# New Golden Ratio (phi) Sequences

This is the golden ratio (phi):

$$\phi = \sqrt{5}/2 - 1/2$$

If we perform alternate subtraction and addition to ascending powers of phi and its reciprocal we get the sequence of numbers known as the 'Lucas numbers'.

$$1 = 1/\phi - \phi$$

$$3 = 1/\phi^2 + \phi^2$$

$$4 = 1/\phi^3 - \phi^3$$

$$7= 1/\phi^4 + \phi^4$$

$$11 = 1/\phi^5 - \phi^5$$

...

If we perform alternate addition and subtraction to ascending powers of phi and its reciprocal we get this strange sequence of numbers.

$$\sqrt{5} = 1/\phi + \phi$$

$$\sqrt{5} = 1/\phi^2 - \phi^2$$

$$(2)\sqrt{5} = 1/\phi^3 + \phi^3$$

$$(3)\sqrt{5} = 1/\phi^4 - \phi^4$$

$$(5)\sqrt{5} = 1/\phi^5 + \phi^5$$

...

As you can see in the following examples we can calculate the square or square-root of phi to any power using simple arithmetic.

If we set the value of x to equal phi, we get the following sequence:

$$x^2 = 1 - x \\ x^3 = x - x^2 \\ x^4 = x^2 - x^3 \\ x^5 = x^3 - x^4 \\ x^6 = x^4 - x^5 \\ x^7 = x^5 - x^6 \\ x^8 = x^6 - x^7$$

...

$$x^2 = 3/2 - \sqrt{5}/2 \\ x^3 = 4/2 - (2) \sqrt{5}/2 \\ x^4 = 7/2 - (3) \sqrt{5}/2 \\ x^5 = 11/2 - (5) \sqrt{5}/2 \\ x^6 = 18/2 - (8) \sqrt{5}/2 \\ x^7 = 29/2 - (13) \sqrt{5}/2 \\ x^8 = 47/2 - (21) \sqrt{5}/2$$

...

If we set the value of x to equal the reciprocal of phi, we get this second sequence:

$$x^2 = 1 + x \\ x^3 = x + x^2 \\ x^4 = x^2 + x^3 \\ x^5 = x^3 + x^4 \\ x^6 = x^4 + x^5 \\ x^7 = x^5 + x^6 \\ x^8 = x^6 + x^7$$

...

$$x^2 = 3/2 + \sqrt{5}/2 \\ x^3 = 4/2 + (2) \sqrt{5}/2 \\ x^4 = 7/2 + (3) \sqrt{5}/2 \\ x^5 = 11/2 + (5) \sqrt{5}/2 \\ x^6 = 18/2 + (8) \sqrt{5}/2 \\ x^7 = 29/2 + (13) \sqrt{5}/2 \\ x^8 = 47/2 + (21) \sqrt{5}/2$$

...

Notice in both cases the similarity to the Fibonacci sequence.

Do theses sequence have a name?

• You have "discovered" one of the linkages between Fibonacci and Lucas numbers. Still a good observation. Jan 28 '18 at 1:26
• Good observation for sure and $+1$ Jan 28 '18 at 4:17

$$L_n = \phi^{n} + (-\phi)^{-n}$$ $$F_n = \frac{1}{\sqrt{5}}\left(\phi^{n} - (-\phi)^{-n}\right)$$
$$\phi^n = \frac{1}{2}(L_n + F_n\sqrt{5})$$