Can a prime number be bigger than the sum of adding the previous twin primes (other than 13)?

Question

A simple heuristic of the first million primes shows that no prime number can be bigger than the sum of adding the previous twin primes.

Massive update: @mathlove made a comment that leaves me completely embarrassed. $13 > 7 + 5$ I don’t know how I missed it and I deeply apologize to everyone!

I ask anyone qualified to suggest any edits for the question.

$7 < 5 + 3$

$11 < 7 + 5$

$17 < 11 + 13$

$23 < 17 + 19$

At larger numbers:

$4886639 < 4886489 + 4886491$

$5389451 < 5388869 + 5388871$

$3155597 < 3154757 + 3154759$

I assume that if it could be proved, it would prove the twin prime conjecture of whether twin primes exist forever.

So I am not exactly seeking for a proof, but rather for possible explanations or references for why it is assumed true (or not)?

Also as the list grows, there seems to be a range for how small or big can a prime be in comparison to the sum of adding the previous twin primes.

As the list grows, a prime is usually never bigger or smaller than slightly above $50\%$ of the sum of the previous twin primes. Any references for such a range will be appreciated too.

*Update: When mentioning "the previous twin primes", I am implying to: $(107, 109), 113, 127, 131, (137, 139)$.

$131 < 107 +109$

Answer 1

Let's modify the OP's observation as follows:

Let $(p,p+2)$ and $(q,q+2)$ be consecutive pairs of twin primes, e.g., $(107,109)$ and $(137,139)$. Then (conjecturally) $q\lt2p+2$.

This is, essentially, a Bertrand's Postulate for twin primes, and it's not hard to confirm that it holds for entries at the outset for the sequence $3,5,11,17,29,41,59,\ldots$. The basic explanation can be found in the heuristic twin-prime "theorem" $\pi_2(x)\sim2C_2x/(\ln x)^2$, although arguing that it (conjecturally) holds for all twin primes, not just for ones that are sufficiently large -- i.e., giving a (heuristic) twin-prime analog of the proof of Bertrand's Postulate -- seems problematic.

Remark: Amusingly, the modification proposed above of the OP's observation is technically agnostic with regard to the twin prime conjecture. Indeed, it would be easiest, in principle, to prove (or disprove) if there were an identifiable last pair of twin primes.

Answer 2

Futher to my comments ...

For any $3$ consecutive primes (regardless of if they contain twin primes or not) $p_{n},p_{n+1},p_{n+2}$, it's true that $$\frac{p_{n+1}+p_{n}}{2}<p_{n+2}<p_{n+1}+p_{n}$$ In fact, $\forall p_{n+k}\geq p_{n+2}$ we have $p_{n+k} > p_{n+1}$ and $p_{n+k} > p_{n}$, thus $$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}$$ For a fixed $k$, we can show that $$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}=\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+1}}=1 \tag{1}$$ This means that for large enough $n$'s we have $$p_{n+k}<2p_{n} \text{ and } p_{n+k}<2p_{n+1}$$ Applying AM-GM yields $$p_{n+1}+p_{n}\geq 2\sqrt{p_{n+1}\cdot p_{n}}=\sqrt{2p_{n+1}\cdot 2p_{n}}> \sqrt{p_{n+k}^2}=p_{n+k}$$ for large enough $n$'s, or $$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}<p_{n+1}+p_{n} \tag{2}$$ The range of the fixed $k$ depends on when $(1)$ gets less than $2$.

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A few words about $(1)$, it is true because of this limit $$\lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$ then $$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}= \lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+k-1}}\cdot \lim\limits_{n\to\infty}\frac{p_{n+k-1}}{p_{n+k-2}}\cdot ...\cdot \lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$ again, for a fixed $k$!