How to calculate $\operatorname{taxicab}(3,8,2)$ (sum of 8 cubes in two different ways)

Question

Could someone explain me how I can calculate $\operatorname{taxicab}(3,8,2)$?

$\operatorname{taxicab}(3,8,2)$ is the smallest natural number that can be written in $2$ different ways as a sum of $8$ powers $3$.

For instance:

\begin{align}\operatorname{taxicab}(4,3,2) &= 2673\\ &= 7^4 + 4^4 + 2^4 (or 2401+256+16) &= 6^4 + 6^4 + 3^4 (or 1296+1296+81) \end{align}

How can I calculate $\operatorname{taxicab}(3,8,2)$?

This notion is a generalization of the notion of Taxicab number, which is mentioned in the famous story about Hardy and Ramanujan. See also Proof that $1729$ is the smallest taxicab number

If you mean the "taxicab" norm (i.e. the $1$-norm), then you simply have to add the absolute value of the coordinates. — NoetherianCheese, Feb 25 '16 at 21:10
isn't standard?? sorry, I want to say: the smallest number that can be written in two different ways as a sum of 8 powers 3. — Mr. Zacarias Satrustegui, Feb 25 '16 at 21:12
sorry, I don't understand you, your answer is a bit short. for examplo: x1^3 + x2^3 + x3^3 + x4^3 + x5^3 + x6^3 +x7^3 +x8^3 = y1^3 + y2^3 + y3^3 + y4^3 + y5^3 + y6^3 +y7^3 + y8^3. I mean this with to say "as a sum of 8 powers 3 or as a sum 8 positive cubes" — Mr. Zacarias Satrustegui, Feb 25 '16 at 21:33
I think it will be better to write cubes instead of powers of $3$ to remove ambiguity. — Saikat, Feb 26 '16 at 02:31
I have added link to to the Wikipedia article. You post has already four close votes and three downvotes. One of the reason might be missing context. You have already improved it by adding the definition. Perhaps adding something about origin of the problem or related problems would be an improvement, too. So I have added information which I was aware of. Feel free to revert the changes if you feel that does not fit your post. — Martin Sleziak, Feb 26 '16 at 09:20

score 5 · Accepted Answer · edited Apr 24 '17 at 04:15

5

$$\eqalign{132=1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 (or 1+1+1+1+1+1+1+125) =1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 (or 1+8+8+8+8+8+27+64)}$$

This can be found by a recursive computation.

Let $N(x,m)$ be the number of different ways to write $x$ as the (unordered) sum of $m$ positive cubes. Then $N(x,m) = M(x,m,\lfloor x^{1/3} \rfloor)$ where $M(x,m,n)$ is the number of different ways to write $x = a_1^3 + \ldots + a_m^3$ with $1 \le a_1 \le \ldots \le a_m \le n$.

$$\eqalign{M(x,0,n) &= \cases{1 & if $x=0$\cr 0 & otherwise}\cr M(x,m,n) &= \sum_{y=1}^{\min(n, \lfloor x^{1/3} \rfloor)} M(x - y^3, m-1, y)}$$

edited Apr 24 '17 at 04:15

Community

1

answered Feb 25 '16 at 21:46

Robert Israel

448,999

Thanks very much. But I don't understand very well. mm.. could you tell me the first steps, please? for example: If "x" we don't know, and "m" is 8 not? so... M(x,8,n) then if "y=1", in the first step, M(x-1³,8-1,1) not? but what is "n"? sorry if I don't understand well. – Mr. Zacarias Satrustegui Feb 25 '16 at 22:06
and.. min(n,⌊x^1/3⌋) I don't understand that, because I don't know what is "n" and "x" is the result. sorry, I am starting in this. – Mr. Zacarias Satrustegui Feb 25 '16 at 22:37
In order to find, say, $M(132,8,5)$, you'll need $M(131, 7,1)$ and $M(124, 7, 2)$ and ... $M(7, 7, 5)$. For each of those you'll need some $M(x,6,n)$, and so on. It looks like a lot of computation, but it's easy with a computer. You could even do it on a spreadsheet. The trick is to remember $M(x,y,n)$ once you compute it, because you'll probably need it for a later step. – Robert Israel Feb 26 '16 at 01:14
How did you get this idea ? – Saikat Feb 26 '16 at 02:34
A standard technique. – Robert Israel Feb 26 '16 at 03:48
yes, I understand this answer but, what is the minimum between "n" and ⌊x1/3⌋? I ask that because in your answer you say M(x,m,⌊x1/3⌋) where M(x,m,n) so.. I think that ⌊x1/3⌋ = n. – Mr. Zacarias Satrustegui Feb 26 '16 at 06:47
You can't take $y > x^{1/3}$ because that would make $x - y^3$ negative. You can't take $y > n$ because $M(x,m,n)$ is supposed to have $a_m \le n$. – Robert Israel Feb 26 '16 at 08:14
ok, thanks so much, If I would like calculate for example taxicab(4,8,2) I only have to change "1/3" by "1/4" or it would has other technique? – Mr. Zacarias Satrustegui Feb 26 '16 at 18:34
And I think another thing. if the taxicab(2) is 1729 because 10^3 + 7^3 = 12^3 + 1^3 for this reasoning the taxicab(3) is 1730 adding up 1^3 both sides and it is not like that, because taxicab(3) is 4104. – Mr. Zacarias Satrustegui Feb 26 '16 at 19:04
Who said taxicab(3) is $4104$? Actually it should be $251 = 2^3 + 3^3 + 6^3 = 1^3 + 5^3 + 5^3$. – Robert Israel Feb 26 '16 at 21:26
$4104$ is not a sum of three positive cubes at all. – Robert Israel Feb 26 '16 at 21:30
take a look http://mathworld.wolfram.com/TaxicabNumber.html – Mr. Zacarias Satrustegui Feb 27 '16 at 07:08

How to calculate $\operatorname{taxicab}(3,8,2)$ (sum of 8 cubes in two different ways)

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