Problem
Let $a_{n> 0}$ be a monotonically increasing integer sequence. $a_1 = 1$, and the rest of $a_n$ is defined through the recursion relation \begin{equation} a_{n+1} = \text{argmax}_{A\in \mathbb{Z}^+} \left\{ S_n(A) = \sum_{i=1}^{n} \frac{1}{(A - a_i)^{\mu}}> \frac{1}{2} \right\} \end{equation} (i.e. $a_n$ is equal to the maximal integer $A$ that satisfies the inequality $S_n(A) > \frac{1}{2} $). Here $0 < \mu < \infty$.
Q: What is the asymptotic behavior of $a_n$ in terms of $n$?
Note: Note $S_n(A)$ is monotonically decreasing for $A \ge a_n + 1$. Since we have $S_n(a_n+1) > 1$ and $S_n( \infty) = 0 $, the maximal integer always exists to define the sequence.
The asymptotic behaivor should be (I think) identical for the real sequence. Let $a_1 = 1$ , and $a_{n+1}$ can be recursively solved from \begin{equation} \sum_{i=1}^{n} \frac{1}{(a_{n+1} - a_i)^{\mu}} = \frac{1}{2} \end{equation}
Partial solution:
I managed to obtain the following leading order behaviors \begin{equation} a_n \sim \left\lbrace \begin{aligned} & n & \quad \mu > 1 \\ & n^{\frac{1}{\mu}} & \quad 0 < \mu < 1 \\ \end{aligned} \right. \end{equation}
Let us start with the ansatz $a_n \sim k n $, where $k$ is a constant. Plugging into the recursion relation \begin{equation} \sum_{n=1}^{N} \frac{1}{( a_{N+1} - a_n )^\mu } \sim \frac{1}{k^{\mu} } \sum_{n=1}^{N} \frac{1}{(N + 1 - n)^\mu} \end{equation} The sum is convergent when $\mu > 1$, so that the LHS can be set to $\frac{1}{2}$ if we adjust $k$. The linear solution works for $\mu > 1$
When $ 0 < \mu < 1$, we try a power function ansatz $a_n \sim n^{\alpha}$. Again we use the recursive relation \begin{equation} \sum_{n=1}^{N} \frac{1}{( a_{N+1} - a_n )^\mu } \sim \frac{1}{N^{\alpha \mu}} \sum_{n=1}^{N} \frac{1}{( ( 1 + \frac{1}{N})^\alpha - (\frac{n}{N} )^{\alpha} )^\mu } \sim \frac{N}{N^{\alpha \mu}} \int_0^1 \frac{1}{( ( 1 + \frac{1}{N})^\alpha - x^{\alpha} )^\mu } dx \end{equation} Since $0 < \mu < 1$, the integral is convergent even at $x \rightarrow $1. For the whole expression to be $\frac{1}{2}$, we must have $\alpha \mu = 1$, thus $\alpha = \frac{1}{\mu}$.
Update v3:
When $\mu = 1$, one guess that $a_n \sim n \ln n$ based on the numerical data and check \begin{equation} \sum_{n= 1}^{N} \frac{1}{( N + 1) \ln ( N + 1 ) - n \ln n} \end{equation} converges in $N\rightarrow \infty$. Following the same trick for $\mu < 1$, one can convert this to an integral \begin{equation} \int_0^1 \frac{1}{( 1 + \frac{1}{N} ) \ln ( N + 1 ) - x \ln (xN) } dx \end{equation}We need to make sure the integral converges at $x = 1$, so we take expand the denominator around $x = 1$ \begin{equation} \frac{\ln N}{ N} + (1 - x) (\ln N + 1 ) + \mathcal{O}(\frac{1}{N} ) \end{equation} If we integrate around $x = 1$, we will get \begin{equation} \frac{- \ln (\ln N / N )}{\ln N +1} \sim 1 \end{equation} when $N \rightarrow \infty$. So the integral is finite, and we have $a_n \sim n \ln n $ when $\mu \rightarrow 1$.
Numerical Test:
The numerical test confirmed the scaling.
(In v1, I stated that the numerical test suggested a logarithmic correction. Not there is not. It was due to an incorrect implementation to numerically search $a_n$. )
Question:
At this stage (v3), the problem has been solved to my satisfaction. But you are welcome to post your solutions.
In the numerical data, I also observed that when $\mu \rightarrow 1$, $a_n$ first behaves as $n \ln n$, and then transits to $n^{1/\mu}$ or $kn$ for larger $n$. I wonder if there is any quantitative way to describe this cross over behavior.
Thanks.