I have two matrices, 


$$A= \left[ \begin{matrix} 5 & 10 & 15 & \cdots \\ 17 & 28 & 39 & \cdots \\ 35 & 52 & 69 & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{matrix} \right]$$


$$B= \left[ \begin{matrix} 6 & 13 & 20 & \cdots \\ 18 & 31 & 44 & \cdots \\ 36 & 55 & 74 & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{matrix} \right]$$
Where
$$
\begin{align\*} 
a\_{i,j}& =3i^2+(6j−3)i−j\\ 
b\_{i,j}& =3i^2+(6j−3)i+(j−1) 
\end{align\*}
$$
As Gerry Myerson commented, the question is: for all positive integers $n$, if $n$ is the sum of a number from $A$ and a number from $B$ (or both from the same set), then $n$ is also the sum of two positive integers, both of which are missing from $A$ and from $B$. 
Does there some related reference or proof of the statement above?



Sum elements in two dimensional arithmetic progression
It seems highly likely that the answer is yes and also quite possible that that it is difficult or even practically impossible to prove.


One could consider the question: can **every** $n \gt 1$ be written as a sum of two numbers missing from $A$ and $B?$ It isn't much different because every $n \gt 38$ can be written $n=5j+m$ for $m \in \{6,13,20,27,34\}$ i.e. $n=a\_{1,j}+b\_{1,t}$ with $t\in\{1,2,3,4,5\}.$


Row $i$ of $A$ is the arithmetic progression $m\_ij+r\_i$ for $(m\_i,r\_i)=(6i-1,6\binom{i}2).$


Row $i$ of $B$ is the arithmetic progression $m\_ij+r\_i$ for $(m\_i,r\_i)=(6i+1,6\binom{i}2-1).$


**Aside** A fact I won't prove, because I don't use it, is that many rows are sub-sequences of earlier rows from one matrix or the other. For example every row with $m\_i$ a multiple of $5$ (which happens $\frac15$ of the time) has $r\_i$ a multple of $5$, so that row is contained in row $1$ of $A.$ Another example: Row $13$ of $A$ is $77j+468$ which is a sub-sequence of both row $2$ of $A:$ $11j+6$ and row $1$ of $B:$ $7j-1.$ More specifically: If $m\_i \mid m\_I$ then $r\_I\equiv r\_i \bmod m\_i,$ i.e. row $I$ is a sub-sequence of row $i.$ This is true when both are from $A$ or both are from $B$ or one is from each.


So let $C$ be the numbers missing from both $A$ and $B.$ The first $50$ members are 


$F=\{1, 2, 3, 4, 7, 8, 9, 11, 12, 14, 16, 19, 21, 22, 23, 24, 26, 29, 32, 33, 37, 38, 42, 43, 46, 47,\\ 49, 51, 53, 54, 56, 58, 63, 64, 66, 67, 68, 71, 73, 77, 78, 79, 81, 84, 87, 88, 91, 92, 98, 99\}$ 


So the density in the first $100$ integers is $0.5.$ The density decreases as one goes further out , in the vicinity of $1,000,000$ the density is about $0.18.$ 


Consider an $n$ with $200 \lt n \lt 1000000.$ Then we have 50 potential sums $n=c+(n-c)$ with $c \in F.$ If we assume that the members of $C$ are random then the chance all those sums would fail is less than $0.82^{50} \lt 0.005\%.$ So we might expect less than $50$ failures with $n \leq 1000000.$ And for each such failure we would have many more potential sums to consider. Of course we do not know that they behave as if they were random, but there are $10$ numbers less than $1000000$ which can not be written as $n=c\_i+(n-c\_i)$ with $i \leq 50.$ Of them $5$ can be so express with $i=51$ and the others with $i=52,54,55,56,62.$


The density does go to $0$ but, as seen above, rather slowly. This can be accounted for simply by virtue of the progressions $mj+r$ with $m$ prime. If from some point on the density was $0$ that would prove that the answer is no. But it does not look like that happens. A conjecture of Erdos and Selfridge is that no finite collection of arithmetic progressions $mj+r$ with all the $m$ odd and distinct can cover all the integers (from some point on).


If an infinite subset $D \subset C$ could be identified with every integer (past $1$ ) being a sum of two members, that would prove that the answer was yes. That seems unlikely.


