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I was messing with some recursive functions and realized it was equivalent to the Fibonacci sequence, but I couldn't figure out why. I then played a little further, and discovered some other interesting patterns. Let's consider functions $A,B$. Let $A(1) = 0$, $B(1) = 1$, $A(n) = B(n-1)$, $B(n) = A(n-1) + B(n-1)$. Then, $B(n) = B(n-2) + B(n-1)$, which causes $A(n) = A(n-1) + A(n-2)$. Of course, now both $A$ and $B$ are the Fibonacci sequence.

I was trying to figure out how this happened precisely, so I went back along the chain. I then discovered $B(n) = 2B(n-2) + B(n-3) = 2B(n-1) - B(n-3)$ and several other similar relations. What causes these relations to happen? Is there a general method of manipulating functions to get equalities?

BrainFRZ
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  • You have for example $$B_n=B_{n-1}+B_{n-2}=B_{n-2}+B_{n-3}+B_{n-2}=2B_{n-2}+B_{n-3}$$ or $$B_n=B_{n-1}+B_{n-2}=B_{n-1}+B_{n-1}-B_{n-3}=2B_{n-1}-B_{n-3}$$ – Peter Sep 12 '17 at 22:53
  • B(n)=A(n−1)+B(n−1) Since $A(n-1)=B(n-2)$ this gives $B(n)=B(n-1)+B(n-2),$. Then if you replace $B(n-1)=B(n-2)+B(n-3),$ you get $B(n) = 2B(n-2) + B(n-3)$ etc. – dxiv Sep 12 '17 at 22:53
  • So this is all just a matter of substitution and moving stuff around? Is there any generalization, or do I have to hunt for it if I want to get to something specific, especially for a more complicated sequence? – BrainFRZ Sep 12 '17 at 23:04

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(Not a complete answer, still maybe relevant.)  The recurrence can be written in matrix form as:

$$ \begin{pmatrix} B_{n} \\ A_{n} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} B_{n-1} \\ A_{n-1} \end{pmatrix} \quad\quad \text{with}\;\; \begin{pmatrix} B_{1} \\ A_{1} \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$

It follows by telescoping (or induction) that:

$$ \begin{pmatrix} B_{n} \\ A_{n} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n-1} \cdot \begin{pmatrix} B_{1} \\ A_{1} \end{pmatrix} $$

But $\displaystyle\; \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n-1} = \begin{pmatrix} F_n & F_{n-1} \\ F_{n-1} & F_{n-2} \end{pmatrix} \;$ (see this for example), so in the end:

$$ \begin{pmatrix} B_{n} \\ A_{n} \end{pmatrix} = \begin{pmatrix} F_n & F_{n-1} \\ F_{n-1} & F_{n-2} \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} F_n \\ F_{n-1} \end{pmatrix} $$

dxiv
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