Can anyone find any rhyme or reason to these sequences? In the following sequence of fractions,
$$21/8,273/32,1333/72,4161/128,10101/200,20881/288$$
the denominators are simply $8n^2$, but can't figure out what the numerator is.  It's not in OEIS.
Also need help with this one if possible:
$$1/2,73/18,601/50,2353/98,6481/162$$
Thanks
EDIT: Robert and Brian yes you are right.  In the second sequence I had the third and fourth terms wrong.  They are now corrected.  I double checked the other ones too. Apologies.
 A: $$16n^4+4n^2+1$$

Here's how. Let $f(n)$ be the sequence of numerators. Define the higher order differences
recursively $\Delta^{[1]}f(n):=\Delta f(n):=f(n+1)-f(n)$, and for all $k>1$ let
$\Delta^{[k+1]}f(n):=\Delta(\Delta^{[k]}f)(n).$ This is the discrete version of higher derivatives. 
For the sequence of numerators $f(n)$ we get that $\Delta^{[4]}f(n)$ is the constant $384$ (but we can only compute it for $n=1,2$). Because $D^4x^4=24$, and $384=16\cdot24$ we recognize this as $\Delta^{[4]}(16n^4)$. For the sequence $g(n):=f(n)-16n^4$ we repeat the exercise, and this time we notice that already the second order differences $\Delta^{[2]}g(n)$ form a constant sequence $8$. This is the second derivative of $4x^2$, and gives the second term. The remaining term of the sequence, $g(n)-4n^2$, is just constant $1$.
A: The numerators are $21=5^2-2^2$, $273=17^2-4^2$, $1333=37^2-6^2$, $4161=65^2-8^2$, $10101=101^2-10^2$, and $20881=145^2-12^2$. A little work with finite differences shows that the sequence $5,17,37,65,101,145$ is given by $4n^2+1$, so it appears that your $n$-th term is $$\frac{(4n^2+1)^2-(2n)^2}{8n^2}=\frac{(4n^2-2n+1)(4n^2+2n+1)}{8n^2}\;,$$ or, if you prefer, $$\frac{16n^4+4n^2+1}{8n^2}\;.$$
Added: In the second sequence there may be a typo in the third term: it may have been supposed to be $2353/98$. With that emendation the numerator of the $n$-th term would be $2n(n-1)d_n+1$, where $d_n$ is the denominator of the $n$-th term. Of course that still leaves the denominators a bit of a puzzle: $d_n=2(2n-1)^2$ would work except for the $40$, which ought by that rule to be $50$. Before spending more time on it, I’d want to be sure that the sequence has been given correctly.
Added2: The corrected sequence, with $601/50$ as third term, fits the pattern that I suspected: the $n$-th term is $$\frac{4n(n-1)(2n-1)^2+1}{2(2n-1)^2}=2n(n-1)+\frac1{2(2n-1)^2}\;.$$
