Proof for form of terms A372010
Michael S. Branicky, Apr 16 2024

Theorem.  a(n) = 1 0^{n-1-h}  9^h, where h = floor(n/2) and ^ represents repeated concatenation.

Proof.

Let k = sum_{i=0}^{n-1} d_i 10^i.

Then R(k) = sum_{i=0}^{n-1} d_i 10^{n-1-i}.

Let F(k) = R(k)/k,

The partial derivative of F(k) with respect to d_i is

p_i = [ k 10^{n-1-i}  - R(k) 10^i ] / k^2.

For n-i >= i, we have p_i > 0 since

k 10^{n-1-i} k = 10k 10^{n-i} > R(k) 10^i,

where the latter is valid since R(k) < 10 k.  Thus, F(k) is maximized by maximizing d_i while n - i >= i or n >= 2i, which means setting d_i = 9 for n >= 2i.

For this part, we may assume R(k) > k in our maximization.  However, we will only use R(k) > k/10.  When 2i > n, we have

10^(2i) R(k) > 10^{n} R(k) > 10^n (k/10),
10^i R(k) > 10^{n-i} k > 10^{n-1-i} k.

Thus, p_i is decreasing with d_i and we need to minimize d_i, which means setting d_{n-1} = 1 and all others to 0 for i > n/2. #