# Are there infinitely many Thâbit ibn Kurrah cousin primes?

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.

$7,11$

$19,23$

$41,47$

$379,383$

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.

• n=34, 64. The next n will be quite large (>500,000). – Michael Stocker Mar 29 '15 at 22:28

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.
• 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. – mick Mar 31 '15 at 22:11
• 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. – mick Mar 31 '15 at 22:25
• 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. – mick Apr 1 '15 at 19:09