Q:

Find the next term in the sequence. $\frac{7}{3},\frac{35}{6},\frac{121}{12},\frac{335}{36},\ldots $

$\dfrac{7}{3},\dfrac{35}{6},\dfrac{121}{12},\dfrac{335}{36},\ldots $
$\bf\text{Answer}$ given is $\dfrac{865}{48}$
I found that $4^{th}$ differencess of the numbers $7,35,121,335\cdots$ are not constant .
and the second differences of the denominator drastically changes,
$3\quad 6\quad 12\quad 36\quad 48\\~\\
\quad 3\quad 6\quad 24\quad \color{red}{12}$
decimal value is also not showing any pattern.
$\frac{7}{3},\ \frac{35}{6},\ \frac{121}{12},\ \frac{335}{36},\ldots $
$2.33,\ 5.83,\ 10.08,\ 9.33,\ldots $

A:

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 $(12-1)^2=121$.

