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With the exception of $4$, every number located between twin primes is divisible by $6$.

This one is obvious, but are there any other properties that can be ascribed to such numbers?

A property may be ascribed either to each number individually or to the entire sequence.

For example, consider the amount of prime factors of such numbers:

  • Is there any known restriction on this amount per each number?
  • Is there any known restriction on this amount as a function of the sequence-index?

The context in which I am asking this question:

What attempts have been made towards proving that there are infinitely many pairs of twin primes, by proving that there are infinitely many numbers located between twin primes?

Bill Dubuque
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barak manos
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    FYI : Average of twin prime pairs (OEIS : A014574) – mathlove Apr 23 '15 at 12:23
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    @mathlove: There are some interesting properties there, and perhaps even more interesting is the fact that they have been determined relatively recently. Thanks. – barak manos Apr 23 '15 at 12:25
  • Maybe useful – Elaqqad Apr 23 '15 at 12:25
  • @Elaqqad: Thank you, but the answer to this question shows that it implies nothing more than these numbers being divisible by $6$ (which as I've mentioned, is obvious). Thanks anyway. – barak manos Apr 23 '15 at 12:27
  • @barak manos, just in case, I asked a related question about twin primes, it might give you another point of view, it is about twin primes in the vicinity of twin primes. http://math.stackexchange.com/questions/1005852/test-total-number-of-twin-primes-in-the-vicinity-of-twin-primes-how-can-i-calc – iadvd Apr 23 '15 at 13:12
  • Conjecturally, amongst the twin prime averages $6x$ there are infinitely many such that $x$ is a prime number that is coprime to $p_1p_2 \cdots p_n$ that was used to sieve (Eratosthenes) to prove that $6x$ is indeed a twin prime average. – Daniel Donnelly Feb 02 '25 at 05:04

2 Answers2

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About the sequence you're after, I would say that it is possibly more interesting to consider the quotients when they're divided by 6. The result is the OEIS sequence A002822.

I'm not sure whether this counts as "special", but there seems to be a way to associate bijectively A002822 to the twin prime pairs. I actually found a few pointers in other questions here, so you could simply check them. In Does proving the following statement equate to proving the twin prime conjecture? you can find a pointer to an arXiv paper from 2011, and in About a paper by Gold & Tucker (characterizing twin primes) you will find a link to a small, 20 years older article which also proves the same bijective relationship and derives a similar formula (even if it's hard to distinguish because of their $G(n)$ function).

My preferred formulation of the theorem is as follows:

Any twin prime pair greater than $(3;5)$ is of the form $(6z-1;6z+1)$ where $z\inℕ$ and satisfies following system of inequalities: $$6xy+5x+5y+4 \neq z$$ $$6xy+7x+5y+6 \neq z$$ $$6xy+7x+7y+8 \neq z$$ for all $(x,y) \in (\mathbb{N}\cup\{0\})^2$; and conversely, for any such integer $z$, the pair $(6z-1;6z+1)$ is a twin prime pair.

In response to your contextual question, you can see above that a few people did try to find a proof using this theorem. And there are regularly similar attempts posted on the arXiv. For the sake of citing an example, I'll mention this one: Twin Prime Sieve.

user3733558
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  • they are Also the only numbers not describable in one of the forms : $6kl+k+l, 6kl-k-l , 6kl+k-l$ for aome $k,l$ where both are positive integers ... – Roddy MacPhee Apr 17 '21 at 23:34
  • Rather than listing each possibility individually, it is easier to say that if $z=6ab\pm a\pm b$ for any $a,b\in \mathbb Z^+$, then $6z\pm 1$ does not represent a twin prime. – Keith Backman Jul 20 '24 at 14:32
  • This is great! I would never have imagined such a theorem. – MJD Jul 20 '24 at 15:09
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An interesting property of $z$ (in my opinion) is that when $6z\pm 1$ is a twin prime, then for all $a<z$, $z^2\not \equiv a^2 \bmod 6a\pm 1$

The proof is simple. If $6z\pm 1$ is a twin prime, then $36z^2-1$ is a semiprime not divisible by any proper factors other than the primes $(6z-1),(6z+1)$. But for any number of the form $6a\pm 1$, the product $36a^2-1$ is divisible by that number. Hence (with the condition that $a<z$) $$36z^2-1 \not \equiv 0 \bmod 6a\pm 1;\ 36a^2-1 \equiv 0 \bmod 6a \pm 1\Rightarrow \\ 36z^2-1 \not \equiv 36a^2-1 \bmod 6a\pm 1 \Rightarrow \\ 36z^2 \not \equiv 36a^2 \bmod 6a\pm 1 \Rightarrow \\ z^2\not \equiv a^2 \bmod 6a\pm 1$$ This result is a different way of describing the fact that if $6z\pm 1$ is a twin prime pair, then $z\ne 6xy\pm x\pm y$, but the connection between the two different statements is a bit convoluted.

J. W. Tanner
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Keith Backman
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