Proportion of nonabelian $2$-groups of a certain order whose exponent is $4$

273 Views Asked by At

Let

$$\displaystyle A(n)=\frac{\text{number of nonabelian 2-groups of order $n$ whose exponent is }4}{\text{total number of nonabelian 2-groups of order $n$}}.$$

Using GAP, I could observe the following:

$$A(16)=\frac{5}{9}=0.5556, A(32)=\frac{21}{44}=0.4773, A(64)=\frac{93}{256}=0.3633, A(128)=\frac{820}{2313}=0.3545, A(256)=\frac{30446}{56070}=0.5430 \text{ and } A(512)=\frac{8791058}{10494183}=0.8377.$$

Can one prove that if $n>4$, then $A(n)>\frac{1}{3}$?