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Let us construct a simple (undirected) graph $T$ as follows:

$\quad$ Let the set of all primes be the vertex set of $T$. For each prime $p$, take the least prime $q>p$ such that $2(p+1)-q$ is prime (such a prime q should exist in view of Goldbach's conjecture), and then set an edge connecting $p$ and $q$.

Clearly the graph $T$ contains no circle. If it is connected then it is a tree.

QUESTION. Is the above graph $T$ a tree?

In Feb. 2013, I constructed the graph $T$ and conjectured that $T$ is indeed a tree. For example, the path connecting $2$ and $191$ is \begin{align*}2&\to 3\to 5\to 7\to 11\to 13\to 17\to 19\to 23\to 29\to 31\to 41, \\ &\to43\to 47\to 53\to 61\to 71\to 73\to 89\to 97\to 107\to 109 \\&\to 113\to 127\to 149\to 151\to 167\to 173\to 181\to 191. \end{align*}

Any ideas towards the solution of the question? Your comments are welcome!

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    $\begingroup$ Could something like this work: we first construct a graph by drawing an edge for all such primes instead of the minimal prime. Using the edge rule we then see two primes are connected by a path of length k iff there exists a solutions to a diophantine equation in k-1 variables whose coordinates are prime. Some conjectures about primes give a solution to this equation, so the graph is connected. Then your graph is a minimal spanning tree If we weight the edges by q. $\endgroup$
    – user156885
    Commented May 5, 2020 at 15:29
  • $\begingroup$ @zz7948 Can you please clarify on spanning tree part? How would repeatedly removing edges and leaving one edge with smallest weight gives still a connected graph? $\endgroup$ Commented May 11, 2020 at 15:24
  • $\begingroup$ Maybe there’s an issue with this approach. We can try using Prim’s algorithm en.m.wikipedia.org/wiki/Prim%27s_algorithm which says locally choosing the minimal weight gives a spanning tree $\endgroup$
    – user156885
    Commented May 11, 2020 at 15:40
  • $\begingroup$ I think the problem is this: The process of leaving one edge is not necessarily the same as Prim's algorithm. $\endgroup$ Commented May 11, 2020 at 17:09

2 Answers 2

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Not an answer, just a drawing of the tree including the OP's $2 \rightarrow 191$ path:


          PrimeTree


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    $\begingroup$ So if we call the trajectory starting with 2 the main branch, the question is whether the trajectory of any prime not on the main branch will eventually hit it. $\endgroup$
    – Wolfgang
    Commented May 3, 2020 at 15:45
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    $\begingroup$ @Wolfgang: Yes. For example, just going a bit further, $223 \rightarrow 251$ is a separate tree, which eventually gets hooked into the main branch. $\endgroup$ Commented May 3, 2020 at 18:08
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    $\begingroup$ Yes, eventually, as expected. I am also wondering: associate with each prime p the number b(p) of branchings the trajectory encounters before arriving in the main branch. So for the small primes above, it is either 0 or 1, but I guess that there should be bigger primes for which there are 2, 3, ... branching points. Or maybe not? Could you easily find some with your implementation? $\endgroup$
    – Wolfgang
    Commented May 3, 2020 at 20:28
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    $\begingroup$ And likewise, as there are numbers like 89 with 3 branches arriving, are there also others with 4 or more? $\endgroup$
    – Wolfgang
    Commented May 3, 2020 at 20:30
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    $\begingroup$ @ Wolfgang: You are asking great questions! I hope @ZhiWeiSun responds. I will think about implementation... [software will not let me @ two users] $\endgroup$ Commented May 3, 2020 at 21:09
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This does not give a complete answer. This provides a strategy for conditional approach.

Given that the nature of the problem is asking for $q$ and $2(p+1)-q$ being simultaneously primes, I think that the question can be approached conditionally (similar to zz7948).

A slight modification is shifting the focus to finding $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime, for given prime $q$. As suggested in zz7948's comment, we extend the edges to include all pairs $(p,q)$ with both $p$ and $2p+2-q$ are primes.

Let us assume the following version of prime $k$-tuple conjecture.

Conjecture 1

There is an absolute constant $C>0$ and a prime $q_0$ such that the number $N(q)$ of primes $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime satisfies $$ N(q)\geq C\frac q{\log^2 q}\geq 1, \ \ \mathrm{ if } \ q\geq q_0. $$

If Conjecture 1 is true, then any prime $q>q_0$ is connected to some smaller prime.

If we can show (computation) that all primes $q\leq q_0$ are connected, then all primes will be connected.

Once we obtain that all primes are connected, we now start to remove the edges to include only the smallest $q$ with $q>p$ and $2p+2-q$ are both primes.

Then, the issue is, whether or not Prim's algorithm of finding spanning tree of a connected graph gives the desired graph.

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  • $\begingroup$ Nice, the converse to this also holds? So this problem is equivalent to a statement about equations in primes? $\endgroup$
    – user156885
    Commented May 9, 2020 at 13:24
  • $\begingroup$ Not completely sure, but I think the statement about equations in primes is stronger statement than being able to connect all primes. $\endgroup$ Commented May 9, 2020 at 13:26
  • $\begingroup$ How about For the converse, by contrapositive If no solution to the equation with prime variables then we get an isolated point $\endgroup$
    – user156885
    Commented May 9, 2020 at 15:34
  • $\begingroup$ As my approach is finding smaller prime $p$ using the solution to the equation (Conjecture 1), the Nonexistence of such solution $p$ does not rule out the possibility of being connected to primes larger than $q$. For example, see 37, 67, 101, etc in Joseph's tree. $\endgroup$ Commented May 9, 2020 at 19:00
  • $\begingroup$ I see, thank you $\endgroup$
    – user156885
    Commented May 9, 2020 at 20:16

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