We have the neat equalities,
I. Group 1
For $k=2,3,4,5,\dots$
$$\sum_{n=1}^{2^k}\epsilon_n(x+n)^k = 2^{\frac{k(k-1)}{2}}k! = 4,\;48,\;1536,\;\color{brown}{122880},\dots$$
for appropriate $\epsilon_n =\pm1.$ See this post.
II. Group 2
However, starting with $k=3$, we can have smaller sums at the cost of a higher number of addends,
$$\sum_{n=1}^{20}\epsilon_n(x+n)^3 = 6$$
$$\sum_{n=1}^{56}\epsilon_n(x+n)^4 = 96$$
$$\sum_{n=1}^{168}\epsilon_n(x+n)^5 = \color{brown}{480}$$
$$\sum_{n=1}^{m}\epsilon_n(x+n)^6 = c\,?$$
Compare the big difference between 122880 and 480. (The cases $k=3,4,5$ are given in this post.)
Question: Anybody knows how to find a relatively small non-zero $c$ for $k=6$?
P.S. There is then a partition of the first 168 $5$th powers into two sets such that their sums $A,\,B$ has $A-B = 480$. This is an optimization version of the partition problem with the constraint that the difference is minimal but does not vanish. (It is related to a previous question I asked on a "Partition problem for consecutive kth powers" where the difference must vanish.)