Integers and Sequences Is there a sequence of integers such that for ∀ k it contains an arithmetic subsequence of length k but it does have no infinitely long arithmetic subsequence?

There's two approaches to constructing a sequence like this.
First, we might try to space out blocks of the sequence sufficiently far. For example,
$$
    1, \qquad 3,4, \qquad 9,10,11, \qquad 23,24,25,26, \qquad 53,54,55,56,57, \dots
$$
where, whenever a block ends at $x$, the next block begins at $2x+1$.
In particular, the gap between the $k^{\text{th}}$ block and the $(k+1)^{\text{th}}$ is wider than the range spanned by the first $k$ blocks. If an arithmetic progression includes at least two terms from the first $k$ blocks, its difference is less than that range, so it can never jump the gap. So there is no infinite arithmetic progression, but there are obviously finite ones of arbitrary length (each block by itself).
Second, we can apply a "just-do-it" construction. For simplicity, we'll consider only increasing sequences of integers.
Enumerate all the infinite arithmetic progressions: this is possible, because there are only countably many pairs $(a,d)$ where $a$ is the first term and $d$ is the difference. Now, to generate our sequence, we alternate the following steps:


*

*Find a term, $x$, of the $k^{\text{th}}$ infinite arithmetic progression which is larger than any term of our current sequence, and write down $x+1$ (ensuring that we will never write down $x$).

*Write down $x+2, x+3, \dots, x+k$, ensuring that we have a finite arithmetic progression of length $k$.


For each infinite arithmetic progression, eventually we will skip one of its terms, and therefore no infinite progression is a subsequence.
