MHB Sequence Challenge: Proving Periodicity of $\left\{x_n\right\}$ (Mod 11)

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The sequence defined by the recurrence relation $x_{n+3} \equiv \frac{1}{3}(x_{n+2}+x_{n+1}+x_n)$ (mod $11$) is shown to be either constant or periodic with a period of 10. This periodicity arises from the properties of modular arithmetic and the specific structure of the recurrence relation. The discussion emphasizes the importance of analyzing the initial conditions and their influence on the sequence's behavior. Participants highlight the mathematical reasoning behind the periodicity and the implications for sequences defined under similar conditions. The findings contribute to a deeper understanding of recurrence relations in modular systems.
lfdahl
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Let the sequence $\left\{x_n\right\}$ of integers (modulo $11$) be defined by the recurrence
relation:

$x_{n+3} \equiv \frac{1}{3}(x_{n+2}+x_{n+1}+x_n)$ (mod $11$), for $n=1,2,..$

Show, that every such sequence $\left\{x_n\right\}$ is either constant or periodic with period $10$.
 
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lfdahl said:
Let the sequence $\left\{x_n\right\}$ of integers (modulo $11$) be defined by the recurrence
relation:

$x_{n+3} \equiv \frac{1}{3}(x_{n+2}+x_{n+1}+x_n)$ (mod $11$), for $n=1,2,..$

Show, that every such sequence $\left\{x_n\right\}$ is either constant or periodic with period $10$.
Since we are working modulo $11$, I understand that $\frac{1}{3}$ means $4$.

This is a linear recurrence relation, whose characteristic polynomial is:
$$x^3 - 4x^2 - 4x - 4$$

Modulo $11$, this polynomial factors as:
$$(x-1)(x-6)(x-8)$$

and this means that the general solution is:
$$x_n = A(1)^n + B(6)^n + C(8)^n\pmod{11}$$

where the constants depend on the initial terms. If $B=C=0$, the sequence is constant; otherwise, since both $6$ and $8$ have multiplicative order $10$ modulo $11$, the sequence has period $10$.
 
castor28 said:
Since we are working modulo $11$, I understand that $\frac{1}{3}$ means $4$.

This is a linear recurrence relation, whose characteristic polynomial is:
$$x^3 - 4x^2 - 4x - 4$$

Modulo $11$, this polynomial factors as:
$$(x-1)(x-6)(x-8)$$

and this means that the general solution is:
$$x_n = A(1)^n + B(6)^n + C(8)^n\pmod{11}$$

where the constants depend on the initial terms. If $B=C=0$, the sequence is constant; otherwise, since both $6$ and $8$ have multiplicative order $10$ modulo $11$, the sequence has period $10$.

Thankyou, castor28!, for your sharp minded contribution and participation, which is highly appreciated!(Yes)
 
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