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Accoring to this question and a linked duplicate, it's been verified empirically up to some number that all twin prime averages greater than six, are the sum of two smaller twin prime averages.

I was curious whether or not these formed a tree (with twin prime pairs at the vertices and the sum relation the edges). In order for these to form a tree, for every twin prime average, we would only be able to find one pair of smaller twin prime averages which summed to it.

Can you find a counterexample, namely any twin prime average which is a sum of two smaller twin prime averages, more than one way?

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    What have you tried? For example $60=30+30=42+18$ or $72=60+12=42+30$ – Henry Jun 23 '23 at 15:49
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    I like your style. Just because the tree is not uniquely defineable in a natural way, don't let that stop you from finding the missing proof! Perhaps many trees is a blessing. And thanks for linking to my post! :) – Daniel Donnelly Jun 26 '23 at 01:01
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    Here's another idea. Form a Poset (partially ordered set) instead, it's kind of like a bunch of trees unioned together because they share nodes. We say $x \leqslant y$ in the poset $P$ if and only if there exists another element $z \in P$ such that $x + z = y$. – Daniel Donnelly Jun 26 '23 at 01:09

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Number experiment finds a lot of counterexamples.

The smallest one is $71 + 73 = (11 + 13) + (59 + 61) = (29 + 31) + (41 + 43)$.

If we allow larger numbers, there can be a lot of ways: $$1877 + 1879 =\\ (5 + 7) + (1871 + 1873) =\\ (179 + 181) + (1697 + 1699) =\\ (269 + 271) + (1607 + 1609) =\\ (599 + 601) + (1277 + 1279) =\\ (827 + 829) + (1049 + 1051) =\\ (857 + 859) + (1019 + 1021)$$

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