Question 113
Let $N$ be a positive integer, and let $\mathbf{a} = (a(1), \dots, a(N))$ and $\mathbf{b} = (b(1), \dots, b(N))$ be sequences of non-negative integers, each written on a circle (so we assume $a(i \pm N) = a(i)$ and $b(i \pm N) = b(i)$). We say $\mathbf{a}$ is **b-harmonic**, if each $a(i)$ is the arithmetic mean of the counterclockwise nearest $b(i)$ numbers, the clockwise nearest $b(i)$ numbers, and $a(i)$ itself; that is,
$$
a(i) = \frac{1}{2b(i) + 1} \sum_{s=-b(i)}^{b(i)} a(i + s). \quad (*)
$$
(A term of **a** may appear more than once in the above sum.) Suppose that neither **a** nor **b** is constant, and that both **a** is **b-harmonic**, and **b** is **a-harmonic**. Prove that more than half of the $2N$ terms across both sequences vanish.
$$
a(i) = \frac{1}{2b(i) + 1} \sum_{s=-b(i)}^{b(i)} a(i + s). \quad (*)
$$
(A term of **a** may appear more than once in the above sum.) Suppose that neither **a** nor **b** is constant, and that both **a** is **b-harmonic**, and **b** is **a-harmonic**. Prove that more than half of the $2N$ terms across both sequences vanish.