I assume that $k$ as fixed. The answer to this problem is closely related to the maximum density $p=p_{k+1}(n)=r_{k+1}(n)/n$ of a subset of $\{1,\ldots,n\}$ without a $(k+1)$-term arithmetic progression.

Indeed, let $S \subset \{1,\ldots,n\}$ be a set of cardinality $r_{k+1}(n)$ without a $(k+1)$-term arithmetic progression. Now construct a random subset $A \subset [4kn]$ as follows. The set $A$ consists of the union of $k$ random translations $S_i$ of $S$, where $S_i=S+d_i$ and $d_i$ for $1 \leq i \leq k$ is picked uniformly at random from the interval $\{4(i-1)n,4(i-1)n+1,\ldots,4(i-1)n+2n-1\}$ of $2n$ integers. It is easy to check that $A$ cannot contain a $(k+1)$-term arithmetic progression. It is also not difficult to check that the expected number of $k$-term arithmetic progressions in $A$ is at least $(cp)^k n^2$ where $c>0$ is an absolute constant. Hence, letting $N=4kn$ and changing $c$ slightly, there is a subset of $[N]$ with no $(k+1)$-term arithmetic progression but the number of $k$-term arithmetic progressions is at least $(cp)^k N^2$.

In the other direction, any subset $B \subset \{1,\ldots,N\}$ with no $(k+1)$-term arithmetic progression has density at most $2p$. A $k$-term arithmetic progression is determined by its first term and by the common difference. The number of possible first terms in a $k$-term arithmetic progression in $B$ is at most $2pN$. The number of possible common differences is at most $N/(k-1)$. Hence, there are at most $2pN \cdot \frac{N}{k-1}=\frac{2}{k-1}pN^2$ total $k$-term arithmetic progressions in $B$.

Therefore, the fraction of $k$-term arithmetic progression in $\{1,\ldots,N\}$ which can be in a set with no $(k+1)$-term arithmetic progression is roughly between $(cp)^k$ and $p$. However, there is still a large gap between Rankin's lower bound and Gowers' upper bound on $p$. Hence, the bound on this problem is closely tied to quantitative bounds for Szemer'edi's theorem.

I assume that $k$ as fixed. The answer to this problem is closely related to the maximum density $p=p_{k+1}(n)=r_{k+1}(n)/n$ of a subset of $\{1,\ldots,n\}$ without a $(k+1)$-term arithmetic progression.

Indeed, let $S \subset \{1,\ldots,n\}$ be a set of cardinality $r_{k+1}(n)$ without a $(k+1)$-term arithmetic progression. Now construct a random subset $A \subset [4kn]$ as follows. The set $A$ consists of the union of $k$ random translations $S_i$ of $S$, where $S_i=S+d_i$ and $d_i$ for $1 \leq i \leq k$ is picked uniformly at random from the interval $\{4(i-1)n,4(i-1)n+1,\ldots,4(i-1)n+2n-1\}$ of $2n$ integers. It is easy to check that $A$ cannot contain a $(k+1)$-term arithmetic progression. It is also not difficult to check that the expected number of $k$-term arithmetic progressions in $A$ is at least $(cp)^k n^2$ where $c>0$ is an absolute constant. Hence, letting $N=4kn$ and changing $c$ slightly, there is a subset of $[N]$ with no $(k+1)$-term arithmetic progression but the number of $k$-term arithmetic progressions is at least $(cp)^k N^2$.

In the other direction, any subset $B \subset \{1,\ldots,N\}$ with no $(k+1)$-term arithmetic progression has density at most $2p$. A $k$-term arithmetic progression is determined by its first two terms. The number of possible first two terms in $B$ is at most ${2pN \choose 2} < 2p^2N^2$. Hence, there are at most $2p^2N^2$ total $k$-term arithmetic progressions in $B$.

In summary, the number of $k$-term arithmetic progression in $\{1,\ldots,N\}$ which can be in a set with no $(k+1)$-term arithmetic progression is between $(cp)^k N^2$ and $2p^2N^2$. However, there is still a large gap between Rankin's lower bound and Gowers' upper bound on $p=r_{k+1}(n)/n$. Hence, the bound on this problem is closely tied to quantitative bounds for Szemerédi's theorem.

I assume that $k$ as fixed. The answer to this problem is closely related to the maximum density $p=p_{k+1}(n)=r_{k+1}(n)/n$ of a subset of $\{1,\ldots,n\}$ without a $(k+1)$-term arithmetic progression.

Indeed, let $S \subset \{1,\ldots,n\}$ be a set of cardinality $r_{k+1}(n)$ without a $(k+1)$-term arithmetic progression. Now construct a random subset $A \subset [4kn]$ as follows. The set $A$ consists of the union of $k$ random translations $S_i$ of $S$, where $S_i=S+d_i$ and $d_i$ for $1 \leq i \leq k$ is picked uniformly at random from the interval $\{4(i-1)n,4(i-1)n+1,\ldots,4(i-1)n+2n-1\}$ of $2n$ integers. It is easy to check that $A$ cannot contain a $(k+1)$-term arithmetic progression. It is also not difficult to check that the expected number of $k$-term arithmetic progressions in $A$ is at least $(cp)^k n^2$ where $c>0$ is an absolute constant. Hence, letting $N=4kn$ and changing $c$ slightly, there is a subset of $[N]$ with no $(k+1)$-term arithmetic progression but the number of $k$-term arithmetic progressions is at least $(cp)^k N^2$.

In the other direction, any subset $B \subset \{1,\ldots,N\}$ with no $(k+1)$-term arithmetic progression has density at most $2p$. A $k$-term arithmetic progression is determined by its first term and by the common difference. The number of possible first terms in a $k$-term arithmetic progression in $B$ is at most $2pN$. The number of possible common differences is at most $N/(k-1)$. Hence, there are at most $2pN \cdot \frac{N}{k-1}=\frac{2}{k-1}pN^2$ total $k$-term arithmetic progressions in $B$.

Therefore, the fraction of $k$-term arithmetic progression in $\{1,\ldots,N\}$ which can be in a set with no $k$-term arithmetic progression is roughly between $(cp)^k$ and $p$. However, there is still a large gap between Rankin's lower bound and Gowers' upper bound on $p$. Hence, the bound on this problem is closely tied to quantitative bounds for Szemer'edi's theorem.

I assume that $k$ as fixed. The answer to this problem is closely related to the maximum density $p=p_{k+1}(n)=r_{k+1}(n)/n$ of a subset of $\{1,\ldots,n\}$ without a $(k+1)$-term arithmetic progression.

Indeed, let $S \subset \{1,\ldots,n\}$ be a set of cardinality $r_{k+1}(n)$ without a $(k+1)$-term arithmetic progression. Now construct a random subset $A \subset [4kn]$ as follows. The set $A$ consists of the union of $k$ random translations $S_i$ of $S$, where $S_i=S+d_i$ and $d_i$ for $1 \leq i \leq k$ is picked uniformly at random from the interval $\{4(i-1)n,4(i-1)n+1,\ldots,4(i-1)n+2n-1\}$ of $2n$ integers. It is easy to check that $A$ cannot contain a $(k+1)$-term arithmetic progression. It is also not difficult to check that the expected number of $k$-term arithmetic progressions in $A$ is at least $(cp)^k n^2$ where $c>0$ is an absolute constant. Hence, letting $N=4kn$ and changing $c$ slightly, there is a subset of $[N]$ with no $(k+1)$-term arithmetic progression but the number of $k$-term arithmetic progressions is at least $(cp)^k N^2$.

In the other direction, any subset $B \subset \{1,\ldots,N\}$ with no $(k+1)$-term arithmetic progression has density at most $2p$. A $k$-term arithmetic progression is determined by its first term and by the common difference. The number of possible first terms in a $k$-term arithmetic progression in $B$ is at most $2pN$. The number of possible common differences is at most $N/(k-1)$. Hence, there are at most $2pN \cdot \frac{N}{k-1}=\frac{2}{k-1}pN^2$ total $k$-term arithmetic progressions in $B$.

Therefore, the fraction of $k$-term arithmetic progression in $\{1,\ldots,N\}$ which can be in a set with no $(k+1)$-term arithmetic progression is roughly between $(cp)^k$ and $p$. However, there is still a large gap between Rankin's lower bound and Gowers' upper bound on $p$. Hence, the bound on this problem is closely tied to quantitative bounds for Szemer'edi's theorem.