While considering the relationship between $6n-1$ (OEIS A002476) and generalized cuban primes(OEIS A007645) I came across something I thought was interesting:

Seems like the description of generalized cuban primes as: "primes of the form $x^2 + xy + y^2$; or primes of the form $x^2 + 3y^2$; or primes equal to 0 or 1 mod 3." is a bit heavy considering:

$$f(x)=\left \lfloor 3x \over 2 \right \rfloor$$Where: $ \left \lfloor x \right \rfloor = floor(x)$

When $x > 3$ and $f(x)$ is prime, it is a generalized cuban prime.

Confirmed terms $n = 1 \to 1000$ against OEIS A007645

Seems a bit more elegant?



$$ f(x) = {3x \over 2}-{1\over 2} = \left \lfloor 3x \over 2 \right \rfloor$$

  • 1
    $\begingroup$ oeis.org/A002476 currently says "primes of the form $6m + 1$". Is that what OP meant? $\endgroup$
    – Mason
    Jul 19, 2018 at 23:08

1 Answer 1


"When $f(x)$ is prime, it is a Cuban prime."

This doesn't seem to be true. For example, take $x=2$. Then $f(x)= \lfloor 3 \rfloor=3$, which is prime. On the other hand, $3$ is not a Cuban prime, as the smallest Cuban prime is $7$.

It appears, by your references, that you are instead considering generalized Cuban primes. That is, whereas Cuban primes are simply those arising as the difference of successive cubes, generalized Cuban primes take the form $$p=(x^3-y^3)/(x-y),$$ i.e. as a difference of powers, divided by one trivial factor (without this division, we would find no generalized Cuban primes that are not Cuban themselves).

Now turning to your problem, suppose that $f(x)$ is prime. If $x=2k$ is even, then $$f(2k)= \lfloor 3(2k)/2 \rfloor = \lfloor 3k \rfloor = 3k,$$ which is composite unless $k=1$ (i.e. $x=2$ and $f(x)=3$). Otherwise, suppose that $x=2k+1$ is odd. Then $$f(2k+1)= \lfloor (3(2k+1)/2\rfloor=\lfloor 3k+3/2 \rfloor =3k+1.$$ Thus $f(x)$ prime implies $p \equiv 0,1 \!\!\mod 3$, so that prime values of $f$ are in fact (generalized) Cuban primes. Of course, since $f$ surjects onto the set of integers congruent to $1$ modulo $3$, the two sets you consider are in fact equal.

Why, then, is your definition non-standard?

  1. The floor function implicitly introduces cases, in situations that may not require casework.
  2. The condition $p \equiv 0,1 \!\!\! \mod 3$ makes the properties of $p$ very clear.
  3. The other two definitions for (generalized) Cuban primes are given in terms of binary quadratic forms. These have widespread use in number theory, and it is useful (yet sometimes non-obvious) to know when a set is related to such functions.

In short, while your definition is elegant (but perhaps less so than $p \equiv 0,1 \!\!\mod 3$), it is far from transparent, which is often more useful in mathematics.

  • $\begingroup$ Forgot to specify that $x$ should be and odd number. $\endgroup$
    – JohnWO
    May 14, 2013 at 3:38
  • $\begingroup$ That is not correct either; See revised question. $\endgroup$
    – JohnWO
    May 14, 2013 at 3:44
  • $\begingroup$ @JohnWO I'm not sure what you've changed with your edit. Is there something in my answer that you disagree with? $\endgroup$
    – awwalker
    May 14, 2013 at 3:54
  • $\begingroup$ Line 7 was edited. It is indeed generalized Cuban primes to which I am referring to. Also, an edit at the end is an example of another way to express $f(x)$ $\endgroup$
    – JohnWO
    May 14, 2013 at 4:02

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