I found a curious pattern... any idea of a deductive formula? - [4, 24, 80, 200, 420, 784, 1344] I found this pattern by taking the absolute value of the difference of both sides of the Cauchy-Schwarz formula with consecutive numbers (I can go into more detail if desired). No worries if this doesn't make sense to you.
I'm looking to find a deductive sequence for this formula (I'm alright if you are able to find a recursive formula too). I can also use a program to generate many more terms of the sequence if you want me to.
Based on intuition, I have a feeling that the answer is going to be a ratio of various factorials.
Let me know if you have any questions... thanks!
 A: It looks to me like being :
$$
a_n = \frac{(n+1)^2((n+1)^2-1)}{3} = \frac{n(n+1)^2(n+2)}{3}$$
starting with $a_1 = 4$.
EDIT : As Gottfried points out below, it is easier to "shift the index" and go with the formula $a_n = \frac{n^2(n^2-1)}{3}$ with $a_2 = 4$ and $a_1 = 0$.
For example : $a_3= \frac{16 \times 15}{3} = \frac{240}{3} = 80$.
Note that you have missed out the term for $n=5$ which is $420$, but all other terms seem to match.

EDIT : Assuming that you have the $420$ in place, so have sufficiently many terms of the sequence to consider, you can use the method of finite differences to check if the terms are polynomial in the index or not.
That is, for a sequence $a_n$ we have the sequence of differences $(\Delta \{a_n\})_k = a_{k+1} - a_k$. Now, if it turns out that $\Delta^t \{a_n\}$ is identically zero, then $a_n$ is in fact a polynomial in $n$ of degree less than $t$.
Once you see this, you can use any $t$ terms of the sequence and try to fit in the coefficients corresponding to the polynomial $a_n  = c_0 + c_1n + ... + c_tn^{t-1}$ to get the coefficients.

For example, considering the sequence of finite differences, we get :
\begin{align}
4 & & 24 & & 80 & & 200 & & 420 & & 784 \\
&\quad \quad 20 & & \quad \quad 56 & & \quad \quad 120 & & \quad \quad 220 & & \quad \quad 364 &  \\
& & 36 & & 64 & & 100 & & 144 && \\
& & & \quad \quad 28 & & \quad \quad \ 36 & & \quad \quad \ 44 &&& \\
& & & & 8 & & 8 & & \\
&&&&&\quad \quad \ \ 0&&&&& 
\end{align}
(note : using $1344$ as well one sees that the fifth row gives a sequence of $0$s)
Thus $a_n$ is a quartic in $n$ , and one can use the first four values of the sequence to get the correct expression.
(Alternately, observe that you can hazard a guess at the third row above , which is just the even squares. From there, you can invert the difference operator twice to get the answer).
I have also explained a similar phenomena for powers in this answer, where I showed how one can find powers of numbers by inverting the difference technique.
