6
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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.)

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  • $\begingroup$ The only hit for $4,6,96,480$ is oeis.org/A052684 $\endgroup$ – Barry Cipra Apr 3 '15 at 2:25
  • $\begingroup$ It is easy to see one can do it with $23040 = \gcd\left(6 \cdot 168\cdot 480,\;6!\cdot2^{6(6-1)/2}\right)$. I doubt this is the actual minimum but this is at least a tighter upper bound. $\endgroup$ – achille hui Apr 3 '15 at 11:36
  • $\begingroup$ @achillehui: Thanks! Would it follow that the upper bound for $k=5$ is $\gcd\left(5\cdot56\cdot96,\;5!\cdot2^{5(5-1)/2}\right) = 3840 = 8\cdot\color{brown}{480}$? $\endgroup$ – Tito Piezas III Apr 3 '15 at 12:07
  • $\begingroup$ @TitoPiezasIII yes. By same logic, 3840 is an upper bound for $k = 5$. $\endgroup$ – achille hui Apr 3 '15 at 12:11
  • $\begingroup$ @achillehui: Using that formula, the upper bounds $U_k$ for $k=3,4,5,6$ are $U_k = 8\cdot\color{brown}6,\,\color{brown}{96},\,8\cdot\color{brown}{480},\,23040$. I was hoping the extraneous factor would be predictable. $\endgroup$ – Tito Piezas III Apr 3 '15 at 12:21

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