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Let $a_x$ be the number of primes $p<x$ starting a constellation $18,18,12$, i.e. $p,p+18,p+36,p+48$ are consecutive primes. Similarly, let $b_x$ and $c_x$ count primes starting a constellation $12,18,18$ and $18,12,18$ respectively. I think that there is strong reason to expect that as $x \rightarrow \infty$ we have $$ \frac{b_x}{a_x} \rightarrow 1$$ $$ \frac{c_x}{a_x} \rightarrow \frac32.$$

I will give my heuristic reasoning for expecting this and some limited computational support. Of course we don't know that the gap $12$ even occurs infinitely often.

Heuristic:

• For $p,p+18,p+36,p+48$ to contain no multiples of $5$ requires $p \bmod 5 \in \{1,3\}$

• For $p,p+12,p+30,p+48$ to contain no multiples of $5$ also requires $p \bmod 5 \in \{1,3\}$

• For $p,p+18,p+30,p+48$ to contain no multiples of $5$ requires $p \bmod 5 \in \{1,3,4\}$

The added condition that the four primes be consecutive seems to be equally restrictive in all three cases.

Computation: Here is a graph,

The top curve is $$\frac{c'_x}{a'_x}$$ and the one below is $$\frac{b'_x}{a'_x}$$ up to $x=3\cdot 10^7$ where $a'_x$ is the number of primes $p<x$ so that $p,p+18,p+36,p+48$ are all primes (but not required to be consecutive.)

Each curve actually has $60$ data points, those for $x$ a multiple of $5000,000.$

Let $a_x$ be the number of primes $p<x$ starting a constellation $18,18,12$, i.e. $p,p+18,p+36,p+48$ are consecutive primes. Similarly, let $b_x$ and $c_x$ count primes starting a constellation $12,18,18$ and $18,12,18$ respectively. I think that there is strong reason to expect that as $x \rightarrow \infty$ we have $$ \frac{b_x}{a_x} \rightarrow 1$$ $$ \frac{c_x}{a_x} \rightarrow \frac32.$$

I will give my heuristic reasoning for expecting this and some limited computational support. Of course we don't know that the gap $12$ even occurs infinitely often.

Heuristic:

• For $p,p+18,p+36,p+48$ to contain no multiples of $5$ requires $p \bmod 5 \in \{1,3\}$

• For $p,p+12,p+30,p+48$ to contain no multiples of $5$ also requires $p \bmod 5 \in \{1,3\}$

• For $p,p+18,p+30,p+48$ to contain no multiples of $5$ requires $p \bmod 5 \in \{1,3,4\}$

The added condition that the four primes be consecutive seems to be equally restrictive in all three cases.

Computation: Here is a graph,

The top curve is $$\frac{c'_x}{a'_x}$$ and the one below is $$\frac{b'_x}{a'_x}$$ up to $x=3\cdot 10^7$ where $a'_x$ is the number of primes $p<x$ so that $p,p+18,p+36,p+48$ are all primes (but not required to be consecutive.)

Each curve actually has $60$ data points, those for $x$ a multiple of $500,000.$

I will suggest below that the running total of constellation $18,18,12,12$ should stay pretty close to $\frac23$ of the running total for $12,18,12,18.$ That should be easy to check up to some limit.

Consider a particular constellation of $k$ integers considered to be roughly of magnitude $x$. For all large enough primes the members are in $k$ distinct congruence classes $\bmod p.$ But for small primes they are not. Knowing the data for all primes should give a conjectured density $\frac{c} {\log^kn} $ with $c$ depending on those small primes . (Well, that is the probability that the $k$ integers are all prime, you need to also consider the requirement that there are no intervening primes)

Before giving a specific example, I pause to say that there is no proof that this frequency is actually occurs, or even that the pattern happens infinitely often (if feasible at all), however there is strong heuristic and computational support.

Consider the constellation $12,18,12,18$ belonging to five integers $a,a+12,a+30,a+42,a+60.$ For $p>7,$ the chance that none is a multiple of $p$ is $\frac{p-5}{p}$ (provided $p \ll x$). And it is $\frac12,\frac23,\frac35,\frac37$ for $p=2,3,5,7.$ That should allow a computation of the (conjectured) probability (with consideration of the "no other primes" requirement.)

Now consider the constellation $18,18,12,12$ belonging to $a,a+18,a+36,a+48,a+60.$ Again the probability that none divides by a given small prime $p >7$ is $\frac{p-5}{p}$. And again $\frac12,\frac23,\frac37$ for $p=2,3,7.$ But it is $\frac25$ for $p=5$.

Accordingly, I would expect what I said at the top.

Computational result:

Considering consecutive primes $p_i,p_{i+1},p_{i+2},p_{i+3}$ with $10^{12}<p_i<p_{i+3}<10^{12}+10^6$ it turns out that the pattern of gaps $12,18,12$ occurs $14$ times while $18,18,12$ and $12,18,18$ occur $9$ and $7$ times respectively. The fact that the second two occur about half as often as the first (I suggest) is because $p,p+12,p+30,p+48$ being primes rules out $p \bmod 5 \in \{2\}$ while $p,p+18,p+36,p+48$ all being prime rules out $p \bmod 5 \in \{2,4\}$

Let $a_x$ be the number of primes $p<x$ starting a constellation $18,18,12$, i.e. $p,p+18,p+36,p+48$ are consecutive primes. Similarly, let $b_x$ and $c_x$ count primes starting a constellation $12,18,18$ and $18,12,18$ respectively. I think that there is strong reason to expect that as $x \rightarrow \infty$ we have $$ \frac{b_x}{a_x} \rightarrow 1$$ $$ \frac{c_x}{a_x} \rightarrow \frac32.$$

I will give my heuristic reasoning for expecting this and some limited computational support. Of course we don't know that the gap $12$ even occurs infinitely often.

Heuristic:

• For $p,p+18,p+36,p+48$ to contain no multiples of $5$ requires $p \bmod 5 \in \{1,3\}$

• For $p,p+12,p+30,p+48$ to contain no multiples of $5$ also requires $p \bmod 5 \in \{1,3\}$

• For $p,p+18,p+30,p+48$ to contain no multiples of $5$ requires $p \bmod 5 \in \{1,3,4\}$

The added condition that the four primes be consecutive seems to be equally restrictive in all three cases.

Computation: Here is a graph,

The top curve is $$\frac{c'_x}{a'_x}$$ and the one below is $$\frac{b'_x}{a'_x}$$ up to $x=3\cdot 10^7$ where $a'_x$ is the number of primes $p<x$ so that $p,p+18,p+36,p+48$ are all primes (but not required to be consecutive.)

Each curve actually has $60$ data points, those for $x$ a multiple of $5000,000.$

I will suggest below that the running total of constellation $18,18,12,12$ should stay pretty close to $\frac23$ of the running total for $12,18,12,18.$ That should be easy to check up to some limit.

Consider a particular constellation of $k$ integers considered to be roughly of magnitude $x$. For all large enough primes the members are in $k$ distinct congruence classes $\bmod p.$ But for small primes they are not. Knowing the data for all primes should give a conjectured density $\frac{c} {\log^kn} $ with $c$ depending on those small primes . (Well, that is the probability that the $k$ integers are all prime, you need to also consider the requirement that there are no intervening primes)

Before giving a specific example, I pause to say that there is no proof that this frequency is actually occurs, or even that the pattern happens infinitely often (if feasible at all), however there is strong heuristic and computational support.

Consider the constellation $12,18,12,18$ belonging to five integers $a,a+12,a+30,a+42,a+60.$ For $p>7,$ the chance that none is a multiple of $p$ is $\frac{p-5}{p}$ (provided $p \ll x$). And it is $\frac12,\frac23,\frac35,\frac37$ for $p=2,3,5,7.$ That should allow a computation of the (conjectured) probability (with consideration of the "no other primes" requirement.)

Now consider the constellation $18,18,12,12$ belonging to $a,a+18,a+36,a+48,a+60.$ Again the probability that none divides by a given small prime $p >7$ is $\frac{p-5}{p}$. And again $\frac12,\frac23,\frac37$ for $p=2,3,7.$ But it is $\frac25$ for $p=5$.

Accordingly, I would expect what I said at the top.

I will suggest below that the running total of constellation $18,18,12,12$ should stay pretty close to $\frac23$ of the running total for $12,18,12,18.$ That should be easy to check up to some limit.

Consider a particular constellation of $k$ integers considered to be roughly of magnitude $x$. For all large enough primes the members are in $k$ distinct congruence classes $\bmod p.$ But for small primes they are not. Knowing the data for all primes should give a conjectured density $\frac{c} {\log^kn} $ with $c$ depending on those small primes . (Well, that is the probability that the $k$ integers are all prime, you need to also consider the requirement that there are no intervening primes)

Before giving a specific example, I pause to say that there is no proof that this frequency is actually occurs, or even that the pattern happens infinitely often (if feasible at all), however there is strong heuristic and computational support.

Consider the constellation $12,18,12,18$ belonging to five integers $a,a+12,a+30,a+42,a+60.$ For $p>7,$ the chance that none is a multiple of $p$ is $\frac{p-5}{p}$ (provided $p \ll x$). And it is $\frac12,\frac23,\frac35,\frac37$ for $p=2,3,5,7.$ That should allow a computation of the (conjectured) probability (with consideration of the "no other primes" requirement.)

Now consider the constellation $18,18,12,12$ belonging to $a,a+18,a+36,a+48,a+60.$ Again the probability that none divides by a given small prime $p >7$ is $\frac{p-5}{p}$. And again $\frac12,\frac23,\frac37$ for $p=2,3,7.$ But it is $\frac25$ for $p=5$.

Accordingly, I would expect what I said at the top.

Computational result:

Considering consecutive primes $p_i,p_{i+1},p_{i+2},p_{i+3}$ with $10^{12}<p_i<p_{i+3}<10^{12}+10^6$ it turns out that the pattern of gaps $12,18,12$ occurs $14$ times while $18,18,12$ and $12,18,18$ occur $9$ and $7$ times respectively. The fact that the second two occur about half as often as the first (I suggest) is because $p,p+12,p+30,p+48$ being primes rules out $p \bmod 5 \in \{2\}$ while $p,p+18,p+36,p+48$ all being prime rules out $p \bmod 5 \in \{2,4\}$