# if $a_{n+1}=xa_{n}+ya_{n-1}$ then have this results :$a_{m+n}=Xa_{m}a_{n+1}+Ya_{m-1}a_{n}$？

it is well known: if $$F_{1}=1,F_{2}=1$$,and $$F_{n+1}=F_{n}+F_{n-1}$$.then we have $$F_{m+n}=F_{m}F_{n+1}+F_{m-1}F_{n}$$ prove 1

and I have know this :if $$a_{0}=0,a_{1}=1$$ and $$x,y$$ be give postive integers,and $$a_{n+1}=xa_{n}+ya_{n-1}$$ then we have $$a_{m+n}=a_{m}a_{n+1}+ya_{m-1}a_{n}$$ prove 2

My question is if the sequence such $$a,b,x,y$$ be give postive integers,and such that $$a_{1}=a,a_{2}=b,a,b\in N^{+}$$,and $$a_{n+1}=xa_{n}+ya_{n-1}$$,then is also have this Similar results $$a_{m+n}=Xa_{m}a_{n+1}+Ya_{m-1}a_{n}?$$ if there exist,Find the $$X,Y$$?Thanks

So, the answer is indeed, yes, such a relation exists (with some caveats), and we can prove it via induction, just as in the 1st answer you have linked in the question.

MAJOR EDITS: My original answer was incorrect, as pointed out by @mathlove. Here is a revised answer.

First, let us prove the induction step. For it, assuming the relation is true upto $$(m,n)$$, we have to prove it for $$(m+1,n)$$ and $$(m,n+1)$$, just as mentioned in the $$1^{st}$$ answer.

For $$(m+1,n)$$, we have \begin{align} a_{m+1+n} &= xa_{m+n}+ya_{(m-1)+n} \\ &= x(Xa_ma_{n+1}+Ya_{m-1}a_n) + y(Xa_{m-1}a_{n+1}+Ya_{m-2}a_n) \\ &= Xa_{n+1}(xa_m+ya_{m-1}) + Ya_n(xa_{m-1}+ya_{m-2}) \\ &= Xa_{m+1}a_{n+1} + Ya_ma_n \end{align}

Similarly we can do so for $$(m,n+1)$$. Thus, the inductive step is proved.

For the base case, as the induction relation involves $$m-1$$ and $$n$$, the lowest values that these can have is $$1$$ (As $$a_0$$ is not defined). So, the base case for the induction must be for $$(m,n) = (2,1)$$.

For the base case to hold, the following relation must hold - \begin{align} a_{2+1} &= Xa_2a_{1+1} + Ya_{2-1}a_1 \\ \implies a_3 &= Xb^2 + Ya^2 \end{align}

From the given recursion relation, we get \begin{align} a_3 &= xa_2 + ya_1 \\ &= xb + ya \end{align}

Hence, the given relation holds iff $$X,Y$$ satisfy $$Xb^2 + Ya^2 = xb + ya$$

Addendum: I added this as this was part of my original answer, and I wanted to show where I originally went wrong.

Now, let us determine what restrictions $$X$$ and $$Y$$ must have, if such a result was to exist.

Let $$m=2$$. Then, the induction gives \begin{align} a_{n+2} &= Xa_2a_{n+1}+Ya_1a_n \\ &= (Xb) a_{n+1} + (Ya) a_n \end{align}

Now, we have already been given that $$a_{n+2}=xa_{n+1}+ya_{n}$$. Thus, we get \begin{align} (Xb) a_{n+1} + (Ya) a_n &= xa_{n+1}+ya_{n}\\ \implies (Xb-x)a_{n+1} &= (y-Ya)a_n \end{align}

Now, if either $$X = \frac{x}{b}$$ or $$Y = \frac{y}{a}$$, then this relation shows that the other must be zero as well. Hence, we have either $$(X,Y) = \left(\frac{x}{b}, \frac{y}{a}\right)$$, or \begin{align} a_{n+1} &= \frac{y-Ya}{Xb-x}a_n \\ \implies a_n &= \left(\frac{y-Ya}{Xb-x}\right)^{n-1} a_1 \end{align}

Note: We can easily see that in both the cases, $$Xb^2 + Ya^2 = xb + ya$$ holds (as it should).

• $X=\frac xb$ and $Y=\frac ya$ don't necessarily hold. For example, $(a,b,x,y,X,Y)=(1,3,2,3,-1,18)$ works where $X\not=\frac xb$ and $Y\not=\frac ya$. Jan 12 at 12:26
• Oh! Of course. Thanks!! Jan 13 at 1:38
• @lshan Deo,so $X,Y$ have lot of value? Feb 14 at 1:11