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According to the Wikipedia's Prime Triplet article,

a prime triplet is a set of three prime numbers in which the smallest and largest of the three differ by $6$. In particular, the sets must have the form $(p, p + 2, p + 6)$ or $(p, p + 4, p + 6)$.

And with respect to this prime triplet's definition, Wikipedia gave a reference from The Prime Glossary: prime triple, which states that,

A prime triple is three consecutive primes, such that the first and the last differ by six.

Also I found on the Prime Constellation article of The Prime Glossary that,

Prime constellations of length $k$ are the shortest admissible $k$-tuples of primes. That is, a $k$-tuple is admissible unless there is a prime $q ≤ k$, which always divides the product of the terms. ...

Suppose we want to find prime constellations of length two. The pattern $(p,$ $p+1)$ is not admissible, because either $p$ or $p+1$ would be even, so one of these would have to be the prime $2$. The pattern $(p, p+2)$ is admissible (both $p$ and $p+2$ can be odd), so prime constellations of length two fit the pattern $(p, p+2)$. ... Of course, we call primes that fit this pattern twin primes.

Now let us find a prime constellation of length three. The primes in our pattern must differ by at least two $($so we are not forcing one of them to be two$)$. So we might try $(p, p+2, p+4)$. But one of these three must be divisible by $3$, so this also is not admissible. Finally, we consider

  • $(p, p+2, p+6)$ and
  • $(p, p+4, p+6)$.

Both of these are admissible, so all prime constellations of length three have one of these forms. $($Examples: $(5,7,11),$ $(7,11,13),$ $(11,13,17),$ $(13,17,19)$ and $(17,19,23)$.$)$ These are prime triples.

After investigating a little bit, I've found on the prime $k$-tuplet article that,

$4$-tuples have the form $\{p,$ $p+2,$ $p+6,$ $p+8\}$. There is a pair of twin primes in every prime $3$-tuple, and a prime $3$-tuple in every prime $4$-tuple $($but not prime $k$-tuple in every prime $(k+1)$-tuple, $7$-tuples do not include $6$-tuples$)$. So some authors use prime $k$-tuplet to mean a prime k-tuple which is not part of a prime $(k+1)$-tuple. They would similarly distinguish prime triplet from prime triple and prime quadruplet from prime quadruple.

Moreover, according to the G. H. Hardy's An Introduction to the Theory of Numbers (chapter I, page 6, sixth edition),

... There are $1{,}224$ such pairs $(p, p + 2)$ below $100{,}000,$ and $8{,}169$ below $1{,}000{,}000$. The evidence, when examined in detail, appears to justify the conjecture

There are infinitely many prime-pairs $(p, p + 2)$.

It is indeed reasonable to conjecture more. The numbers $p,$ $p + 2,$ $p + 4$ cannot all be prime, since one of them must be divisible by $3$; but there is no obvious reason why $p,$ $p + 2,$ $p + 6$ should not all be prime, and the evidence indicates that such prime-triplets also persist indefinitely. Similarly, it appears that triplets $(p,$ $p + 4,$ $p + 6)$ persist indefinitely. We are therefore led to the conjecture

There are infinitely many prime-triplets of the types $(p,$ $p + 2,$ $p + 6)$ and $(p,$ $p + 4,$ $p + 6)$.

Such conjectures, with larger sets of primes, may be multiplied, but their proof or disproof is at present beyond the resources of mathematics.

One important thing to notice here is that, unlike the mentioned articles above, Hardy didn't say that, prime-triplets must have the form $(p,$ $p + 2,$ $p + 6)$ or $(p,$ $p + 4,$ $p + 6)$. But as far as I know, he also didn't say that, $(p,$ $p + 2,$ $p + 4)$ is also a type of prime triplet or prime triple, did he?

But I have seen a couple of Mathematics Stack Exchange questions (for instance, MSEQ-2055623, MSEQ-549458, MSEQ-1653536, etc.), where the statements are like this:

there are no prime triplet other than $3,$ $5,$ $7$ .

There are some variations also. For example, the statement on the first quotation of the MSEQ-522585:

the only prime triple $p,$ $p + 2,$ $p + 4$ is the triple $3,$ $5,$ $7$.

In a nutshell, I am confused about the definition of Prime Triplet and Prime Triple. Any genuine reference(s) or citation(s) would be appreciated.

Aleph-null
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    Words and phrases can be used for multiple different things. It should be easy to see that the only triple of the form $(a,a+2,a+4)$ such that all three entries are prime is $(3,5,7)$ (since by simple modular arithmetic arguments one can see that one of $a,a+2,a+4$ must be divisible by three and the only way to have that be true and these all be prime is to have $a=3$ itself). Since all triples of primes of this category $(a,a+2,a+4)$ are so easily understood, we can move on to discussing triples of primes of other forms... in particular $(a,a+2,a+6)$ or $(a,a+4,a+6)$. – JMoravitz Feb 01 '22 at 15:26
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    Why we'd choose to look at $(a,a+2,a+6)$ or $(a,a+4,a+6)$? Because it is of interest to see primes so closely bunched together and these categories describe the next most closely bunched triples of primes beyond $(3,5,7)$. As for the naming scheme... "triple" or "triplet" is merely what you call an ordered collection of three numbers... – JMoravitz Feb 01 '22 at 15:28

1 Answers1

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The crucial condition to create interesting entries for tuples is that there is no forced prime factor of one of the entries. A tuple satisfying this criterion is called "admissable".

$(p,p+2,p+4)$ is not admissable since one of the numbers must be divisible by $3$ , but $(p,p+2,p+6)$ is admissable since there is no prime necessarily dividing one of the entries. The difference between the largest and smallest entry is called the diameter.

If no element can be added within the original span of the tuple without destroying the admissibility , then the tuple is called a prime constellation. So, $(p,p+6)$ is no prime constellation because we can either add $p+2$ or $p+4$ (but not both).

The conjecture is now that every admissable tuple produces infinite many prime tuples and every prime constellation therefore produces infinite many so-called "prime-clusters". This is a special case of the Schinzel hypothesis.

Peter
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