TITLE: On those integers $n>1$ such that there exists a commutative ring with identity with exactly $n$ ideals
QUESTION [0 upvotes]: Let $n>1$ be an integer; we call $n$ a "ring number" if there exists a commutative ring $R$, with identity, having exactly $n$ ideals (including $\{0\}$ and $R$); now since for every $n>1$, $\mathbb Z_n$ is a commutative ring with unity having exactly $d(n)$ (no. positive divisors of $n$) ideals and $d(n)$ is unbounded as $n$ grows large, so we know that there are infinitely many ring numbers. 
My question: 

Is it true that all ring numbers are of the form $d(n)$ for some $n$ ? If not all ring numbers are of the form $d(n)$ then how many exceptions are there; finitely many or infinitely many? Are there infinitely many positive integers which 'are not' ring numbers? Can we characterize ring numbers in any way? 
  Is it true that if $n$ is a ring number then there is a commutative ring with identity $R$ which is also a PIR, having exactly $n$ ideals?

Answer or reference to any of the questions would be highly appreciated. Please help. Thanks in advance.

REPLY [1 votes]: Yes, $d(n)$ can attain any positive integer for some $n$, consider $n=2^k$. Hence all ring numbers are of the form $d(n)$.
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