### 4.1 Case (1)

In this section we study the following special case of Equation 1

{x}_{n+1}=\frac{{x}_{n}{x}_{n-1}}{{x}_{n}-{x}_{n-1}},

(7)

where the initial conditions *x*
_{-1}, *x*
_{0} are arbitrary positive real numbers.

#### Theorem 5

Let {\left\{{x}_{n}\right\}}_{n=-1}^{\infty} be a solution of Equation 7. Then for *n* = 0, 1,...

{x}_{n}=\frac{{\left(-1\right)}^{n}hk}{{F}_{n-1}k-{F}_{n-2}h},

where *x*
_{-1} = *k*, *x*
_{0} = *h* and *F*
_{
n-1
}, *F*
_{
n-2}are the Fibonacci terms.

**Proof:** For *n* = 0 the result holds. Now suppose that *n* > 0 and that our assumption holds for *n*-1, *n*-2. That is;

{x}_{n-2}=\frac{{\left(-1\right)}^{n-2}hk}{{F}_{n-3}k-{F}_{n-4}h},\phantom{\rule{2.77695pt}{0ex}}{x}_{n-1}=\frac{{\left(-1\right)}^{n-1}hk}{{F}_{n-2}k-{F}_{n-3}h}.

Now, it follows from Equation 7 that

\begin{array}{lll}\hfill {x}_{n}& =\frac{{x}_{n-1}{x}_{n-2}}{{x}_{n-1}-{x}_{n-2}}=\frac{\left(\frac{{\left(-1\right)}^{n-1}hk}{{F}_{n-2}k-{F}_{n-3}h}\right)\left(\frac{{\left(-1\right)}^{n-2}hk}{{F}_{n-3}k-{F}_{n-4}h}\right)}{\left(\frac{{\left(-1\right)}^{n-1}hk}{{F}_{n-2}k-{F}_{n-3}h}-\frac{{\left(-1\right)}^{n-2}hk}{{F}_{n-3}k-{F}_{n-4}h}\right)}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\frac{\left(\frac{{\left(-1\right)}^{n-1}hk}{{F}_{n-2}k-{F}_{n-3}h}\right)\left(\frac{-1}{{F}_{n-3}k-{F}_{n-4}h}\right)}{\left(\frac{1}{{F}_{n-2}k-{F}_{n-3}h}+\frac{1}{{F}_{n-3}k-{F}_{n-4}h}\right)}=\frac{{\left(-1\right)}^{n}hk}{\left({F}_{n-2}k-{F}_{n-3}h+{F}_{n-3}k-{F}_{n-4}h\right)}\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ =\frac{{\left(-1\right)}^{n}hk}{{F}_{n-1}k-{F}_{n-2}h}.n\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}

Hence, the proof is completed.

For confirming the results of this section, we consider numerical example for *x*
_{-1} = 11, *x*
_{0} = 4 (see Figure 1), and for *x*
_{-1} = 6, *x*
_{0} = 15 (see Figure 2), since the solutions take the forms {6, -12, 4, -3, 1.714286, -1.090909, .6666667, -.4137931, .2553191,....}, {-60, 10, -8.571428, 4.615385, -3, 1.818182, -1.132075, .6976744,...}.

### 4.2 Case (2)

In this section we study the following special case of Equation 1

{x}_{n+1}=\frac{{x}_{n-1}{x}_{n-2}}{{x}_{n-1}-{x}_{n-2}},

(8)

where the initial conditions *x*
_{-2}, *x*
_{-1}, *x*
_{0} are arbitrary positive real numbers.

### Theorem 6

Let {\left\{{x}_{n}\right\}}_{n=-2}^{\infty} be a solution of Equation 8. Then {x}_{1}=\frac{rk}{k-r}, for *n* = 1, 2,...

{x}_{n+1}=\frac{hkr}{{g}_{n-4}hk+{g}_{n-3}kr+{g}_{n-2}hr},

where *x*
_{-2} = *r*, *x*
_{-1} = *k*, *x*
_{
0
} = *h*, {\left\{{g}_{m}\right\}}_{m=0}^{\infty}=\left\{1,\phantom{\rule{2.77695pt}{0ex}}-2,0,3,\phantom{\rule{2.77695pt}{0ex}}-2,\phantom{\rule{2.77695pt}{0ex}}-3,\phantom{\rule{2.77695pt}{0ex}}\dots \right\}, i.e., *g*
_{
m
} = *g*
_{
m-2}+ *g*
_{
m-3}, *m* ≥ 0, *g*
_{-3} = 0, *g*
_{-2} = -1, *g*
_{-1} = 1.

**Proof:** For *n* = 1, 2 the result holds. Now suppose that *n* > 1 and that our assumption holds for *n* - 1, *n* - 2. That is;

{x}_{n-2}=\frac{hkr}{{g}_{n-7}hk+{g}_{n-6}kr+{g}_{n-5}hr},
{x}_{n-1}=\frac{hkr}{{g}_{n-6}hk+{g}_{n-5}kr+{g}_{n-4}hr}. Now, it follows from Equation 8 that

\begin{array}{lll}\hfill {x}_{n+1}& =\frac{{x}_{n-1}{x}_{n-2}}{{x}_{n-1}-{x}_{n-2}}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\frac{\left(\frac{hkr}{{g}_{n-6}hk+{g}_{n-5}kr+{g}_{n-4}hr}\right)\left(\frac{hkr}{{g}_{n-7}hk+{g}_{n-6}kr+{g}_{n-5}hr}\right)}{\left(\frac{hkr}{{g}_{n-6}hk+{g}_{n-5}kr+{g}_{n-4}hr}-\frac{hkr}{{g}_{n-7}hk+{g}_{n-6}kr+{g}_{n-5}hr}\right)}\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ =\frac{hkr}{\left({g}_{n-7}hk+{g}_{n-6}kr+{g}_{n-5}hr-{g}_{n-6}hk+{g}_{n-5}kr+{g}_{n-4}hr\right)}\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ =\frac{hkr}{{g}_{n-4}hk+{g}_{n-3}kr+{g}_{n-2}hr}.\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}

Hence, the proof is completed.

Assume that *x*
_{-2} = 8, *x*
_{-1} = 15, *x*
_{0} = 7, then the solution will be {17.14286, -13.125, 11.83099, 7.433628, -6.222222, -20, 3.387097, -9.032259,...}(see Figure 3).

The proof of following cases can be treated similarly.

### 4.3 Case (3)

Let *x*
_{-2} = *r*, *x*
_{-1} = *k*, *x*
_{0} = *h*, \prod _{i=0}^{-1}{A}_{i}=1 and *F*
_{2i-1}, *F*
_{2i
}, *F*
_{2i+1}(where *i* = 0 *to n*) are the Fibonacci terms. Then the solution of the difference equation

{x}_{n+1}=\frac{{x}_{n-1}{x}_{n-2}}{{x}_{n}-{x}_{n-2}},

(9)

is given by

{x}_{2n}=\frac{h\prod _{i=0}^{n-1}\left({F}_{2i-1}h-{F}_{2i}r\right)}{\prod _{i=0}^{n-1}\left({F}_{2i+1}r-{F}_{2i}h\right)},\phantom{\rule{0.3em}{0ex}}{x}_{2n+1}=\frac{kr\prod _{i=0}^{n-1}\left({F}_{2i+1}r-{F}_{2i}h\right)}{\prod _{i=0}^{n}\left({F}_{2i-1}h-{F}_{2i}r\right)},\phantom{\rule{0.3em}{0ex}}n=0,1,...\phantom{\rule{2.77695pt}{0ex}}.

Figure 4 shows the solution when *x*
_{-2} = 9, *x*
_{-1} = 12, *x*
_{0} = 17.

### 4.4 Case (4)

Let *x*
_{-2} = *r*, *x*
_{-1} = *k*, *x*
_{0} = *h*. Then the solution of the following difference equation

{x}_{n+1}=\frac{{x}_{n-1}{x}_{n}}{{x}_{n}-{x}_{n-2}}

(10)

is given by

{x}_{2n-1}={\left(\frac{h}{h-r}\right)}^{n}k,\phantom{\rule{1em}{0ex}}{x}_{2n}=\frac{{h}^{n+1}}{{r}^{n}},\phantom{\rule{2.77695pt}{0ex}}n\phantom{\rule{2.77695pt}{0ex}}=0,1,\phantom{\rule{2.77695pt}{0ex}}\dots \phantom{\rule{2.77695pt}{0ex}}.

Figure 5 shows the solution when *x*
_{
- 2}= 21, *x*
_{
- 1}= 6, *x*
_{0} = 3.

### 4.5 Case (5)

Let *x*
_{
- 2}= *r*, *x*
_{
- 1}= *k*, *x*
_{0} = *h*. Then the solution of the following difference equation

{x}_{n+1}=\frac{{x}_{n-1}{x}_{n}}{{x}_{n-1}-{x}_{n-2}},

(11)

is given by

{x}_{4n}=\frac{h{\left(hk\right)}^{2n}}{{\left(rk\left(h-k\right)\left(k-r\right)\right)}^{n}},\phantom{\rule{0.3em}{0ex}}{x}_{4n+1}=\frac{{\left(hk\right)}^{2n+1}}{{\left(rk\left(h-k\right)\right)}^{n}{\left(k-r\right)}^{n+1}},

{x}_{4n+2}=\frac{h{\left(hk\right)}^{2n+1}}{{\left(\left(h-k\right)\left(k-r\right)\right)}^{n+1}{\left(rk\right)}^{n}},\phantom{\rule{0.3em}{0ex}}{x}_{4n+3}=\frac{{\left(hk\right)}^{2n}}{{\left(r\left(h-k\right)\left(k-r\right)\right)}^{n+1}{k}^{n}},n\phantom{\rule{2.77695pt}{0ex}}=0,1,\phantom{\rule{2.77695pt}{0ex}}\dots \phantom{\rule{2.77695pt}{0ex}}.

Figure 6 shows the solution when *x*
_{
- 2}= 9, *x*
_{
- 1}= 5, *x*
_{0} = 4.

Figure 7 shows the solution when *x*
_{
- 2}= .9, *x*
_{
- 1}= 5, *x*
_{0} = .4.

### 4.6 Case (6)

Let *x*
_{
- 2}= *r, x*
_{
- 1}= *k, x*
_{0} = *h*, Then the solution of the following difference equation

{x}_{n+1}=\frac{{x}_{n-2}{x}_{n}}{{x}_{n}-{x}_{n-2}},

(12)

is given by

{x}_{n}=\frac{hkr}{{u}_{n-3}hr+{u}_{n-2}hk+{u}_{n-1}kr},\phantom{\rule{2.77695pt}{0ex}}n\phantom{\rule{2.77695pt}{0ex}}=0,1,\phantom{\rule{2.77695pt}{0ex}}\dots \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}},

Where {\left\{{u}_{m}\right\}}_{m=0}^{\infty}=\left\{-1,1,0,\phantom{\rule{2.77695pt}{0ex}}-1,2,\phantom{\rule{2.77695pt}{0ex}}-2,1,1,\phantom{\rule{2.77695pt}{0ex}}-3,\phantom{\rule{2.77695pt}{0ex}}\dots \right\} i. e. *u*
_{
m
} = u_{
m-1}- u_{
m-3}, *m* ≥ 0, *u*
_{-3} = 0, *u*
_{-2} = 0, *u*
_{-1} = 1.

Figure 8 shows the solution when *x*
_{-2} = 11, *x*
_{-1} = 6, *x*
_{0} = 17.

### 4.7 Case (7)

Let *x*
_{-2} = *r, x*
_{-1} = *k, x*
_{0} = *h* and F_{
n-1}F_{, n-2}, *F*
_{
n
} are the Fibonacci terms.

Then the solution of the following difference equation

{x}_{n+1}=\frac{{x}_{n-2}{x}_{n}}{{x}_{n-1}-{x}_{n-2}},

(13)

is given by

{x}_{2n}\phantom{\rule{2.77695pt}{0ex}}=\frac{hkr}{\left({F}_{n-2}k-{F}_{n-1}r\right)\left({F}_{n-2}h-{F}_{n-1}k\right)},

{x}_{2n+1}=\frac{hkr}{\left({F}_{n-1}k-{F}_{n}r\right)\left({F}_{n-2}h-{F}_{n-1}k\right)},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}n\phantom{\rule{2.77695pt}{0ex}}=0,1,\phantom{\rule{2.77695pt}{0ex}}\dots \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}.

Figure 9 shows the solution when *x*
_{
- 2}= 8, x_{
- 1}= 5, *x*
_{0} = 0.9.