Fibonacci Sequence, Golden 
Ratio and Wythoff's Game 
LOEKY HARYANTO 
Mathematics Department, Hasanuddin University, 
email: L.Haryanto@unhas.ac.id 
HaryantoL@outlook.com 
GSM’s #: +6281342127598
A Relationship among Fibonacci 
Sequence, Fibonacci Word, P-positions 
of Wythoff’s game and the Golden Ratio 
   
1 5 
2 
USE THIS PRESENTATION AS A NEW STRATEGY FOR STUDENT-CENTERED 
LEARNING (SCL) METHOD. 
The Maplet copies here were created to make students firstly being familiar with (not 
necessarily mastering the theory of) combinatorial game and its related topics before 
the students being introduced with the theoretical parts of the subject; e.g. before 
they were given some formal theories which were written in the next page! 
By the way, since mathematics is a language which is full of written symbols, without 
visual and ‘seemingly’ interactive presentations, most of students tend to sleep in abstract 
algebra classes. Nevertheless, IMO most strategies proposed for the SCL method by 
experts in education are not appropriate for math classes, or even worse than the common 
usual (old) teaching method.
Relationships: 
 Fibonacci word (automata theory): 
Alphabet: A = {a, b}; morphism  on monoid A* 
(a) = ab, (b) = a and (u,v  A*)[(uv) = (u)(v)] 
 Fibonacci Sequence (discrete math, combinatorics) 
F= 1, F= 2 and (n > 2)[F= F+ F] or 
0 1 n n  1 n  2nn 
using the Binet’s formula F= 
  (  ) 
n5 
where  = 1 5 
is the golden ratio 
 Complementary Beatty Sequence (recreational math) 
{[n(1 + 1/])}n1  {([n(1 + )]}n1 =  
& 
{[n(1 + 1/])}n1  {([n(1 + )] }n1 =  
 
2 
More relationships (graph theory, fractal theory, dynamical theory, etc)
The next Maplet simulates 12 iterations of the 
morphism : 0(a), 1(a), …, 12(a). 
It shows: 
1. the word produces at each morphic iteration; 
2. the sequence of word lengths produced by the 
done morphic iterations; 
3. the P-positions of the Wythoff’s game.
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Complementary Beatty’s Sequences 
 Particular: P-Positions of Wythoff’s game 
{[n(1 + 1/])}n1  {([n(1 + )]}n1 =  
& 
{[n(1 + 1/])}n1  {([n(1 + )] }n1 =  
15 
 
where  =  1.6180339887… is the golden ratio 
2 
 General: Given irrational x  (0, 1) 
{([n(1 + x)]}n1  {[n(1 + 1/x])}n1 =  
& 
{([n(1 + x)] }n1  {[n(1 + 1/x])}n1=  
Advantages of Maplet’s observations, e.g. for any irrational x with 0 < x < 1, 
n x n n x n     
lim  (1  )   2  1 & lim  (1  1/ )  
 
2 
x  x 
   
 1 1 
   
Next Maplet Output gives 
 1. The first twenty-one P-positions (ai, bi) of Wythoff’s game where 
ai = floor(i*(1 + 1/rho)), bi = floor(i*(1 + rho)), 
 2. The first twenty-one elements of Beatty’s sequence (Ai, Bi) where 
Ai = floor(i*(1 + 1/x)), Bi = floor(i*(1 + x)), 
for some x = 1 + y > 1. 
Theorem 
Let x > 0 be an irrational number and y = 1/x, then the two sequences positive integers 
{floor(i(1 + x))}i  1 and {floor(i(1 + y))}i  1 are two complementary sequences; 
i.e. 
1. {floor(i(1 + x))}i  1  {floor(i(1 + y))}i  1 = . 
2. {floor(i(1 + x))}i  1  {floor(i(1 + y))}i  1 = N.
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Nothing extra ordinary on the second pair of sequences. 
The Maplet outputs only illustrate that if {xi} is a sequence 
of positive irrational numbers sastifying lim xi = 0 then the 
previous observation works. It is equivalent with 
So those 
1   
1 lim (1 ) 2 1 & lim (1 (1 )) 2 
i i i x i n n n x n 
  
         
The irrational properties of xi cannot be dropped as the 
following Maplet examples show that in case xis are 
rationals, the two resulting sequences may not be disjoint. 
In the first example, they share the integer 53 (and many 
other integers) whereas in the second example, they share 
the integer 116.
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
The Wythoff’s Game 
P-Positions of Wythoff’s game are the pairs ([[n(1 + 1/], n(1 + )]). 
Theorem: 
{[n(1 + 1/])}n1  {([n(1 + )]}n1 =  
& 
{[n(1 + 1/])}n1  {([n(1 + )] }n1 =  
15 
 
where  =  0.6180339887… is the golden ratio. 
2 
 Wythoff’s game is played by two players that alternatingly remove some tokens from 
two heaps according to the following rules: 
1. Each player can remove any positive number of tokens from one of the two heaps. 
2. Each player can remove the same number of tokens from both heaps. 
3. A player is lost the game if there is no token left for, or to be removed by the player. 
Equivalently, a player who takes the last tokens is the winner.
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Fibonacci  Sequence, Golden Ratio and Wythoff's Game
Notice that once the computer has made automatic moves, 
you were forced to respond the following pairs of tokens’ 
numbers: 
1. (74, 120); 
2. (27, 44); 
3. (4, 7); 
4. (1, 2); 
5. (0, 0). 
These pairs belong to the P-positions of Wythoff’s game. 
The pair (27, 44) can be seen at the Maplets’ 7-8th morphic 
iterations whereas the pair (74, 120) can be seen at the 
Maplet’s 9th morphic iteration.

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Fibonacci Sequence, Golden Ratio and Wythoff's Game

  • 1. Fibonacci Sequence, Golden Ratio and Wythoff's Game LOEKY HARYANTO Mathematics Department, Hasanuddin University, email: L.Haryanto@unhas.ac.id HaryantoL@outlook.com GSM’s #: +6281342127598
  • 2. A Relationship among Fibonacci Sequence, Fibonacci Word, P-positions of Wythoff’s game and the Golden Ratio    1 5 2 USE THIS PRESENTATION AS A NEW STRATEGY FOR STUDENT-CENTERED LEARNING (SCL) METHOD. The Maplet copies here were created to make students firstly being familiar with (not necessarily mastering the theory of) combinatorial game and its related topics before the students being introduced with the theoretical parts of the subject; e.g. before they were given some formal theories which were written in the next page! By the way, since mathematics is a language which is full of written symbols, without visual and ‘seemingly’ interactive presentations, most of students tend to sleep in abstract algebra classes. Nevertheless, IMO most strategies proposed for the SCL method by experts in education are not appropriate for math classes, or even worse than the common usual (old) teaching method.
  • 3. Relationships:  Fibonacci word (automata theory): Alphabet: A = {a, b}; morphism  on monoid A* (a) = ab, (b) = a and (u,v  A*)[(uv) = (u)(v)]  Fibonacci Sequence (discrete math, combinatorics) F= 1, F= 2 and (n > 2)[F= F+ F] or 0 1 n n  1 n  2nn using the Binet’s formula F=   (  ) n5 where  = 1 5 is the golden ratio  Complementary Beatty Sequence (recreational math) {[n(1 + 1/])}n1  {([n(1 + )]}n1 =  & {[n(1 + 1/])}n1  {([n(1 + )] }n1 =   2 More relationships (graph theory, fractal theory, dynamical theory, etc)
  • 4. The next Maplet simulates 12 iterations of the morphism : 0(a), 1(a), …, 12(a). It shows: 1. the word produces at each morphic iteration; 2. the sequence of word lengths produced by the done morphic iterations; 3. the P-positions of the Wythoff’s game.
  • 18. Complementary Beatty’s Sequences  Particular: P-Positions of Wythoff’s game {[n(1 + 1/])}n1  {([n(1 + )]}n1 =  & {[n(1 + 1/])}n1  {([n(1 + )] }n1 =  15  where  =  1.6180339887… is the golden ratio 2  General: Given irrational x  (0, 1) {([n(1 + x)]}n1  {[n(1 + 1/x])}n1 =  & {([n(1 + x)] }n1  {[n(1 + 1/x])}n1=  Advantages of Maplet’s observations, e.g. for any irrational x with 0 < x < 1, n x n n x n     lim  (1  )   2  1 & lim  (1  1/ )   2 x  x     1 1    
  • 19. Next Maplet Output gives  1. The first twenty-one P-positions (ai, bi) of Wythoff’s game where ai = floor(i*(1 + 1/rho)), bi = floor(i*(1 + rho)),  2. The first twenty-one elements of Beatty’s sequence (Ai, Bi) where Ai = floor(i*(1 + 1/x)), Bi = floor(i*(1 + x)), for some x = 1 + y > 1. Theorem Let x > 0 be an irrational number and y = 1/x, then the two sequences positive integers {floor(i(1 + x))}i  1 and {floor(i(1 + y))}i  1 are two complementary sequences; i.e. 1. {floor(i(1 + x))}i  1  {floor(i(1 + y))}i  1 = . 2. {floor(i(1 + x))}i  1  {floor(i(1 + y))}i  1 = N.
  • 31. Nothing extra ordinary on the second pair of sequences. The Maplet outputs only illustrate that if {xi} is a sequence of positive irrational numbers sastifying lim xi = 0 then the previous observation works. It is equivalent with So those 1   1 lim (1 ) 2 1 & lim (1 (1 )) 2 i i i x i n n n x n            The irrational properties of xi cannot be dropped as the following Maplet examples show that in case xis are rationals, the two resulting sequences may not be disjoint. In the first example, they share the integer 53 (and many other integers) whereas in the second example, they share the integer 116.
  • 33. The Wythoff’s Game P-Positions of Wythoff’s game are the pairs ([[n(1 + 1/], n(1 + )]). Theorem: {[n(1 + 1/])}n1  {([n(1 + )]}n1 =  & {[n(1 + 1/])}n1  {([n(1 + )] }n1 =  15  where  =  0.6180339887… is the golden ratio. 2  Wythoff’s game is played by two players that alternatingly remove some tokens from two heaps according to the following rules: 1. Each player can remove any positive number of tokens from one of the two heaps. 2. Each player can remove the same number of tokens from both heaps. 3. A player is lost the game if there is no token left for, or to be removed by the player. Equivalently, a player who takes the last tokens is the winner.
  • 40. Notice that once the computer has made automatic moves, you were forced to respond the following pairs of tokens’ numbers: 1. (74, 120); 2. (27, 44); 3. (4, 7); 4. (1, 2); 5. (0, 0). These pairs belong to the P-positions of Wythoff’s game. The pair (27, 44) can be seen at the Maplets’ 7-8th morphic iterations whereas the pair (74, 120) can be seen at the Maplet’s 9th morphic iteration.