# Partition problem for consecutive $k$th powers with equal sums (another family)

This is the partition problem as applied to a special set, namely the first $n$ $k$th powers. Assume the notation,

$$[a_1,a_2,\dots,a_n]^k = a_1^k+a_2^k+\dots+a_n^k$$

I. Family 1

The following infinite family is quite well-known,

$$[1,\color{blue}{4}]^1 = [2,3]^1$$

$$[1,4,6,7]^2 = [2,3,5,\color{blue}{8}]^2$$

$$[1,4,6,7,10,11,13,\color{blue}{16}]^3 = [2,3,5,8,9,12,14,15]^3$$

and so on, for a partition of the $k$th power of the first $\color{blue}{4,8,16,32,64} = 2^{k+1}$ positive integers. (See also this post.)

II. Family 2

However, it can be noticed that,

$$[1, \color{brown}{4}]^1 = [2, 3]^1$$

$$[1, 2, 4, \color{brown}{7}]^2 = [3, 5, 6]^2$$

$$[1, 2, 4, 8, 9, \color{brown}{12}]^3 = [3, 5, 6, 7, 10, 11]^3$$

$$[1, 6, 9, 11, 12, 14, 15, 16, 17]^4 = [2, 3, 4, 5, 7, 8, 10, 13, 18, \color{brown}{19}]^4$$

$$[1, 5, 7, 8, 12, 14, 17, 18, 19, 20, 22, 24, 25, 26]^5 = [2, 3, 4, 6, 9, 10, 11, 13, 15, 16, 21, 23, 27, \color{brown}{28}]^5$$

and so on, with the sequence $\color{brown}{4,7,12,19,28} = k^2+3.$ (For $k=3$, see this post.)

Questions:

1. Is it true that there is a partition of the $k$th power of the first $k^2+3$ positive integers such that the sum of each set is equal?
2. Is $k^2+3$ the minimum number of consecutive terms, or can there be a third family with a smaller number of terms for higher $k$?
3. What is the partition of the first $n$ $6$th powers with $n\leq6^2+3$?

$\color{red}{Update:}$

It turns out $k^2+3$ is not the minimum number $n$ for higher $k$. I used Mathematica to check,

$$\sum_{m=1}^n \epsilon\, m^k = 0$$

where $\epsilon = \pm1$. For $k=1,2,3$, we find $n = 4,7,12$ as in the second family. However, $k=4$ has a smaller solution at $n=16$,

$$[1, 2, 3, 4, 8, 9, 10, 11, 12, 16]^4 = [5, 6, 7, 13, 14, 15]^4$$

and $k=5$ also has one at $n=24$,

$$[1, 2, 5, 7, 10, 13, 14, 15, 19, 20, 21, 23]^5 = [3, 4, 6, 8, 9, 11, 12, 16,17, 18, 22, 24]^5$$

Presumably for $k=6$, it should be $28\leq n<<39$, but my Mathematica code is too slow for $n>28$. (The Wikipedia link above gives a pseudo-polynomial time algorithm, maybe some interested soul might be interested to implement it.)

It seems there might be several families then, and it would be nice to find a unifying principle like in the first family.

## 1 Answer

There are $2^n$ subsets of the first $n$ positive integers, and for each subset the sum is bounded below by zero and above by $\sum_{i=1}^n i^k = O(n^k/k)$. The problem of finding a partition with equal sums is equivalent to finding a subset with a particular sum in this range, specifically $\frac 12 \sum_{i=1}^n i^k$. If we think these subset sums are distributed somewhat uniformly in this range, then this heuristic suggests the minimum $n$ should grow like $k\log k$, and for $n=k^2+3$ there should be increasingly many possible partitions.

Note that in addition to your examples there are at least 15 other partitions with $k=4,n=19$, e.g. \begin{align} \frac 12 \sum_{i=1}^{19} i^4 &= [1, 3, 4, 8, 10, 11, 14, 17, 19]^4 \\ &= [1,2,3,6,7,14,15,17,18]^4 \\ &= [1,2,4,9,11,12,15,17,18]^4 \\ & = [1,2,5,7,8,11,14,15,16,18]^4 \end{align} and two others for $k=5,n=24$, and at least five more for $k=5,n=28$. So it seems to me you won't be able to unify a sequence of solutions into a family based solely on the values of $n$.

I found these for $k=6,n=32$ \begin{align} \frac 12 \sum_{i=1}^{32} i^6 & = [1, 2, 3, 4, 6, 8, 9, 10, 15, 16, 17, 21, 23, 28, 31, 32]^6 \\ & = [1, 2, 3, 8, 11, 13, 14, 16, 17, 18, 21, 24, 26, 27, 29, 32]^6 \\ &= [1, 2, 7, 8, 9, 12, 14, 17, 18, 19, 20, 23, 24, 29, 30, 31]^6 \end{align} but I only looked for balanced partitions, so I haven't verified there are none with $n<32$.

• Ah, so there's that elusive creature, $$\small[1, 2, 3, 4, 6, 8, 9, 10, 15, 16, 17, 21, 23, 28, 31, 32]^6 = [5, 7, 11, 12, 13, 14, 18, 19, 20, 22, 24, 25, 26, 27, 29, 30]^6$$ My primitive code would have taken $2^{31}$ steps to find it. Thanks! (Yes, for the $k < 6$, I did notice there were multiple solutions.) – Tito Piezas III Apr 4 '15 at 6:57
• For $k=7$, what do you think should be $n$ for $$\sum_{m=1}^n \pm m^7 = 0$$ I assumed $n=40$, rigged a faster code, ran it for more than hour before giving up. It can't be $n=41,42$ as the positive sum is odd, so is it $44<n<49$? In other words, for $k>3$, is $n\leq k^2$ a conservative bound? – Tito Piezas III Apr 16 '15 at 12:27