The sequence $a_{n}a_{n+1}=a_{n+2}$

Question

The product of two corresponding terms in a sequence $a_n$ determines the next term. Find the general solution.

My approach:

$$x^{b+c}=x^bx^c$$

Let $b=F_n$ and $c=F_{n+1}$ then the sum $b+c=F_{n+2}$. $F_n$ is the n-th Fibonacci number then, $a_n=x^{F_n}$.

However this doesn't seem to be the solution because if we start with $a_0=1$, $a_1=2$, then $a_3=2$ ... $a_4=4$....$a_5=8$. And I can't find any $x$ that will make this true.

May someone help.

Answer 1

If the $a_n$ are positive, we can use a new letter, maybe $$ d_n = \log a_n, $$ so that $$ a_n = e^{d_n}. $$ Then $$ d_{n+2} = d_{n+1} + d_n. $$ It follows that there are real constants, call them $A,B,$ not necessarily positive, so $$ d_n = A F_n + B L_n, $$ where $L_n$ are the Lucas numbers. Then $$ a_n = e^{A F_n} e^{B L_n}. $$ So, if we tkae $e^A = G > 0,$ and $e^B = H > 0, $ we do get $$ a_n = G^{F_n} H^{L_n} $$

If you prefer to stick with Fibonacci numbers, with a new positive real $J$ you can write $$ a_n = G^{F_n} J^{F_{n-1}} $$

Answer 2

Take logarithms.

$a_n=k^{F_n}$

Then $a_{n+2}=k^{F_{n+2}}=k^{F_{n+1}+F_n}=k^{F_{n+1}}k^{F_n}=a_{n+1}a_n$

Answer 3

Suppose the first two terms are $a$ and $b$, respectively. Then - breaking into powers of $a$ and $b$ - the terms are:

• $a^1b^0$

• $a^0b^1$

• $a^1b^1$

• $a^1b^2$

• $a^2b^3$

• $a^3b^5$

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Does this help show the general pattern? HINT: You're right to be thinking about Fibonnaci numbers . . .