If I have a sequence, $\{t_1, \,t_2, \,t_1+t_2,\, t_1+2t_2,\cdots\}$ I know that the formula for it is:

$T_1=t_1, T_2=t_2, T_{n+2}=T_{n+1}+T_n.$

If there are two sequences, $A_n, B_n$ such that $A_1=1, A_2=0, A_{n+1}+A_n =A_{n+2}, B_1=0, B_2=1, B_{n+2}=B_{n+1}+B_n$

Then $A_n=F_{n-1}$ and $ B_n=F_{n-1}$ with $F_n$ being the Fibonacci numbers and $F_1=1, F_2=1$.

So the sequence becomes $T_n =t_1 F_{n-2} +t_2F_{n-1}$

The characteristic equation, which comes from $T_{n+2}=T_{n+1}+T_{n}$ is $x^2-x-1=0$ which has the roots $\frac{1\pm \sqrt{5}}{2}$. This is the same characteristic equation as the Fibonacci sequence.

Is it possible for two different sequences to have the same characteristic equation?

  • $\begingroup$ What does $t_1,t_2,t_1+t_2,t_1+2t_2$ mean? $\endgroup$ – lulu May 21 at 21:08
  • $\begingroup$ that is the sequence given $\endgroup$ – user130306 May 21 at 21:08
  • $\begingroup$ Well, it doesn't make any sense. What does the notation mean? What, say, does $t_2, t_1$ mean? $\endgroup$ – lulu May 21 at 21:08
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    $\begingroup$ To your broader question, of course two sequences can have the same characteristic equation. They just need to have different initial conditions. $\endgroup$ – lulu May 21 at 21:10
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    $\begingroup$ A recurrence of the form $a_n=pa_{n-1}+qa_{n-2}$ is obviously completely determined by any two consecutive values. In fact it is determined by any pair of values. This means that if two sequences match in two places they are identical. The same recurrence can have different initial values and these will generate a different sequence. $\endgroup$ – Mark Bennet May 21 at 21:22

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