Positive integers of the form $3 * 2^n - 1$ are called Thâbit ibn Kurrah numbers.

and if those numbers are prime they are called Thâbit ibn Kurrah primes.

Now if for a fixed positive integer $n$ , $3 * 2^n - 1$ and $3 * 2^n - 5$ are both prime then we have a Thâbit ibn Kurrah cousin prime.

Conjecture : there are infinitely many Thâbit ibn Kurrah cousin primes.





Are there more known ??

Although its not reasonable to expect a proof for the conjecture since prime twins and cousin primes are not proven to be infinite sequences , it might be possible to disproof the conjecture.

But I was not able to find a reason why the conjecture would fail.

I know that there are only a finite amount of prime twins of type $2^n + 3$,$2^n + 5$ because from working with mod $24$ you can show that at least one of the two must be divisible by $3,5,7,11$ or $13$. ( and actually $11$ can be dropped ).

But I was not able to find such an argument for the Thabit ibn Kurrah cousins.

  • 2
    $\begingroup$ n=34, 64. The next n will be quite large (>500,000). $\endgroup$ Mar 29, 2015 at 22:28

1 Answer 1


The following is a heuristic based on the Borel-Cantelli lemma that suggests that there will only be finitely many such primes. I believe that actually proving such a result would be very difficult.

Assuming independence, the probability that both $3*2^n-1$ and $3*2^n-5$ are prime is $$\approx \frac{C_1}{(\log(2^n))^2}=\frac{C_2}{n^2}$$ for some constants $C_1,C_2$. Now, the sum $$\sum_{n=1}^\infty \text{Pr}(E_n)=\sum_{n=1}^\infty \frac{C_2}{n^2}$$ converges, and so by the Borel-Cantelli lemma we expect that there are only finitely many such cousins.

  • $\begingroup$ Well , not so fast. :) From the first Hardy-Littlewood conjecture we know that $C_1$ is about $1.33$. Therefore $C_2 = 1.33 / ln(2)^2 $. From that it follows that the expected amount is $ C_2 \zeta(2) $ or equivalently $ (1.33 \pi^2) / (6 ln(2)^2) $. This gives us about $4.55$. Even if I would round this $4.55$ to $5$ , I see that there are 6 pairs known. (Last 2 given in the comment by Michael Stocker and primality checked by me). So this seems to contradict the Borel-Cantelli lemma. $\endgroup$
    – mick
    Mar 31, 2015 at 22:11
  • $\begingroup$ But despite that and the fact that I was aware of the Borel-Cantelli lemma I give you a +1 anyway :) $\endgroup$
    – mick
    Mar 31, 2015 at 22:15
  • $\begingroup$ On the other hand if we multiply (from above comments) the $4.55$ with $(1-2/3)^{-1} = 3$ we get $13.66$ which is larger then the $6$ pairs we found so far. That multiplication might be justified because neither term can ever be divisible by $3$. Well ... maybe. $\endgroup$
    – mick
    Mar 31, 2015 at 22:25
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    $\begingroup$ @mick, you should read how Borel-Cantelli works - what you are saying is not correct. $\endgroup$ Apr 1, 2015 at 17:35
  • $\begingroup$ Notice I use the distributive property , $ 3 ( P(a) + P(b) + P(c) ) = 3 P(a) + 3 P(b) + 3 P(c) $ So that multipliying the whole by $3$ is just as good as doing (multiplying) all probabilities seperately. So its not so naive as it looks. $\endgroup$
    – mick
    Apr 1, 2015 at 19:09

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