Asymptoptic density of prime k-tuples The first Hardy-Littlewood conjecture concerns the asymptotic density of prime k-tuples. Assuming that the tuple $\{p, p+2m_1, \ldots, p+2m_k\}$, where all the elements are primes and $m_i$ for all $i$ is a positive integer, occurs infinitely often, then the number of such primes $p$ below a given positive integer $n$, denoted by $\pi_{m_1, \ldots, m_k}(n)$ is (taken from Wolfram):
$$
\pi_{m_1,\ldots, m_k}(n)∼C(m_1,\ldots,m_k)\int_2^n \frac{dt}{\log^{k+1}t}, 
$$
where the quantity $C(m_1,\ldots,m_k) $ is defined as:
$$
C(m_1,\ldots,m_k)=2^k \prod_{q} \frac{1-\frac{w(q;m_1,\ldots,m_k)}{q}}{(1-\frac 1 q)^{k+1}}, 
$$
and $w(q;m_1,\ldots,m_k)$ denotes the number of distinct residues of $0, m_1, \ldots, m_k \ \ (\text{mod} \ \ q)$.
However, I have seen in several places (here and here) other formulae that seem to suggest the existence of something along the lines of:
$$
\pi_{m_1,\ldots, m_k}(n)∼K\prod_{p>k}\frac{p^{k-1}(p-k)}{(p-1)^k}\int_2^n \frac{dt}{\log^{k+1}t},
$$
where the value of $K$ seems to be a bit mysterious to me. For instance, for the tuple $\{p,p+2,p+6\}$ we have $K=\frac 9 2$, for $\{p,p+2,p+6,p+8\}$ we have $K=\frac{27}{2}$, for $\{p,p+4,p+6,p+10\}$ we have $K=27$, $\ldots$.
I (naïvely) prefer the second formulation of $\pi_{m_1,\ldots, m_k}(n)$ since (it fits into a single line) has an elegant form and does not directly involve computing residues within a product. Unfortunately, I have not found any documentation anywhere that would give a formula for the value of $K$, and I am having trouble deriving the second formula from the first due to the residue calculation within the infinite product. Can anyone give a formula for $K$ or provide a demonstration of how one achieves the second formula from the first?
 A: I think there's some unfortunate confusion about the meaning of $k$ here. The MathWorld article about the $k$-tuple conjecture that you link to is inconsistent in that it speaks of $k$-tuples but defines $k$ such that the tuples are actually $(k+1)$-tuples; whereas the other two pages you link to use $k$ to mean what one would expect it to mean, the number of primes in the $k$-tuple.
I'm afraid the resulting confusion has infected your two expressions for the density. If we consistently use $k$ to denote the number of primes in the tuple, including the one with offset $0$, then I believe all exponents in the denominators should be $k$, both for the logarithm and for the prime-dependent factors. Then the two expressions can be transformed into each other by noting that
$$
\frac{p^{k-1}(p-k)}{(p-1)^k}=\frac{1-\frac kp}{\left(1-\frac1p\right)^k}
$$
and that $w(q;m_1,\ldots,m_k)=k$ for all but finitely many primes $q$, so that the factors in the products differ only for finitely many primes; the factor $K$ is there to account for these finitely many discrepancies, as well as the factor $2^k$.
Edit in response to the comment:
To see how to determine the factor $K$ in the second form from the first form, let's do it for two of the examples you gave. For $p$, $p+2$, $p+6$, we have $w(q;0,2,6)=3$ for $q\ge5$, but $w(3;0,2,6)=2$. Thus, the factor missing in the second form is
$$
2^2\cdot\frac{1-\frac23}{\left(1-\frac13\right)^3}=\frac92\;.
$$
For $p$, $p+4$, $p+6$, $p+10$, we have $w(q;0,4,6,10)=4$ for $q\ge7$, but $w(3;0,4,6,10)=2$ and $w(5;0,4,6,10)=3$. For some reason, they chose to start the product at $5$ and not at $7$, so in this case we have to not only account for the missing factors but also divide by the incorrect factor included for $p=5$, resulting in
$$
K=2^3\cdot\frac{1-\frac23}{\left(1-\frac13\right)^4}\cdot\frac{1-\frac35}{\left(1-\frac15\right)^4}\cdot\left(\frac{1-\frac45}{\left(1-\frac15\right)^4}\right)^{-1}=2^3\cdot\frac{1-\frac23}{\left(1-\frac13\right)^4}\cdot\frac{1-\frac35}{1-\frac45}=27\;.
$$
