This post is about “Taxi cab numbers” specifically the “Ramanujan-Hardy number”, 1729. This specific taxi cab number is so-called because it is the smallest positive number that can be written as a sum of two cubes in two ways.

The real definition of a Taxi cab number(Wiki) T(n) is “a number that can be expressed as the sum of cubes in \(n\) distinct ways”. So, given that definition, 1729 is T(2).

Worth mentioning is that according to *Fermat’s last theorem*, \(x^n+y^n\neq z^n\) for each \(n\geq 3\). This means that \(\sqrt[n]{x^n+y^n}\) is a non-integer. If the theorem was false, the taxi cab numbers would not be so special, there would be lots of them. Although it seems as though the definition of a taxi cab number is restrictive, there are many solutions of which the sum of two cubes are only expressed in exactly **one** way for instance, as opposed to three ways of which there are very few, if any (that I know of).

The origins of the taxi cab number, 1729, comes from the Indian mathematician Srinivasa Ramanujan and Godfrey H. Hardy. As Hardy recalls:

I remember once going to see him when he (Ramanujan) was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. “No”, he replied, “it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways.”

Ok, so we can do this in a very simple way: We wish to find \(a^3 + b^3 = 1729\), where \(a,b\) are integers. We can do this with two nested for-loops and try all a and b etc. Unfortunately, this solution is inefficient, we can do better. First of all, we don’t know the number 1729, so we have to find it using a loop. Instead of trying all \(a\) and \(b\), we can use the fact that \(a^3 = 1729-b^3\). So, we can try for instance **b=1** and we get 1729-1, so **a** must be an integer and it should have the value **1728^(1/3)**.

We can do this for each **b** from 1 to any number **num** and just check which of these **b** giving us an integer solution to **a**.

In Matlab this can be achieved using a vector

**b = 1:(num)^(1/3)**

and

**a = (num-b.^3).^(1/3)**

We only need to check **b** from **1** to **num^(1/3)** because by inspecting the last element of b we have **(num-(num^1/3)^3)** which is **0** (for larger **b** we get complex answers).

How do we know for which **a** we have an integer expression?

We need to be careful, for instance

**1728^(1/3) == 12^3** returns **0**

We have that

**1728^(1/3)** is **11.999999999999998** in Matlab.

This is because of the machine accuracy for numbers. In Matlab they are by default double precision, which means we have an accuracy of ~15 decimal places, so we need to round off a little.

To be on the safe side, we round the numbers in **b** to 10 decimal places (**digits=10**) and compare with the integers, I store these in a vector **cands**.

**cands = round(b) == round((10^digits*b))/(10^digits)**

**cands** is now a binary vector. To get the candidates we write

**terms = find(cands)**

To make sure we only find **1729**, we add the condition

**if sum(cands~=0)==num_sums**

The number of terms in **cands** are (ways to sum)\(^2\), because we have two candidates for each.

We get

**cands = 1 0 0 0 0 0 0 0 1 1 0 1**

**terms = find(cands) ans = 1 9 10 12**

The solutions are thus \(1^3 + 12^3\) and \(9^3 + 10^3\). In order to print them, I arranged the answers using **reshape** and took the outer values of the vector **terms** and work inwards

**tmp = reshape(terms, 2, length(terms)/2); tmp(2,:) = tmp(2, end:-1:1)**

Please note: We will always have an even number of elements in ‘terms’.

For this run I compute 1729, T(2), and then I continue to find the larger sums for fun. These numbers are not taxi cab numbers, because they are larger than 1729.

1^3 + 12^3 = 1729

10^3 + 9^3 = 1729

**2^3 + 16^3 = 4104 15^3 + 9^3 = 4104**

**2^3 + 24^3 = 13832 20^3 + 18^3 = 13832**

**…**

**9^3 + 58^3 = 195841 57^3 + 22^3 = 195841**

The code takes about 5 seconds to run with this example.

S. Ramanujan war ein indischer Mathematiker, der für sein Gespür für Zahlen berühmt war. So existiert folgende Anekdote über ihn: Einmal war der englische Mathematiker G. H. Hardy auf dem Weg um Ramanujan im Krankenhaus zu besuchen. Das Taxi, in dem er saß um zum Krankenhaus zu gelangen, hatte die Nummer 1729. Bei Ramanujan angekommen, bezeichnete Hardy die Nummer als “rather dull number” was so viel “eher langweilige Zahl” bedeutet. Ramanujan widersprach ihm und sagte, dass diese Nummer eine sehr interessante Zahl sei, da sie die kleinste Zahl ist, die als Summe von zwei kubischen Zahlen auf zwei unterschiedlichen Wegen ausgedrückt werden kann.

Schreiben Sie ein Programm, das zu einer Zahl n ? N alle Zahlen z ? n ausgibt, die auf mindestens zwei Arten als Summe von zwei kubischen Zahlen ausgedrückt werden können. D.h. finden Sie alle Zahlen z, für die gilt: ?a 6 = b 6 = c 6 = d : z = a 3 + b 3 = c 3 + d 3 , a, b, c, d ? N Verwenden Sie dazu vier verschachtelte for-Schleifen.

I could not have said it better in German myself.