How to maximize min.entropy from a bounded log normal distribution? Background:
I have random source that can be measured and histogrammed over many samples as:

My measuring instrument can only measure a fixed range of $0 -1023$, so that axis must always stay the same.  All samples are discrete integer values.  I have full control over the mean and standard deviation of the samples, so that I can spread out the distribution.  I can spread it out so far that it would exceed a reading of $1023$.  So I might have the following, where $V_{max} = 1023$:

You can see that the missing part of the distribution exceeds the allowable range and it's samples just pile up in the blue bar at $V_{max}$.  As I stretch the standard deviation further, $P(V_{max})$ increases and $P(V_{mode})$, yellow bar,  decreases.
min.entropy is defined to be $H = -log(P_{max})$, where $P_{max}$ is the most likely value in a distribution.  
Question: What shape would the histogram have to be to maximize $H$?  
I think that there has to be a sweet spot somewhere during scaling of the standard deviation.  I suspect that H is maximal when $P(V_{mode}) = P(V_{max})$ but I'm not sure.  This means the yellow and blue bars would be identical heights. So achieving something like this:
$$H_{max} = -\log \left( \min\left.\begin{cases}\max(P_v) & 0 \le v < V_{max} \\ P_v & v=V_{max} \end{cases} \right\} \right)$$
 A: This answer is valid for $H$ as defined by the OP. It is also valid for $H=p\ln{p}$, because the solution will still occur where $P(V_{mode})=P(V_{max})$, the factor of $p$ will drop out.
This answer assumes that the discrete (integer-valued) distribution in the question is "similar enough" to a log-normal that the following two approximations are valid:


*

*$P(V_{mode})$ is proportional to the probability density at the mode of the log-normal.

*$P(V_{max})$ is proportional to the probability mass of the log normal that lies to the right of $V_{max}$


Therefore, with (positive) proportionality constants ($\alpha_1$ and $\alpha_2$), we have that 
$$P(V_{mode})=\alpha_1 \frac{e^{-\frac{(\ln{(e^{\mu-\sigma^2})} - \mu)^2}{2\sigma^2}}}{e^{\mu-\sigma^2}\sigma\sqrt{2\pi}}=
\alpha_1 \frac{e^{\frac{\sigma^2}{2}-\mu}}{\sigma\sqrt{2\pi}}$$ and 
$$P(V_{max})=\alpha_2\left[\frac{1}{2}-\frac{1}{2} \mathrm{erf}\left(\frac{\ln{(V_{max})}-\mu}{\sqrt{2}\sigma} \right)\right]$$
where erf is the error function (the first equation comes from plugging in the mode of the lognormal to the PDF for the log normal, and the second comes from plugging in $V_{max}$ to one minus the CDF of the log-normal).
Note that for any value of $\sigma$, $P(V_{mode})$ monotonically decreases in $\mu$.  Also note that the erf is monotonically increasing, so for any value of $\sigma$, $P(V_{max})$ is monotonically increasing in $\mu$.  
Note further that $P_{max}$ occurs either at $P(V_{mode})$ or at $P(V_{max})$. So for any value of $\sigma$, we maximize $H$ by varying $\mu$ until $P(V_{mode})=P(V_{max})$. At this value of $\mu$, increasing $\mu$ would increase $P(V_{max})$ and decreasing $\mu$ would increase $P(V_{mode})$.  
Thus, it is certainly the case that $H$ is maximized where $P(V_{mode})=P(V_{max})$.
Furthermore, given the constraint $P(V_{mode})=P(V_{max})$ (and $\sigma >= 0$), it is possible to pick a value of $\sigma$ that maximizes $H$.  This optimization problem could be solved numerically if it were possible to write down the actual probability mass function for this discrete "log-normalish" distribution.
