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Question edited in view of the comments below

By Yamada's paper we can conclude that if $n>e^{e^{36}}$ be an even number then it can always be written as the sum of a prime and a semi-prime.

My question is,

Question 1. Let $p_1,\ldots,p_k$ be the first consecutive odd primes and $n>\max(e^{e^{36}},p_k)$ be an even integer such that for all $i\in \{1,\ldots,k\}$, $n-p_i$ is composite. Is it possible that $n=p_r+p_sp_t$ (where $p_r,p_s,p_t\in\{p_1,\ldots,p_k\}$)?

Note that it is easy to show that if $n>p_k+p_k^2$ then the answer to the above question is "yes". So we only need to check whether the answer to the above question is "yes" for all even integers $n\le p_k+p_k^2$.

Following Gerhard Paseman's suggestion below I think a possible approach of attacking the problem will be to know more about the integers $n$ for which the hypothesis of the question holds. In this regard I have the following specific question in mind,

Question 2. Let $p_1,\ldots,p_k$ be the first consecutive odd primes and $n$ be an even integer such that for all $i\in \{1,\ldots,k\}$, $n-p_i$ is composite. What is known about the lower bound of $n$ (in terms of $p_1,\ldots,p_k$ satisfying such hypothesis?

Based upon my calculations, my best guess is that that $n> p_{k+1}$ but I couldn't prove it.

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    $\begingroup$ You mean that you look for triples which additionally satisfy $\max(q, r) >p$ abs ask whether they can be always found? $\endgroup$ – Fedor Petrov Sep 20 '18 at 14:47
  • $\begingroup$ @FedorPetrov: I am not sure I interpret your above comment correctly but it may happen that both $q=p_s$ and $r=p_t$ for some $s,t\in\{1,\ldots,k\}$ but we also have $\max(p_s,p_t)>p_j$. It's not what I am asking. $\max(q,r)$ needs to satisfy the inequality $\max(q,r)>p_k$. $\endgroup$ – user 170039 Sep 20 '18 at 14:54
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    $\begingroup$ I am confused with quantifiers. Is $k$ fixed? If yes, how? $\endgroup$ – Fedor Petrov Sep 20 '18 at 15:09
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    $\begingroup$ No it is not clear. What if $n=p_1+p_{k-1}p_{k-2}$? $\endgroup$ – Fedor Petrov Sep 20 '18 at 15:17
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    $\begingroup$ First, you changed the game on me, by adding the composite condition. Second, if such small n exists with n minus the primes composite, then my comment still holds. You need to ask the right question if you want the right comments. Gerhard "Can't Go Back To Edit" Paseman, 2018.09.20. $\endgroup$ – Gerhard Paseman Sep 20 '18 at 15:44

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