Is the following producing a list of all prime numbers without skipping in a consecutive order?

To add to Henry's answer: your claim that the first appearances of primes (other than 3) come in increasing order is also correct. More specifically, for all primes $p > 3$, the first occurrence of $p$ in the last column of the table is next to either $A = 2p-1$ or $A = 2p+1$. Proof: since $B = (A-1)(A+1)/6$ (and $p$ doesn't divide 6), $p$ divides $B$ if and only if it divides either $A-1$ or $A+1$. The first such odd numbers $A > 1$ with this property are $A = 2p \pm 1$. So $p$ will first appear next to either $A = 2p-1$ (if this isn't a multiple of 3) or $A = 2p+1$ (otherwise).

In effect you are asking whether there exists a prime $p$ which is not a factor of $\frac{A^2-1}{6}$ for any $A$ not divisible by $2$ or $3$

Any $A$ not divisible by $2$ or $3$ will be of the form $A=6n+1$ or $6n-1$ for some $n$.

• If $A=6n+1$ then $\frac{A^2-1}{6} = 2n(3n+1)$ which is divisible by $n$
• If $A=6n-1$ then $\frac{A^2-1}{6} = 2n(3n-1)$ which is also divisible by $n$

So just let $n=p$ and $A=6p\pm1$ (your choice): you will then have $\frac{A^2-1}{6}$ divisible by $p$.

For example for $p=2$, $A=11$ or $13$ works; for $p=3$, $A=17$ or $19$ works; for $p=5$, $A=29$ or $31$ works; for $p=7$, $A=41$ or $43$ works; for $p=11$, $A=65$ or $67$ works; and so on. There will always be such examples fitting this pattern on top of many other cases. And you do not need to restrict $n$ to be prime: for example for $n=4$, $A=23$ or $25$ leads to $\frac{A^2-1}{6}$ being divisible by $4$.

So no, there are no such primes.