Find the next term in the sequence. $\frac{7}{3},\frac{35}{6},\frac{121}{12},\frac{335}{36},\ldots $
If your fourth term is $\dfrac{335}{\color{red}{24}}$ instead of $~\dfrac{335}{\color{red}{36}}~,~$ then the pattern you're looking for is
a recurrence relation of the form $~a_{n+1}~=~6n+1-\dfrac{a_n}2~,$ with $~a_1=\dfrac73~.$ This idea
came to me while trying to approximate each term with its nearest integer; in particular,
by noticing that $~6^2=36\simeq35$, and $(12-1)^2=121$.