1
$\begingroup$

Let's say that a given $n\in\mathbb{N}$ is writable as sum of cubes in $k$ consecutive ways if it can be written as sum of $j,j+1,\ldots, j+(k-1)$ nonzero cubes, for some $j\geqslant 1$.

For example, $26$ is writable as sum of squares in $4$ consecutive ways, for \begin{align} 26 &= 1+25 &&\text{($2$ squares)}\\ 26 &= 1+9+16 &&\text{($3$ squares)}\\ 26 &= 4+4+9+9 &&\text{($4$ squares)} \\ 26 &= 1+1+4+4+16 &&\text{($5$ squares)} \end{align}

I'm trying to show that there is $n\in\mathbb{N}$ such that $n$ is writable as sum of cubes in $9$ consecutive ways, i.e. that there is $j\geqslant 1$ such that $n$ is the sum of $j,j+1,\ldots,j+8$ cubes.

I'm using that every number can be written as sum of $9$ or fewer cubes. (Or at least that there is some $g(3)\in\mathbb{N}$ such that every number is the sum of $g(3)$ or fewer cubes. I prefer this one because this problem seems to ask to be generalized to Waring bases $P_k=\{n^k:n\in\mathbb{N}\}$ in general.)

The example I found for $26$ for the case of squares was calculated manually, but I believe that there must be some general strategy to tackle this problem. I'm not much interested in to find an specific $n$ that is sum of cubes in $9$ consecutive ways, but in the method of finding one (if there is such).

I believe it holds for any $k$ in any Waring basis, but the case of cubes for $k=9$ would be good enough haha.

Any hints and tips would be much appreciated!

$\endgroup$
  • 1
    $\begingroup$ $n=1072$ is a sum of $2,3,4,...,20$ cubes, as I found out with PARI/GP , but I do not have an idea for a concrete construction. $\endgroup$ – Peter Nov 3 '15 at 0:06
  • $\begingroup$ @Peter That's nice! I found the $n=26$ for squares using Python. I suspect that there must some kind of pattern in the representations of such numbers. Anyway, if this pattern exists it must be nontrivial, for if $n$ is the least number writable as sum of cubes in $9$ (or $19$) consecutive ways, then no cube appears in all $9$ (or $19$) representations (otherwise there would be a smaller number with this property). $\endgroup$ – Alufat Nov 3 '15 at 0:12
  • $\begingroup$ @Peter Oh, but are you considering negative cubes?? $\endgroup$ – Alufat Nov 3 '15 at 0:15
  • 1
    $\begingroup$ @Peter Wow o.O, $1072$ must be a really amazing number then hahaha. Would you mind post these representations as an answer? I'm really curious!! (and a bit too lazy to open Python right now :P) $\endgroup$ – Alufat Nov 3 '15 at 0:20
  • 1
    $\begingroup$ @vadim123 Maybe it's special in that it begins with 2. (Fermat guarantees that it can't begin with 1, since no cube is the sum of two cubes.) $\endgroup$ – Akiva Weinberger Nov 3 '15 at 1:43
1
$\begingroup$

$1072$ represented by the sum of $k$ cubes $k=2,...,30$

? n=1072;s=10;for(k=2,30,gef=0;forvec(z=vector(k,m,[1,s]),if(gef==0,if(sum(j=1,k
,z[j]^3)==n,gef=1;print(k,"    ",z))),1))
2    [7, 9]
3    [2, 4, 10]
4    [1, 6, 7, 8]
5    [1, 1, 5, 6, 9]
6    [1, 3, 4, 5, 7, 8]
7    [1, 1, 1, 5, 6, 6, 8]
8    [1, 2, 2, 4, 6, 6, 6, 7]
9    [1, 1, 1, 2, 5, 5, 5, 7, 7]
10    [1, 1, 1, 1, 1, 1, 1, 1, 4, 10]
11    [1, 1, 1, 1, 2, 2, 2, 4, 5, 7, 8]
12    [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 10]
13    [1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 7, 7, 7]
14    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 3, 10]
15    [1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 6, 6, 6, 7]
16    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 5, 5, 7, 7]
17    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 8, 8]
18    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 4, 5, 7, 8]
19    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 10]
20    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 7, 7, 7]
21    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 10]
22    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 6, 6, 6, 7]
23    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 5, 6, 7, 7]
24    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 8, 8]
25    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 5, 7, 8
]
26    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3
, 10]
27    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 3, 3, 6
, 6, 8]
28    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 5
, 5, 6, 8]
29    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2
, 2, 2, 8, 8]
30    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
, 4, 4, 4, 7, 8]
$\endgroup$
  • 1
    $\begingroup$ That's a really strange property at first sight haha. Thanks! :D $\endgroup$ – Alufat Nov 3 '15 at 0:27
  • 1
    $\begingroup$ Alas, Fermats last theorem forbids a cube being the sum of two cubes :( So, a cube cannot have this property. $\endgroup$ – Peter Nov 3 '15 at 0:31
1
$\begingroup$

For $k=31,...,38$, there are still representations :

? n=1072;s=10;for(k=31,40,gef=0;forvec(z=vector(k,m,[1,s]),if(gef==0,if(sum(j=1,
k,z[j]^3)==n,gef=1;print(k,"    ",z))),1))
31    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
, 1, 4, 4, 4, 5, 9]
32    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
, 1, 1, 1, 4, 5, 7, 8]
33    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
, 1, 1, 1, 1, 2, 2, 3, 10]
34    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
, 1, 1, 2, 3, 3, 4, 4, 7, 8]
35    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
, 1, 1, 1, 1, 1, 4, 5, 5, 6, 8]
36    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
, 1, 1, 1, 1, 1, 1, 1, 2, 2, 8, 8]
37    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
, 1, 1, 1, 1, 1, 1, 2, 3, 3, 5, 5, 9]
38    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 10]
$\endgroup$
  • $\begingroup$ I posted another answer because the of the overview . $\endgroup$ – Peter Nov 3 '15 at 0:43
  • 1
    $\begingroup$ I'm really confused HAHAHA. Either it's a common property among large integers or $1072$ is a VERY special number. I'll continue to seek for a general solution, but your answer was far more satisfactory then I expected xD $\endgroup$ – Alufat Nov 3 '15 at 0:48
  • 1
    $\begingroup$ I am glad about your enthusiasm :) $\endgroup$ – Peter Nov 3 '15 at 0:50
  • $\begingroup$ The situation becomes much more interesting, if we only consider representation without repeating cubes. I am currently searching $k=51$ and did not find a representation so far ... $\endgroup$ – Peter Nov 3 '15 at 0:55
  • 1
    $\begingroup$ oeis.org/… $\endgroup$ – Peter Nov 3 '15 at 1:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.