Does anybody spot anything familiar in this integer sequence? $0,3,9,21,40,67,106,154,220,298,395,510,644,\dots$
These are the maxima of the distances between permutations of length $n$ up to $n=13$ according to a modified version of Spearman's footrule number calculated by
$\sum_{i=1}^{n} |i-p_i| (n-i+1)$
I cannot find any formula for the above sequence. I've looked up the sequence at OEIS but nothing came up. I'm not a mathematician and don't even know where to start for finding a formula, so any help would be greatly appreciated. For reference, here is the output of my program:

n=2     max=3    at (2 1) 
n=3     max=9    at (3 1 2) 
n=4     max=21   at (4 3 1 2) 
n=5     max=40   at (5 4 1 2 3) 
n=6     max=67   at (6 5 4 1 2 3) 
n=7     max=106  at (7 6 5 1 2 3 4) 
n=8     max=154  at (8 7 6 1 2 3 4 5) (8 7 6 5 1 2 3 4) 
n=9     max=220  at (9 8 7 6 1 2 3 4 5) 
n=10    max=298  at (10 9 8 7 1 2 3 4 5 6) 
n=11    max=395  at (11 10 9 8 7 1 2 3 4 5 6) 
n=12    max=510  at (12 11 10 9 8 1 2 3 4 5 6 7) 
n=13    max=644  at (13 12 11 10 9 8 1 2 3 4 5 6 7) 

Strange thing at $n=8$ as well.
 A: Here's a partial solution, assuming for now that the permutation exhibiting the maximum distance looks like $(n, n-1, \ldots, n-a+1, 1, 2, \ldots, n-a)$.  (You can probably prove that a local maximum must have this form.)  The score can be computed in closed form for such a permutation; you get
$$S(n,a)=\frac{1}{6} \left(4 a^3-9 a^2 n-3 a^2+6 a n^2+3 a n-a\right)+\frac{1}{2} a
   (a-n-1) (a-n).$$
To find the optimal $a$, let $a=\alpha n$, and assume first that $\alpha\le1/2$. Define
$$f(x)=\begin{cases}
  (1-2x)(1-x),& 0\le x\le \alpha\\
  \alpha(1-x),& \alpha < x \le 1
  \end{cases}
$$
Then the score is the Riemann sum for $f$, up to scaling.  But 
$$\int_0^1 f(x)\,dx = \frac{7 \alpha^3}{6}-\frac{5 \alpha^2}{2}+\frac{3 \alpha}{2};$$
it's easy to see that the integral is maximized at $\alpha=3/7$.  (A similar argument shows that you don't do as well for $\alpha>1/2$.)  Accordingly, we expect the score to be maximized at $a=\langle \frac{3n}7\rangle$ (the closest integer to $3n/7$. Indeed, the expression $S(n, \langle \frac{3n}7\rangle)$ produces the correct sequence of values:
$$0,3,9,21,40,67,106,154,220,298,395,510,644,803,980,1190,1421,1684,1976,\ldots$$
