# Combinations of four consecutive primes in the form $10n+1,10n+3,10n+7,10n+9$

Here $n$ is some natural number. For example, among the primes $< 1000$ I found four such combinations:

\begin{array}( 11 & 13 & 17 & 19 \\ 101 & 103 & 107 & 109 \\ 191 & 193 & 197 & 199 \\ 821 & 823 & 827 & 829 \end{array}

Using Mathematica I was able to move further, so the sequence $n_k$ starts with:

$$\{n_k\}=\{1,10,19,82,148,187,208,325,346,565,943,\dots\}$$

A question already exists on this topic, however there is not a lot of information there.

I would like to know if the sequence $n_k$ was studied before, and what can we tell about the distribution of $n_k$ among the natural numbers?

Distances $n_{k+1}-n_k$ seem to grow on average, but 'close' quadruples still exist even for large $n_k$, for example:

$$n_{872}=960055,~~~n_{873}=960058$$

The plot of all the distances for $n_k<10^6$ is provided below (there are $898$ of them):

As the author of the linked question stated, every $n_k$ has the form $3m+1$, so the distances are all divisible by $3$.

Found OEIS A007811 with some information

It is conjectured that the number of such $n_k$ up to $X$ is of the size (ignoring leading constants) $$\frac{X}{\log^4 X},$$ and similarly that the number of $k$-tuples up to $X$ in fixed, admissible configurations is of the size $$\frac{X}{\log^k X}.$$
Your tuple is $(10n+1, 10n+3, 10n+7, 10n+9)$. If you were to consider the $8$-tuple $(10n+1, 10n+3, 10n+7, 10n+9, 10n+91, 10n+93, 10n+97, 10n+99)$, (which I think is admissible but I didn't actually check), then it is conjectured that the number of such $8$-tuples up to $X$ is of the size $$\frac{X}{\log^8 X}.$$ Notice that this is actually two of your $4$-tuples separated by $90$. So conjecturally we believe there should be infinitely "smallish" gaps between $4$-tuples of your shape.
More distribution-style statements can be made along these lines. You'll get very far by looking up the prime $k$-tuple conjecture and studying its progress and results.
• What does 'fixed, admissible configurations' mean? What about something like $1000n+111,1000n+333,1000n+777,1000n+999$? – Yuriy S Mar 30 '16 at 19:01
• Right! The word "admissible" means that it doesn't trivially not have infinitely many primes. For instance, $n, n+1$ is not "admissible." – davidlowryduda Mar 30 '16 at 19:15
• How come the known result $\frac{X}{\log^k(X)}$ is such a poor estimate for $k>1$? For example, let $R_k = \frac{\text{estimate}}{\text{actual}}$ and $X=10^6$. Then, rounded off, we have, $$R_1 = \frac{72400}{78500} = 0.92\\ R_4 = \frac{27}{898} = 0.03$$ Only the first is a respectable prediction. Are there formulas for $k>1$ with better estimates? – Tito Piezas III Mar 30 '16 at 20:12
• @TitoPiezasIII, you are confusing numbers here. $10^6$ is the boundary for $n$, while $X$ would be around $10^7$ – Yuriy S Mar 30 '16 at 20:20
• @TitoPiezasIII, this function doesn't give approximate values for a number of quadruples, it describes how this number grows on average with $X$. And it seems to be doing a good enough job. – Yuriy S Mar 30 '16 at 21:13