Admissible k-tuples and primorials

Question

Let $ (a_{1},\cdots,a_{k}) $ an admissible $ k $ -tuple and $ P_{k} $ the product of the first $ k $ primes. Do we have a conjectural expression for the number of positive integers $ n $ not exceeding $ x $ such that $ (n+a_{1},\cdots,n+a_{k})\in\mathbb{P}^{k}\Longrightarrow(n+P_{k}+a_{1},\cdots, n+P_{k}+a_{k})\in\mathbb{P}^{k} $ in function of $ x $ or an inconditional upper bound thereof ?

Answer 1

Isn't this just the question of the number (up to $x$) of prime tuples corresponding to the admissible tuple $A \cup A+P_k$ (where $A=\{a_1,\cdots,a_k\}$ )?

Note that $A \cup A+P_k$ could have as few as $k+1$ and as many as $2k$ elements. The first Hardy-Littlewood conjecture gives a formula for that. I am quite fond of this article by Polya discussing the heuristics supporting a simple case. Replacing $P_k$ by some multiple $Q$ of $P_k$ with the same radical should not affect the density. For that matter, the question makes as much sense for $A \cup A+Q$ with $|A|$ perhaps greater than the number of primes dividing $Q.$

For $A=\{a_1,a_2,\cdots,a_k\}$ a set of integers and prime $q,$ let $w(q,A)$ be the number of distinct residues $\bmod q$ in $A.$ We can (and will) assume that $a_1=0$ and that the $a_i$ are increasing. Call $A$ a pattern and consider the constant $$C_A=\prod_q\frac{1-\frac{w(q,A)}{q}}{(1-\frac1q)^{k}}$$

where the product is over all primes. Note that $w(q,A)=k$ provided that $q \gt a_k$ so there are only finitely many numerators affected by the actual choice of $A.$ Note also that $C_A=0$ in the event that $w(A,q)=q$ for some prime.

Let $\pi_A(x)$ denote the number of primes $p \lt x$ such that the members of $p+A=\{p+a_1,p+a_2,\cdots,p+a_k\}$ are all prime. Then it is conjectured that, asymptotically, $\pi_A(x)=C_A\int_2^x\frac{dt}{\ln^kt}.$ (I believe I am quoting that value correctly.)

$A$ is called admissible if $C_A>0.$ The conjecture is rather trivially true in case $A$ is not admissible. For $A$ admissible it is widely believed but even the weaker conjecture that the number of such $p$ is infinite is essentially open (It is known that not every pattern occurs finitely often.)

Given a set $A$ and an integer $Q$ let $A+Q$ be the pattern $\{a+Q\mid a\in A\}.$ The set of $p\lt x$ with $p+A$ all prime such that also $p+(A+Q)$ are all prime is simply the set of $p \lt x$ such that $p+\big(A \cup (A+q)\big)$ are all prime whose size $\pi_{A \cup (A+Q)}(x)$ has a conjectured (asymptotic) value. Note that this depends entirely on $A$ and the set of prime divisors of $Q.$

One might ask for conditions on $A$ and $Q$ such that $A \cup (A+Q)$ is admissible provided that $A$ is. In the case that $Q=P_9=2\cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23$ or any other integer with these prime divisors, it is sufficient that $|A| \leq 9$ but also that $|A| \leq 14:$ For prime $q \leq 23,\ \ w(q,A \cup (A+Q))=w(A,q) \lt q$ and for prime $q \gt 23$ we have $q \geq 29$ so $w(q,A \cup (A+Q)) \leq |A \cup (A+Q)| \leq 28 \lt q.$

There are stronger conjectures as well. There is a second Hardy-Littlewood conjecture Which seems plausible but is now doubted because it can't be true if the first one is.