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Statement. Put $b_n=(n+1)!+n$ for $n\in\mathbb N_0$, and $B=\{b_n:n\ge0\}$. Then $B$ is Sidon and meets every infinite arithmetic progression in $\mathbb N_0$, so its complement contains none. The set $B$ therefore answers Problem 198 negatively.

Source. The public problem page attributes this explicit construction to AlphaProof. Its discussion links an AI-assisted formalization. The source record distinguishes that report from the natural-language proof below and from a reproduced build.

Proof. Every $b_n$ is positive, and for $n\ge0$,

$$b_{n+1}-2b_n =(n+2)!+(n+1)-2((n+1)!+n) =n\bigl((n+1)!-1\bigr)+1>0.$$

Thus the doubling-gap lemma makes $B$ a Sidon set, including uniqueness of sums with equal summands.

Let $P(a,d)=\{a+td:t\ge0\}$ with $a\ge0$ and $d\ge1$. Set $n=a+d$. Since $1\le d\le n+1$, the integer $d$ divides $(n+1)!$. Hence

$$b_n=a+\left(\frac{(n+1)!}{d}+1\right)d\in P(a,d).$$

This point also belongs to $B$, so $P(a,d)$ cannot be contained in the complement. The case $a=0$ is included, and all elements of $B$ are positive. $\square$

Method. The index fixes a residue class while factorial divisibility removes the nonlinear term modulo any prescribed step. Rapid growth prevents collisions of two-term sums. This is an explicit instance of the same separation mechanism as the enumeration construction.

Bears on. Problem 198: the set $B$ is a Sidon set whose complement contains no infinite arithmetic progression, a negative answer.