Indices in recursion

Number Theory Level 3

Define the sequence $\langle a_{n} \rangle_{n}$ recursively by $a_{1}=2$ , $a_{2}=5$ , and

$a_{n+1}=(2-n^{2})a_{n}+(2+n^{2})a_{n-1}$

for $n \geq 2$ . Do there exist indices $p,q,r$ such that $a_{p}a_{q}=a_{r}$ ?

No Yes

1 solution

Aditya Khurmi
Nov 2, 2017

I claim that $a_{n} \equiv 2 \pmod {3}$ $\forall n \in N$ .

We proceed by induction.

This is true for $n=1,2$ . Now, let it hold for for some $n=k,k-1$ . Then

$a_{k+1}=(2-k^{2})a_{k}+(2+k^{2})a_{k-1} \equiv (2-0)(2)+(2+0)(2)$ or $(2-1)(2)+(2+1)(2) =8 \equiv 2 \pmod {3}$

where the 'or' condition arises depending on whether $k^{2} \equiv 0$ or $1 \pmod{3}$ , but in both the cases, we get that $a_{k+1} \equiv 2 \pmod {3}$ completing the induction.

Now, $a_{p}a_{q} \equiv (2)(2) \equiv 1 \pmod {3}$ whereas $a_{r} \equiv 2 \pmod {3}$ , and hence this equality can not hold. Hence, the answer is $\boxed{No}$ .

Indices in recursion

1 solution

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