Prove that two different $m$ for $a_{n-m} + a_{n-m+1} + \dots + a_{n+m} = a_n \cdot (2m + 1)$ form an arithmetic progression

Question

Define $\{a_n\}$ (a sequence of real numbers) as satisfying the H(m) condition if, for all integers $n$, the following recurrence relation holds:

$$ a_{n-m} + a_{n-m+1} + \dots + a_{n+m} = a_n \cdot (2m + 1) $$

In other words, the average of the $2m + 1$ consecutive terms from $a_{n-m}$ to $a_{n+m}$ is $a_n$.

The task is to prove the following conjecture: If a sequence satisfies both the H(p) and the H(q) condition for integers $p$ and $q$ $(p \neq q, p>1,q > 1)$, then the sequence ${a_n}$ must be an arithmetic progression. In other words, it satisfies H(1) condition.

What I’ve Tried:

I’ve explored using a method similar to the Euclidean algorithm to compute small cases for specific values of $p$ and $q$.

Then I suspect that this problem could be transformed into one involving the roots of the characteristic equation associated with the recurrence relations, where the only common root between the equations is 1. This related to a question left unsolved here.

I've also tried: $$ x^{n-m} + x^{n-m+1} + \dots + x^{n+m} = x^n \cdot (2m + 1) $$ $$ \frac {x^{m+1} - x^{-m}}{x-1} = 2m + 1 $$ let $x = e^{2i\omega}$, ($x \in \mathbb{C}$, so $\omega \in \mathbb{C}$) $$\implies \frac {e^{(2m+2)i\omega} - e^{(-2m)i\omega}}{e^{2i\omega}-1} = 2m + 1 $$ $$\implies \frac {e^{(2m+1)i\omega} - e^{-(2m+1)i\omega}}{e^{i\omega}-e^{-i\omega}} = 2m + 1 $$ $$\implies \sin((2m+1)\omega) = (2m+1)\sin\omega $$

Any guidance would be greatly appreciated!

Answer 1

DTFT of $a_n = A(z) = \sum_n a_n z^{-n} $

DTFT of $b_n = \left[\frac{1}{(2m+1)},...,\frac{1}{(2m+1)} - 1,...,\frac{1}{(2m+1)} \right] = B_m(z) = \frac{1}{2m+1} \frac{z^{-(m+0.5)}-z^{m+0.5}}{z^{-0.5}-z^{0.5}} - 1$

$$H(m): \text{ convolution}(\{a_n\},\{b_n\}) = 0 = A(z) \times B_m(z) = 0 \implies (B_m(z) \neq 0 \implies A(z) = 0),(A(z) \neq 0 \implies B_m(z) = 0)$$

$$S_m = \{z : B_m(z) = 0 \}$$

Now the condition is true for $m = p,q$. $$z \in S_p \cap S_q \implies \frac{1}{2p+1} \frac{z^{-(p+0.5)}-z^{p+0.5}}{z^{-0.5}-z^{0.5}} - 1 = \frac{1}{2q+1} \frac{z^{-(q+0.5)}-z^{q+0.5}}{z^{-0.5}-z^{0.5}} - 1 = 0$$ We want to prove: $$S_1 \supseteq S_p \cap S_q$$.

Above equations imply: For $z = e^{j\omega}$, $$\omega = 2k\pi \in S_p \cap S_p \text{ but we know that } 2k\pi \in S_1$$

So probably for $z = e^{j \omega}$, $\{a_n\} = a[1,1,....,]$ is the only solution.

Solve for other $z \in S_p \cap S_q$, $z \in \mathbb{C}$ and check those cases.

There could be mistakes in above, please comment below if you find any.