Filling two gaps without a four-term progression

Take the list $3,4,5,a,b,30,40,50$ with $a,b$ integers. How many ordered pairs $(a,b)$ make the list strictly increasing while ensuring that no four of its entries, taken in any order, form an arithmetic progression?

$228$. We need $5<a<b<30$. Values $6$ and $20$ are excluded (they complete $3,4,5,6$ and $20,30,40,50$), leaving $\binom{22}{2}=231$ pairs; the pairs $(7,9)$, $(12,21)$ and $(16,28)$ create progressions $3,5,7,9$; $3,12,21,30$; $4,16,28,40$, so $231-3=228$ (confirmed by brute force).