Let's say we have an unknown random variable whose entropy is $1.75$ Our job is to find minimum distributions for this random variable.
What I wrote was: $$ p_1 \log_2\Big(\frac 1 {p_1}\Big) + \ldots + p_n \log_2\Big(\dfrac 1 p_n \Big)= \log_2\Bigg[\Big(\frac{1}{p_1}\Big)^{p_1} \cdot \Big(\frac{1}{p_2}\Big)^{p_2} ...\Big(\frac{1}{p_n}\Big)^{p_n}\Bigg]= 1.75$$
So $2^{1.75} = \Big(\dfrac 1 p_1\Big)^{p_1} \cdot \Big(\dfrac 1 p_2\Big)^{p_2} \cdot \ldots \cdot \Big(\dfrac 1 p_n\Big)^{p_n}$ which by using AM-GM, can be found that $n \ge 2^{1.75} \implies n = 4$ as min value.
However, finding these probability distribution is proving to be quite tricky.
Any suggestions?
For a four state system, if all probabilities are equal the entropy is $2$ bits. If $p_1=1,$ and the others are all zero the entropy is $0$. by continuity, there must be a value $p$ such that $p_1=1-3p,p_2=p_3=p_4=p$ that has an entropy of $1.75$. You can find it numerically, if nothing else. Alpha finds two solutions. Of course, other distributions are possible.