's say there are Golden Absolutes. The Golden Absolutes hold together numerous formulas which can be specified, independant of images. FOR INSTANCE 旼컴컴컴컴컴컴컴컴 \� (any number) + 1 = X1 EQ. 0 旼컴컴컴컴컴컴컴컴 \� (X1) + 1 = X2 旼컴컴컴컴컴컴컴컴 \� (X2) + 1 = X3 sufficiently repeated 旼컴컴컴컴컴컴 \� (X(n...) + 1 = G = 1.618033988875... 읕컴훒volves toward컴컴켸 INEVITABLY: The square root of the sum of (any number plus 1), is the start of a repeat which inevitably homes towards a result having a converging value that closes upon a constant value historically known as the Golden Section, (herein called `G'), which is historically has been defined as: (��5) + .5 = g = 1.61803398875 ... (i.e. G) ; in which the number value G is thought to be irrational, locked to the rational values of 5 and .5 , as well as to the proportionate arms of a 5 sided absolute geometry figure, and to the sides of a Golden Rectangle. In effect the irrationality of G can be said to be co-incidental to the decimal digital system used to express it in base 10. The Golden Rectangle (and its Golden Ratio) is thought to have many sidepaths. In fact it has been said that new (previously unpublished) permutations may be found by almost anyone who looks for such, in an almost infinite array of variables. Such is the motility of the number G . Another motility example is that G x �5 = G + 2 in which digits after a decimal point recurr. 1.61803398875 + 2 = 3.61803398875. G x 2.23606797750 = 3.61803398875. SOME OF THE MOTILITY CAN BE GROUPED INTO A SINGLE SOURCE SERIES FOR THE GOLDEN SECTION, WHICH GENERATES FIBONNACI NUMBERS; AS FOLLOWS: TABLE 1 SINGLE SOURCE SERIES FIBONNACI NUMBERS 컴컴컴� 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴 1 G = 1G + 0 커 � + 1 main 旼 2 켸 whole � 3 + 5 = 8 G = 1G + 1 커 ........begins .... integer as in � 5 + 8 = 13 � + 0 digit � 8 + 13 = 21 3 켸 series 읕 G = 2G + 1 커 .....Continue � + 1 4 켸 G = 3G + 2 커 � + 1 This source series as shown is 5 켸 not absolute when represented G = 5G + 3 커 F in terms of base 10. � + 2 I 6 켸 B It would be absolute if shown in G = 8G + 5 커 O a geometry figure having parts � + 3 N expanding at the said rate of 7 켸 N G = 13G + 8 커 A 1 1+1 1+n... � + 5 C G G .... G ; 8 켸 I G = 21G + 13 커 and each expansion is shown to � + 8 N be equal to other geometry parts 9 켸 U whose total of summed multiple G = 34G + 21 커 M parts equals � + 13 B 1+n... 10 켸 E G G = 55G + 34 커 R in which � + 21 case, then, the Fibonnaci digits 11 켸 S represent multiples of geometric G = 89G + 55 커 U pieces which needn't have decimal � + 34 M measures to be counted. Golden 12 켸 S Rectangles leading to a Golden G = 144G + 89 커 Spiral would be such a geometry � + .....Continue figure, for instance. Golden etc. ... ... 켸 Rectangles welling up and down within larger and smaller 1st 2nd 3rd pentagrams would be another series series series example. Fibonnaci Fibonnaci Fibonnaci An obvious observation is that any power of G can be predicted in advance, given only the existence of Fibonnaci Numbers. For instance in EXAMPLE 0 following, a higher powers of G is determined from a list of Fibonnaci numbers per se, via the following reasoning. Any Fibonnaci Term (X) can be determined as the result of G times the next Fibonnaci TERM above (X) , plus the Fibonnaci Term of (X) itself. For instance with a list of Fibonnaci Numbers (such as LIST 1 below); the wish is to calculate the 27th power of G . This is done by a combination of a simple multiple and addition. Specifically, G is multiplied by the value of the 28th Fibonnaci Term, and to this is added the value of the 27th Fibonnaci Term, to yield the value of the 27th power of G . Find a Power of G by MULTIPLYING AND ADDING EXAMPLE 0 旼컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� � The 27th Power of G = 439204.000004 ..... � 읕컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� Step 1. 196418 = Value of the 28th Fibonnaci Term Step 2. 317811.000002 = G times the 28th Term (196418) Step 3. + 121393 = Value of the 27th Fibonnaci Term sum 컴컴컴컴컴컴� 439204.000002 = The 27th power of G which is Step 3 added to Step 2. LIST 1 旼컴컴컴� Term � 旼컴컴컴� Fibonnaci Number � � 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴 32 Fibonnaci Numbers are listed in sequence 1. 0 9. 21 17. 987 25. 46368 2. 1 10. 34 18. 1597 26. 75025 3. 1 11. 55 19. 2584 27. 121393 4. 2 12. 89 20. 4181 28. 196418 5. 3 13. 144 21. 6765 29. 317811 6. 5 14. 233 22. 10946 30. 514229 7. 8 15. 377 23. 17711 31. 832040 8. 13 16. 610 24. 28657 32. 1346269 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴 ..... Continue A second level observation is already historically well known. That is, each successive or preceeding Fibonnaci Term in the lineal series of Fibonnaci numbers can be determined merely by adding or subtracting proceeding or preceeding Fibonnaci Numbers from the second next higher or lower Term. For example, to calculate the value of Fibonnaci Term 27 , add the the value of Fibonnaci Term 25 to Fibonnaci Term 26. Or conversely, subtract the value of Fibonnaci Term 28 from Fibonnaci Term 29. For instance, using data from LIST 1 immediately above, it is obvious that: Find a Fibonnaci Number by ADDING 旼컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� � Fibonnaci Term 27 = 121393 ..... � 읕컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� Fibonnaci Term 25 = 46368 Fibonnaci Term 26 = 75025 + 컴컴컴컴 Sum: 121393 ..... Hence: Term 27 is calculated by adding Term 25 to Term 26. Find a Fibonnaci Number by SUBTRACTING 旼컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� � Fibonnaci Term 27 = 121393 ..... � 읕컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� Fibonnaci Term 29 = 317811 Fibonnaci Term 28 = 196418 - 컴컴컴컴 Difference: 121393 ..... Hence: Term 27 is calculated by subtracting Term 28 from Term 29. Other aspects of Fibonnaci Numbers and powers of the Golden Ratio are immediately obvious as seen in TABLE 1 (further above). For instance it is found that: (n) In TABLE 1 any G ......... divided by the Fibonnaci Number Sum of its: 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� 旼� 컴� � st nd � � 1 series term + 2 series term � 읕� 컴� approaches 旼 커 � 1 � � 컴컴컴 � + 1 = 1.38196601125 = �5 � G � G + 1 � 읕컴컴컴컴컴� 읕 켸 � � � 읕컴컴컴 = .38196601125 � � � For instance, given ; � � � 12 旼 커 � G = � 144 G + 89 � � 읕 켸 � � � then ; 144 + 89 = 233 ; � � � 12 � and since ; G = 321.9968943801 ; � � � then ; � 旼 12 커 � � G � � � 컴컴컴 � = 1.38196020 <컴컴컴컴� � 233 � 읕 켸 .... approaching a convergence 旼컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� � FURTHER GOLDEN MOTILITY � 읕컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� Further Golden Section motility can be grouped into a Multiple Source Series, in which FIBONNACI Numbers (hence Golden Harmonics) are generated from LUCAS Numbers (and visa versa); as follows: TABLE 2 LUCAS NUMBERS MULTIPLE SOURCE SERIES 컴컴컴� 컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴� 旼컴 These are basic approximated FIBONNACI Numbers; times G � � 旼컴 FIBONNACI Numbers � � � � 旼컴 FIBONNACI Numbers � � � � � � 旼컴 LUCAS Numbers � � � � � � � � 旼컴 LUCAS Numbers � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � 1 G = 1G + 0 = 1/2 ( 1�5 + 1) 커 � 2 2 켸__________ begin ______ main G = 1G + 1 = 1/2 ( 1�5 + 3) 커 whole � 1 integer 3 켸 digit G = 2G + 1 = 1/2 ( 2�5 + 4) 커 LUCUS Sum � 3 series 4 켸 G = 3G + 2 = 1/2 ( 3�5 + 7) 커 � 4 5 켸 旼컴컴컴커 G = 5G + 3 = 1/2 ( 5�5 + 11) 커 � Where: � � 7 읕컴컴컴켸 6 켸 An example for finding G = 8G + 5 = 1/2 ( 8�5 + 18) 커 a LUCAS Number (n) � 11 is as follows: 7 켸 G = 13G + 8 = 1/2 (13�5 + 29) 커 When: � 18 7 8 켸 G = 13G+8 = 1/2(13�5+(n)) G = 21G + 13 = 1/2 (21�5 + 47) 커 � 29 9 켸 Then: G = 34G + 21 = 1/2 (34�5 + 76) 커 7 � 47 2G - (13�5) = (n) = 29 = 10 켸 a LUCAS Number G = 55G + 34 = 1/2 (55�5 + 123) 커 � 76 Because: 11 켸 G = 89G + 55 = 1/2 (89�5 + 199) 커 2 (13G+8) - (13�5) = (n) � 123 읕컴컴컴컴켸 12 켸 G = 144G + 89 = 1/2 (144�5 + 322) 커 Note the similar digits: � 199 13 켸 13�5 = 29.068883707 G = 233G + 144 = 1/2 (233�5 + 521) 커 13G+8 = 29.034441853 커 � 322 13G+8 - (13�5) � Etc. ... ... 1/2 (n�n + n) 켸 = .034441853 켸 .....Continue 旼컴컴컴커 � And: � 읕컴컴컴켸 An example of finding a next LUCAS Number: | 6 | given G = | 8G + 5 = 1/2 (8�5 + 18) | 7 | given G = | 13G + 8 = 1/2 (13�5 + 29) 컴� | + � 18 8 | 컴� find G = | 21G + 13 = 1/2 (21�5 + (n)) (n) = 47 | � | 읕컴컴커 | � Therefore the next LUCAS Number (n) = 47 since 18 + 29 = 47 旼컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴커 � And also LUCAS and FIBONNACI Numbers can find FINONNACI Numbers: � 읕컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴켸 旼횸ucas 旼횸ucas 旼컴Fibonnaci � � 旼컴Fibonnaci � � � � 旼컴훞ext Fibonnaci 6 | G = 8G + 5 = 1/2 (8�5 + 18) | 18 - 5 = 13 as in 5, 8, 13 | -- 7 | G = 13G + 8 = 1/2 (13�5 + 29) | 29 - 8 = 21 as in 8, 13, 21 | -- Both Single, and Multiple, SOURCE SERIES are formed in a major numbers system. But from there on things get busier and busier. Noticable are that the even numbered square powers of G (the Golden Harmonic Ratio 1.6180339887) approximates every 2nd value of Lucas Numbers, with the Golden Ratio values converging toward its Lucus value through greater powers of n . G 7 For instance, G is 29.034441853 , and the associated LUCAS NUMBER (from TABLE 2) is 29. 13 Whereas the number G is 521.001919378 whose associated LUCAS NUMBER (from TABLE 2) is 521 . It is clear that the approximation toward the absolute LUCAS NUMBER's value is greater for 521 than for number 29 . TABLE 3 GOLDEN RATIO POWERS 컴컴컴� In another series (with every next 6th power of G highlighted) an obvious progression occurs; as follows: 旼컴컴� � 1 � 1 � G � = 1.6180339887 and 1/G = .6180339887 읕컴컴� Approximates 1st Lucas Number = 1 旼컴컴컴컴컴컴 (n) terms of 6 x G . � 2 넬컴컴컴컴� � G = 2.6180339887 납 Where: � regards � 냅컴컴컴컴� every 6th � 3 旼컴컴컴� = �5 + 2 � power of G � G = 4.236067977 � � Approximates 3rd Lucas Number = 4 � (n)x6 (n)2 � 4 � G = (�5 +2) � G = 6.854101966 � is an inner series � � (identified by the � 5 � whole no. integers � G = 11.090169943 � in the Left margin). � Approximates 5th Lucas Number = 11 � � 6 2 � The Decimal parts 1 G = 17.944271910 = (�5 + 2) � of EVEN numbered � powers of (n) 7 � G G = 29.034441853 � settle toward 1. Approximates 7th Lucas Number = 29 � 8 � The Decimal parts G = 46.978713763 � of ODD numbered � powers of (n) 9 � G G = 76.013155617 � settle toward 0. Approximates 9th Lucas Number = 76 � 10 � In particular, G = 122.991869381 � the reciprocal � of each of the 11 � (n) G = 199.005024998 � ODD powers of G Approximates 11th Lucas Number = 199 � is the part after 12 = 2 x 6 2x2=4 � the decimal. 2 G = 321.996894380 = (�5 + 2) � � 1 15 13 � See G and C G = 521.001919378 � for examples. Approximates 13th Lucas Number = 521 � 14 � G = 842.998813758 � 旼컴컴� 15 � 15 � = 1364.000733137 and 1/G = .000733137 � G � 읕컴컴� NOTE: that the reciprocal of an ODD power is the decimal portion of its number value. 16 G = 2206.99954689 � 17 � The part after the G = 3571.00028003 � decimal of each EVEN � powered (n) is: 18 = 3 x 6 3x2=6 � G 3 G = 5777.99982693 = (�5 + 2) � � 旼컴컴 컴컴컴컴컴� 19 � � 1 � G = 9349.00010696 � � 컴컴컴컴컴컴컴컴 � � � 旼� 컴� � 20 � � � 1 � � G = 15126.9999339 � � � 컴컴컴컴컴컴 � � � � � 旼 (n) 커 � + 1 � 21 � � � � G - 1 � � � G = 24476.0000409 � � � 읕� 컴� � � � � 읕컴컴 컴컴켸 � 22 � 읕컴컴컴� 컴컴컴컴컴컴� G = 39602.9999747 � � As for example: 23 � G = 64079.0000155 � 18 � (G - 1) = 5776.99982693 24 = 4 x 6 2x4=8 � 4 G = 103681.999990 = (�5 + 2) � 18 � 1/(G - 1) = .000173100 25 � = X1 G = 167761.000006 � 1 � � 컴컴컴컴 = .999826929 ---------------�----------------- � (X1 + 1) = X2 � � 30 10 � 18 5 G = 1860498.00000 = (�5 + 2) � (G - X2) = 5777.000000000 � � � � � 읕� Approximates 25th Lucas Number = 167761 so convergence toward zero (thus total equality) is now plain to see. (n) And the reciprocal of each EVEN powered G is the difference between the real value of (n) (n) rounded off to the nearest whole number. G and G This is no accident, because there is actually: 1. 旼 (n) odd (n) odd 커 (n) odd � G - 1/G � = The LUCAS Number of that G 읕 켸 2. 旼 (n) even (n) even 커 (n) even � G + 1/G � = The LUCAS Number of that G 읕 켸 For LUCAS Numbers as found in the LUCAS series of TABLE 2. Digital trunkation becomes a problem when attempting to proove the reality of the equation immediately above for higher EVEN powers of G . MORE EXAMPLES: 13 [odd] | 13 1. G = 521.00191378 | 1/G = .001919378 | | 旼 13 13 커 | 13 旼컴컴� L � G - 1/G � = 521 | and G = 1/2 (233 �5 + 521) � U 읕 켸 | � C | � A -------------------0--------------------- � S | � 14 [even] | 14 � n 2. G = 842.999813758 | 1/G = .001186241 � u | � m | � b 旼 14 14 커 | 14 � e � G + 1/G � = 843 | and G = 1/2 (377 �5 + 843) � r 읕 켸 | 읕컴컴� s 幡賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽� 넴같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같겡� 白複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複� With these preceding informations, it is obvious that the historically known Golden Section (i.e Golden Harmonic Ratio) rather than offering stale mysteries ad infinitum, is a perfectly coherent interdependant system of number sums and ratios, which is both definative and predictive. For instance, unknown FIBONNACI terms in the source series of TABLE 1 can be predicted given only (n) . G (n) Since any source series factor contains 2 unknowns; given only G these unknowns can be identified as: (n) 旼 커 旼 커 G = G � n � + � n � where n and n are the unknowns. 읕 f 켸 읕 i 켸 f i To understand their predictiveness requires something of a keen insight into the differences between: A. An approximate value of a number (i.e. an approached value). B. A rounded off value of a number. C. And the real value of a number. Such insight is provided in the following news (as well as throughout the preceding information above). Contained (in a sample segment) are the following number series, patterns, and ratios: TABLE 4 SERIES, PATTERNS, and RATIOS 컴컴컴� 旼 LUCAS GROUP 旼� FIBONNACI GROUP � � � 旼 Lucas � � � Numbers � � � 18 G = 5777.99982693 = (2584 G + 1597) = 1/2 (2584 �5 + 5778) 19 G = 9349.00010696 = (4181 G + 2584) = 1/2 (4181 �5 + 9349) 20 G = 15216.9999339 = (6765 G + 4181) = 1/2 (6765 �5 + 15127) (n) Note that each Lucas Number is the G value rounded to the nearest whole number. 21 G = 24476.0000409 = (10946 G + 6765) = 1/2 (10946 �5 + 24476) 22 G = 39602.9999747 = (17711 G + 10946) = 1/2 (17711 �5 + 39603) 23 G = 64079.0000155 = (28657 G + 17711) = 1/2 (28657 �5 + 64079) � � � 旼컴컴컴커 � � � Approximate � RATIOS � include: � � � 읕컴컴컴켸 times G times G times G � � � 24 � � � G = 103681.999990 = (46368 G + 28657) = 1/2 (46368 �5 + 103682) � 읕컴컴컴컴컴컴컴컴컴켸 읕컴컴컴켸 읕컴컴컴컴컴컴� � �5 is approximated x G is approximated (approx. x �5) 읕컴컴컴컴컴컴� (approx. x G) The horizontal ratios are uniform for any (n) G 旼 (n) 커 � Also any G � � x G = � Ratios between like parts are uniform � (n)+1 � for any two adjacent powers. � vertical G � 읕 켸 旼컴컴커 � SUMS � 읕컴컴켸 旼 커 � (n) (n)+1 � (n)+2 Any � G + G � = G 읕 켸 旼 커 � 20 21 � 22 For example, the SUM of � G + G � = G 읕 켸 Also, SUMS in non lineal combinations are instate. For instance: TABLE 5 INSTATE SUMS EXAMPLES 컴컴컴� 컴컴컴컴컴컴컴컴컴컴� 18 21 23 17 G + G = 1/2 (G - G ) -------------------------------RANDOMLY CHOSEN TEST CASE 20 22 24 18 G + G = 1/2 (G - G ) -------------------------------RANDOMLY CHOSEN TEST CASE BEGIN..... 18 20 22 16 RANDOMLY CHOSEN AS G + G = 1/2 (G - G ) A SERIES. INDICATED IS PROGRESSIVE INCREMENTATIONS 18 21 23 17 IN (n) SUM PARTS. G + G = 1/2 (G - G ) G 18 22 24 18 17 G + G = 1/2 (G - G - 2G ) 18 23 25 19 18 17 G + G = 1/2 (G - G - 2G - 2G ) 18 24 26 20 19 18 17 G + G = 1/2 (G - G - 2G - 2G - 2G ) Predicted and correct. 26 19 = 1/2 (G - 4G ) But this is news of a different order. NOTE HOWEVER THAT: 旼 커 � (n) (n)+(n)... � from the look of it � G + G � in a series lacks simple made 읕 켸 all inclusive statements suitable for predicting any given values of (n). 幡賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽賽� 넴같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같겡� 넴같같 ROOT 5 AND THE GOLDEN RATIO (PERMUTATIONS) 같같같같같같같같枋 넴같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같같겡� 白複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複複� A permutive Golden Section Series is also formed upon the values of G and �5 . This series has expression for EVEN powered (n) as follows: G (n) TABLE 6 FOR EVEN POWERS OF G � �5 컴컴컴� 旼 커 � 2 � 旼 커 (n) � 컴컴컴컴� � 넬 (n) � � G � (n) � 납 G � � LOGO: 컴컴컴 THEN: � G � = 10 납 컴컴� � - (n ) � �5 � 컴컴� � 납 �5 � f � � �5 � 냅 � � 읕 켸 읕 켸 AND SO: WHEN: THEN: HENCE: A partial 2 � 2 � Fibonnaci G /�5 = 1.170820393 � 2/(G /�5) = 1.70820393 (1/10 + 0) � Numbers � actually x (10 - 10) � Series 4 � 4 � using G /�5 = 3.0652475841 � 2/(G /�5) = .652475841 (1/10 + 3) � every � � second 6 � 6 � Fibonnaci G /�5 = 8.024922359 � 2/(G /�5) = .24922359 (1/10 + 8) � number. � � 8 � 8 � These terms G /�5 = 21.009519494 � 2/(G /�5) = .09519494 (1/10 + 21) � correspond � � to the 10 � 10 � 1st series G /�5 = 55.003636123 � 2/(G /�5) = .0363123 (1/10 + 55) � found in � � the Single 12 � 12 � Source G /�5 = 144.001388875 � 2/(G /�5) = .01388875 (1/10 + 144)� Series of � TABLE 1. � � 읕컴컴컴컴컴컴컴컴컴컴컴컴컴컴컴켸 re-occuring digits .............................. (n) G /�5 Continued . . . In intepretation; when given an EVEN numbered (n) as in 旼 커 � (n) � � G /�5 � ; then: 읕 켸 旼 커 2 � 旼 커 旼 커 � 컴컴컴컴컴 � � (n) � � � � 旼 커 10 � � G /�5 � - � n � � = � (n) � � 읕 켸 읕 f 켸 � � G /�5 � 읕 켸 읕 켸 where n is the Fibonnaci Number (n) associated with (n)G f of a given (n) in the G 1st series, shown in TABLE 1. (n) 6 For instance, when G + G ; then in the Single Source 6 Series of TABLE 1, G = (8G + 5) ; having (8G) in the 旼 커 6 1st Series; hence the � n � of G in: 읕 f 켸 旼 커 � 旼 커 旼 커 � � � (n) � � � � 10 � � G /�5 � - � n � � is 8 . � 읕 켸 읕 f 켸 � 읕 켸 Herein; the Single Source Series can be included as: 旼컴 Single source factor � 旼 커 2 � 旼 커 � 컴컴컴컴컴� � � 8G + 5 � � 旼 커 10 � � 컴컴컴 � - 8 � = � 6 � � � �5 � � � G � � 읕 켸 � � 컴컴컴� � 읕 켸 � �5 � 읕 켸 But this is only one of the ways a given expression in the Permutations Series of TABLE 6 can be numerically described. Another way is even unwieldier: 旼� 컴� � 2 � � 컴컴컴컴컴컴 � 6 � 旼 커 � G � � 8G + 5 � � + 8 = 컴컴컴컴 EQ. 1 � � 컴컴컴컴 � � �5 � � �5 � � � 읕 켸 � 읕� 컴� 컴컴컴컴컴컴컴컴 10 Which can be even more unwieldy, yet still be correct: 旼� 컴� � 2 � � 컴컴컴컴컴컴컴컴컴컴 � (n) � 旼 커 旼 커 � 旼 커 G � � n � G + � n � � + � n � = 컴컴컴컴 EQ. 2 � � f � � i � � � f � �5 � 읕 켸 읕 켸 � 읕 켸 � 컴컴컴컴컴컴컴컴컴컴 � � �5 � 旼 커 旼 커 읕� 컴� where � n � and � n � are 컴컴컴컴컴컴컴컴컴컴컴컴컴 � f � � i � 10 읕 켸 읕 켸 the two adjacent Fibonnaci Numbers associated with any given (n) G in the Single Source Series summarized in TABLE 1. 旼 커 � n � Otherwise: � f � is the 1st Series Term, 읕 켸 旼 커 � n � � i � is the 2nd Series Term of TABLE 1. 읕 켸 Suppose (F) stands for a Fibonnaci Number; and (F1), (F2), (F3), are three in a consecutive Series such that: (F3) = (F1) + (F2) where n can be (F3) or (F2) (F2) = (F3) - (F1) f (F1) = (F3) - (F2) and n can be (F2) or (F1). i (n) Then given only G ; therefore having no prior knowledge of (F1), (F2), (n) (n) or (F3) associated with G ; n is found by dividing G by �5 f and rounding the result to the nearest whole number. And n is found by i dividing n by G and rounding the result to the nearest whole number. f 12 For instance: G = 321.996894379 = (144 G + 89) 12 and G = 144.001388875 컴컴컴 �5 which rounds off to 144; which is n of f (n) 旼 커 旼 커 G = � n � G + � n � ; � f � � i � 읕 켸 읕 켸 and 144 = 88.996894379 which rounds off to 89; 컴컴컴 to become n so that: G i 旼 커 旼 커 (n) � (n) � � (n) � G = � G � � + � G � � � where the arrows � 컴컴컴컴 � � � 컴컴컴컴 � � � indicate each � �5 � <켸 � �5 � <켸 � rounding off. 읕 켸 읕 켸 � 컴컴컴컴컴컴 <켸 G This approach - approximating to find n and n as otherwise f i (n) unknowns given only G - works because of the RATIOS pointed out in TABLE 4 ; where for instance in the LUCAS Group, each 旼 커 LUCAS Number is a rounded off value of � (n) � � and is also � G <� � 읕 켸 旼 커 closely approached by � n � �5 . So that, for example: � f � 읕 켸 旼컴 컴커 � 旼 커 � 23 � � 旼 커 � 23 � � G = 1/2 � � � n � �5 � + G <� � � � � f � � � � � 읕 켸 � � EQ. 3 � 읕 켸 � 읕컴 컴켸 旼 커 旼 커 旼 커 However; since the � n � value of � n � = � (n) � � f � � f � � G � � 읕 켸 읕 켸 � 컴컴컴컴 � � � �5 � <켸 읕 켸 instates as a rounded off value; then an attempt such as the equations: 旼컴컴 컴컴커 � 旼컴 컴커 � � � 旼 커 � 旼 커 � (n) � � � (n) � � � (n) � � � G = 1/2 � � � G � � �5 � + � G � <� � � � � 컴컴컴컴 � � � 읕 켸 � � � � �5 � <켸 � � � � 읕 켸 � � � 읕컴 컴켸 � 읕컴� 컴컴켸 旼컴 컴커 � 旼 커 旼 커 � � � (n) � � � (n) � � � = 1/2 � � G � <� + � G � <� � � 읕 켸 읕 켸 � 읕컴 컴켸 OR 旼컴 컴커 � 旼 커 旼 커 � (n) � � (n) � � (n) � � � G = 1/2 � � G � + � G � <� � have to be innacurate, � 읕 켸 읕 켸 � as well as awkward. 읕컴 컴켸 In a brief summary, in Golden Section Motility: when (F) are Finbonnaci Numbers and (L) are Lucas Numbers THEN: FIBONNACI GROUP LUCAS GROUP 旼� 컴� SERIES (n) 旼 커 旼 커 � 旼 커 旼 커� FACTORS G = � n � G + � n � = 1/2 � � n � �5 + � L 납 � f � � i � � � f � � f 납 읕 켸 읕 켸 � 읕 켸 읕 켸� 읕� 컴� 1st 2nd 1st 1st F F F L series series series series AND: SYSTEM FACTORS (n) � � G = (F3) G + (F2) 커 = 1/2 � (F3) �5 + (L2) � 커 � � � � + � (F1) + � (L1) (n)+1 � � � � G = (F4) G + (F3) 켸 = 1/2 � (F4) �5 + (L3) � 켸 � � 1st 2nd 3rd 1st 1st 2nd F F F F L L series series series series series series ---------------------------------------------------------------- ALSO: (n) 旼 커 G � 1 � 컴컴컴컴컴컴컴� approaches � 컴컴컴� � + 1 = (2 - G) + 1 旼 커 旼 커 � G + 1 � � n � + � n � 읕 켸 = 1.3819660112 � f � � i � 읕 켸 읕 켸 � This number has � been shown further 읕 above in facts (n) preceding TABLE 2. G 旼 커 컴컴컴컴 approaches � n � �5 � f � 읕 켸 旼 커 FOR � n � SOURCE � f � 旼 커 SERIES 읕 켸 approaches � n � FACTORS 컴컴컴컴 � i � OF G 읕 켸 TABLE 1 AND: 旼 커 旼 커 � n � + � n � 旼 커 � f � � i � approaches � n � 읕 켸 읕 켸 � f � 컴컴컴컴컴컴컴� 읕 켸 G ------------------------------------------------------------------- ALSO: AND: (n) for any G 旼 커 旼 커 旼 커 � n � + � L � � L � � f � � f � � f � 읕 켸 읕 켸 읕 켸 � � 컴컴컴컴컴컴컴컴� = 2 ; 컴컴컴컴 approaches 쿒 (�5) = G + 2� 旼 커 旼 커 旼 커 � � � n � + � n � � n � � f � � i � � i � 읕 켸 읕 켸 읕 켸 AS WELL: AND: (n)+1 (n)+2 G 1 G 2 컴컴컴컴 = G 컴컴컴컴 = G (n) (n) G G SO THAT: 旼컴컴커 旼컴컴커 旼컴컴커 旼컴컴커 퀁atios� 퀁atios� 퀁atios� 퀁atios� 읕컴컴켸 읕컴컴켸 읕컴컴켸 읕컴컴켸 (n)+3 커 커 커 커 G � 1 � � � 쿒 � � � � � � � (n)+2 캑 � � � � G � 1 � � � � � G � � 2 � � � � � G � 3 � (n)+1 캑 켸 � � G � 4 G � 1 � � � G � G � � � � � � � (n) 켸 켸 켸 켸 G AND ALSO: 旼컴커 旼컴커 퀂ums� 퀂ums� 읕컴켸 읕컴켸 (n)+5 커 G � (n)+3 - = � G 커 � � (n)+1 (n)+4 孤 - = � G G � (n)+2 � - = � G 孤 � � (n) (n)+3 孤 - = � G G � (n)+1 � - = � G 孤 � � (n)+2 孤 � 1 G � (n) - = � - = � G � � 켸 (n)+1 켸 G etc.... ALSO WHEN: EXERCISE 1. (n) � z0 = G (�y) (n)+1 where (y) is any � z1 = G (�y) number, such as �5, or � (n)+2 � z2 = G (�y) (n)+3 � z3 = G (�y) THEN: EXERCISE 2. 旼컴컴커 旼컴컴커 旼컴컴커 旼컴컴커 퀁atios� 퀁atios� 퀁atios� 퀁atios� 읕컴컴켸 읕컴컴켸 읕컴컴켸 읕컴컴켸 z0 커 커 커 � 2 � � � � G � � � z1 孤 � 커 � � � 2 � � � � � G � � 4 � � z2 孤 켸 � G � 6 � � 2 � � G � etc.... � G � � � z3 켸 켸 켸 However, EXERCISES 1 and 2 are merely tautologies, tossed in as examples to show that Golden Ratio Numbers can show up in seemingly many mysterious ways but when investigated, such occurrances are found to be very ordinary. FINISHED IN DIVINE ORDER IN LOVE AND SERVICE PEACE JOY AND HAPPINESS