# Construction of a cyclic sequence re the Golden Ratio

1. Sep 6, 2007

### ramsey2879

The fractal sequence http://www.research.att.com/~njas/sequences/A054065
is of interest because it provides permutations of the numbers 1-n such
that the decimal part of k*tau (k = {1,2,3,...n} is ordered from the
lowest possible value to the highest. For instance if n = 3 the
permutation is 2,1,3 since .2360 < .6180 <.8541. (3*(1 + sqrt5)/2 =
4.8541..). However, other than calculating the decimal parts for each
k*tau and sorting the list no algorithm is provided for obtaining the
proper permutation of the numbers 1-n.
I have the algorithm to share with you.
Let F(odd) be the highest fibonacci number less or equal to n and F(2b)
be the highest even fibonacci number less than n.
start with k = F(odd) since of the numbers 1-n, k=F(odd) provides the
lowest possible decimal part for k*tau. Now follow the two rules below
in the order listed to get the subsequent terms in the permutation:
1. add F(odd) if the resulting value is less than F(odd+2)
2. subtract F(2b)
If the result of the first operation yields a number higher then you want in the permutation omit it. It would have been a part of the permutation
if n was chosen to be as larger such as F(odd+2)-1, however, so you can choose
to leave it in if you prefer to increase the number of terms in the
permutation instead. But even if you omit it continue the algorithm as
though it was included to get the next n in the permutation.
I said the permutation was cyclic because in fact you could start
with any number up to n and follow the above 2 rules to get all n
numbers in the permutation as though you were going around a clock only
you started at 6 instead of at 1.
I hope I didn't bore you with this.

Last edited: Sep 6, 2007
2. Sep 14, 2007

### ramsey2879

Last edited: Sep 14, 2007
3. Sep 16, 2007

### ramsey2879

Even better
given $$0 \leq i \leq F_{2n}$$ where $$F_{2n}$$ is an even term in the Fibonacci sequence, then the fractional parts of i*tau denoted as {i*tau} when put in ascending order have each term put into the position $$i*F_{2n-1} \mod F_{2n}$$
Thus for n = 3, F(2n) = 8 F(2n-1) = 5 Then mod 8
0*5 mod 8 = 0 {0*tau} = .000..... 0*5-0*3 =0
5*5 mod 8 = 1 {5*tau} = .090..... 5*5-3*8 = 1
2*5 mod 8 = 2 {2*tau} = .236..... 2*5-1*8 = 2 ...6-5 = 1
7*5 mod 8 = 3 {7*tau} = .326..... 7*5-4*8 = 3 ...9-5 = 4
4*5 mod 8 = 4 {4*tau} = .472..... 4*5-2*8 = 4 ...12-10= 2
1*5 mod 8 = 5 {1*tau} = .61803..1*5-0*8=5 ...15 - 15 = 0
6*5 mod 8 = 6 {6*tau} = .708.....6*5-3*8=6 ...18-15 = 3
3*5 mod 8 = 7 {3*tau} = .854.....3*5-1*8=7 ...21-20 = 1
8*5 mod 8 = 8 {8*tau} = .944.....8*5-4*8=8 ...24 -20= 4

Since it is known that F(2n-2)*F(2n+2) + 1 = F(n-1)^2 and tau is sightly greater than F(2n)/F(n-1) the above relation follows from multiplying this equation by i = 1,2,3,4... and reducing mod 8 and mod 5 respectively

As n increases $$\{F_{2n}*tau \}$$ approaches .999... . For instance F(16) = 987
{987*tau} = .99954..