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Prove that $a_{n+1}=\frac{a_n^2+4}{a_{n-1}}$ recursion formula is equivalent to $\frac{a_{n+1}+a_{n-1}}{a_n}=6 \hspace{0,5cm}$

if $a_1=1, a_2=5$ $\forall n \gt1$

I spent a couple of hours with the task, which can be found on the below link, when I noticed the equivalence between the two recursion formulas, unfortunately I could not prove it.

Finding the general formula of $a_{n+1}=\frac{a_n^2+4}{a_{n-1}}$ with $a_1=1$ and $a_2=5$

Could someone help with the proof, thanks in advance.

JV.Stalker
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2 Answers2

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Starting from the second recusion formula $a_{n+1}=6a_n-a_{n-1}$, we get for all $n\ge2$ $$ \left[\begin{array}{cc}a_{n+2}&a_{n+1}\\a_{n+1}&a_{n}\end{array}\right]=\left[\begin{array}{cc}6&-1\\1&0\end{array}\right]\left[\begin{array}{cc}a_{n+1}&a_n\\a_n&a_{n-1}\end{array}\right]. $$ Define $\displaystyle M_n=\left[\begin{array}{cc}a_{n+1}&a_n\\a_n&a_{n-1}\end{array}\right]$ and $\displaystyle A=\left[\begin{array}{cc}6&-1\\1&0\end{array}\right]$. Observe that we have $$ \det(M_{n+1})=\det(A)\det(M_n)=\det(M_n)=\cdots=\det(M_2)=4. $$ Since $\det(M_n)=a_{n+1}a_{n-1}-a_n^2=4$ is equivalent to the first recursion formula, we can see that the two formulae give the same sequence because the first two terms determine the whole sequence.

Note: It can be noted that this relationship can be generalized to $a_{n+1}= ta_n-a_{n-1}$ and $a_{n+1}a_{n-1}-a_n^2=rst-r^2-s^2$ with the common first two terms $a_2=s, a_1=r$ and arbitrary $t$.

Myunghyun Song
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  • thank you for the clear answer, the method is new for me, regards. – JV.Stalker Mar 14 '19 at 14:08
  • You're welcome! I hope it will help you :) – Myunghyun Song Mar 14 '19 at 14:10
  • @ Song I'm afraid there's some flaw in your proof, since the equivalence breaks down for other initial values. Take $a_1=1$, $a_2=2$ for instance. The linear sequence starts 1, 2, 11, 64, 373, the non-linear one 1, 2, 8, 34, 145. – Dr. Wolfgang Hintze Mar 14 '19 at 14:39
  • @Dr.WolfgangHintze As you can see in my last comment, if we choose different first two terms, then corresponding recursion formula should be changed accordingly; in this case, $a_{n+1}a_{n-1}-a_n^2 =rst-r^2-s^2=1\times 2\times 6-1^2-2^2=7$. I hope this makes it clear. (You can check that 7 indeed works.) Note that the two formulae in the OP are equivalent only when conditioned on certain initial conditions. – Myunghyun Song Mar 14 '19 at 14:50
  • @ Song It can be easily shown that equivalence holds if $a_{2} = 3 a_{1} \pm 2 \sqrt{2a_{1}^2-1}$ – Dr. Wolfgang Hintze Mar 14 '19 at 15:26
  • @Dr.WolfgangHintze That is obvious, if we set $t=6$ and solve $rst-r^2-s^2=-r^2+6st-s^2=4$ for $a_1=r$ and $a_2=s$. So what is your point really? – Myunghyun Song Mar 14 '19 at 15:29
  • I have given an explicit formula for the eqivalence to hold. That's all. – Dr. Wolfgang Hintze Mar 14 '19 at 15:31
  • @Dr.WolfgangHintze But you said there's some flaw in my answer. And did my last comment not give this relationship? – Myunghyun Song Mar 14 '19 at 15:32
  • I said cautiously, I'm afraid. With your final remark there's no flaw left, of course. I was focusing on your nice determinant approach and didn't realise this remark. For the same reason I provided independently the necessary condition for equivalence to hold. There might be some parallel thinking around ... ;-) – Dr. Wolfgang Hintze Mar 14 '19 at 15:38
  • @Dr.WolfgangHintze Ah, I got it. Thank you for your comment ! – Myunghyun Song Mar 14 '19 at 15:39
  • @Song, you might actually want to stop at $\det(M_2)$, since there is no $a_0$. – Barry Cipra Mar 14 '19 at 16:03
  • @Barry Cipra Thank you for pointing out :) – Myunghyun Song Mar 15 '19 at 00:02
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To give an induction proof, suppose that

$$a_{n+1}={a_n^2+4\over a_{n-1}}=6a_n-a_{n-1}$$

which holds for the base case, $n=1$ when $a_0=1$ and $a_1=5$, since $(5^2+4)/1=29=6\cdot5-1$. We want to prove that $(a_{n+1}^2+4)/a_n=6a_{n+1}-a_n$, in which case we can say both are equal to $a_{n+2}$.

From $a_{n+1}=6a_n-a_{n-1}$ we have

$$0=a_{n+1}(a_{n+1}-6a_n+a_{n-1})=a_{n+1}^2-6a_na_{n+1}+a_{n+1}a_{n-1}$$

From $a_{n+1}a_{n-1}=a_n^2+4$, this becomes

$$0=a_{n+1}^2-6a_na_{n+1}+a_n^2+4=a_{n+1}^2+4-a_n(6a_{n+1}-a_n)$$

and thus $(a_{n+1}^2+4)/a_n=6a_{n+1}-a_n$, as desired.

Barry Cipra
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