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For $n>0$, let $c_n$ be the number of local rings $R$ such that $R^\ast$ is cyclic of order $n$.

Note that $c_1 =1$. (A local ring $R$ such that $R^\ast = \{1\}$ has precisely two elements. See Is there a classification of local rings with trivial group of units? .)

Note that $c_2 = 2$. (In this case $R= \mathbf{Z}/3\mathbf{Z}$ or $R=\mathbf{Z}/4\mathbf{Z}$.)

Question. Can one give an upper bound for $c_n$ in terms of $n$ better than the following one?

For any $x$ in $R$, we have that $x+(1-x) = 1$. Therefore, $x$ or $1-x$ is a unit. Therefore, the number of elements in $R$ is bounded from above by $2n$. Thus, $c_n$ is bounded from above by the number of rings with $2n$ elements. This is then bounded from above by the number of groups of order $2n$. Conclusion: $c_n \leq$ (number of groups of order $2n$).

Question. Can one give an explicit formula for $c_n$?

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