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Example 1. Taking $r = 1$ and $p=2$, to find a number field $K$ such that $\mathcal O_K$ needs at least two generators as a $\mathbf Z$-algebra ($K$ is not "monogenic") it is sufficient to find a $K$ such that $[K:\mathbf Q] > 2$ and $2$ splits completely in $\mathcal O_K$, such as a cubic field in which $2$ splits completely. Dedekind found the first example of such a field: $\mathbf Q(\alpha)$ where $\alpha^3 - \alpha^2 - 2\alpha - 8 = 0$. A method of constructing infinitely many such $K$ is to use the cubic subfield of the cyclotomic field $\mathbf Q(\zeta_p)$ for primes $p$ such that $p \equiv 1 \bmod 3$ and $2^{(p-1)/3}\equiv 1 \bmod p$; that means $p$ splits completely in $\mathbf Q(\sqrt[3]{2},\zeta_3)$, and there are infinitely many such $p$ (and hence infinitely many such cubic fields) since the density of such $p$ is $1/6$ by the Chebotarev density theorem. The first few such $p$ are $31$, $43$, $109$, and $127$. For example, using PARI and Galois theory, the cubic subfield of $\mathbf Q(\zeta_{31})$ is $\mathbf Q(\alpha)$ where $\alpha$ is a root of $$x^3 + x^2 - 10x - 8$$ and $2$ splits completely in this cubic field (e.g., PARI says this cubic polynomial splits completely over $\mathbf Q_2$) so the ring of integers of this cubic field needs at least $2$ generators as a $\mathbf Z$-algebra. (This cubic field is different from Dedekind's, e.g., Dedekind's cubic has discriminant -2012 -- too bad I didn't write about this 6 years ago! -- while this cubic has discriminant $3844=62^2$.)

Example 1. Taking $r = 1$ and $p=2$, to find a number field $K$ such that $\mathcal O_K$ needs at least two generators as a $\mathbf Z$-algebra ($K$ is not "monogenic") it is sufficient to find a $K$ such that $[K:\mathbf Q] > 2$ and $2$ splits completely in $\mathcal O_K$, such as a cubic field in which $2$ splits completely. Dedekind found the first example of such a field: $\mathbf Q(\alpha)$ where $\alpha^3 - \alpha^2 - 2\alpha - 8 = 0$. A method of constructing infinitely many such $K$ is to use the cubic subfield of the cyclotomic field $\mathbf Q(\zeta_p)$ for primes $p$ such that $p \equiv 1 \bmod 3$ and $2^{(p-1)/3}\equiv 1 \bmod p$; that means $p$ splits completely in $\mathbf Q(\sqrt[3]{2},\zeta_3)$, and there are infinitely many such $p$ (and hence infinitely many such cubic fields) since the density of such $p$ is $1/6$ by the Chebotarev density theorem. The first few such $p$ are $31$, $43$, $109$, and $127$. For example, using PARI and Galois theory, the cubic subfield of $\mathbf Q(\zeta_{31})$ is $\mathbf Q(\alpha)$ where $\alpha$ is a root of $$x^3 + x^2 - 10x - 8$$ and $2$ splits completely in this cubic field (e.g., PARI says this cubic polynomial splits completely over $\mathbf Q_2$) so the ring of integers of this cubic field needs at least $2$ generators as a $\mathbf Z$-algebra. (This cubic field is different from Dedekind's, e.g., Dedekind's cubic field has $1$ real embedding while this cubic field has $3$ real embeddings, of Dedekind's cubic field has discriminant $-503$ while this cubic field has discriminant $961 = 31^2$.)

Update: This proof extends to the relative case. If $E/F$ is an extension of number fields then a sufficient condition for $\mathcal O_E$ to need more than $r$ generators as an $\mathcal O_F$-algebra is that there is a (nonzero) prime ideal $\mathfrak p$ in $\mathcal O_F$ such that (i) $[E:F] > ({\rm N}\mathfrak p)^r$ and (ii) $\mathfrak p$ splits completely in $\mathcal O_E$. In the above proof, "ring homomorphism" has to be replaced by "$\mathcal O_F$-algebra homomorphism".

I hope from these examples you see the pattern by which, for each $r$, you can use a subfield of degree $d$ in $\mathbf Q(\zeta_p)$ for infinitely many primes $p \equiv 1 \bmod d$ to get infinitely many number fields whose ring of integers require more than $r$ generators as a $\mathbf Z$-algebra.

It is time to prove (i) and (ii) above are sufficient conditions for $\mathcal O_K$ to require more than $r$ generators as a $\mathbf Z$-algebra, let $K$ be a number field such that $\mathcal O_K$ has at most $r$ generators as a $\mathbf Z$-algebra. We will show either condition (i) or (ii) has to break down, or more simply if (ii) holds then (i) does not: if a prime $p$ splits completely in such an $\mathcal O_K$ then we will show $[K:\mathbf Q] \leq p^r$.

I hope from these examples you see the pattern by which, for each $r$, you can use a subfield of degree $d$ in $\mathbf Q(\zeta_p)$ for infinitely many primes $p \equiv 1 \bmod d$ to get infinitely many number fields whose ring of integers requires more than $r$ generators as a $\mathbf Z$-algebra.

It is time to prove (i) and (ii) above are sufficient conditions for $\mathcal O_K$ to require more than $r$ generators as a $\mathbf Z$-algebra. Let $K$ be a number field such that $\mathcal O_K$ has at most $r$ generators as a $\mathbf Z$-algebra. We will show either condition (i) or (ii) has to break down, or more simply if (ii) holds then (i) does not: if $\mathcal O_K$ has at most $r$ generators as a $\mathbf Z$-algebra and a prime $p$ splits completely in $\mathcal O_K$ then we will show $[K:\mathbf Q] \leq p^r$.