Primes in the binomial transform of $ [1, 1, 2, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, ...]$. This question is related to this sequence A139482. A commentator gives the following formula for $a_m$
$$a_m = {3m^2-9m+10 \above 1.5pt 2}$$
I have that you should consider the sequence $b_n =3n+2$ and arrange the terms as follows:
$$B_3 = \begin{matrix}
    &&&&&65&\ldots         \\
    &&&&47&62&\ldots       \\
    &&&32&44&59&\ldots     \\
    &&20&29&41&56&\ldots   \\
    &11&17&26&38&53&\ldots \\
    5&8&14&23&35&50&\ldots 
    \end{matrix}$$
Then $a_m$ appears to be the the sequence on the uppermost slope of $B_3$ for $m>2$. In particular $a_m$  can be written as $b_n$ for some suitable $n$. Dirichlet tells us there are infinitely many primes in $B_3$. The obvious question is are there infinitely many primes on the uppermost slope? If that is too much to ask can we show that every column or row of $B_3$ has at least one prime number?
Update: I believe I have the differences of consecutive terms on the uppermost slope: $a_{m+1}-a_m=3(m+1)$. So that $a_{m+1} = 3(m+1)+a_m$. I also have that $(3,a_m) =1$ for every $m$. Would not Dirichlet's theorem tell us there are infinitely many primes of the form $3(m+1)+a_m$? <--- The answer to this question is NO!
 A: Your question about primes in the diagonal is whether or not the integer-valued polynomial
$$
\frac{3x^2 - 9x + 10}{2} = 3 \binom{x-1}{2} + 2
$$
is infinitely often prime.
It likely is. However, the question is similar to Landau's fourth problem of $x^2 + 1$, and is a special case of famous conjectures such as Schinzel's hypothesis H. To my knowledge, there is no quadratic integer-valued polynomial that has been proven to generate infinitely many prime values.
(A related question.)
A: Not quite the answer but I can show that the collection of all uppermost slopes of $B_a$ with $a$ odd will contain infinitely many primes. Without showing any work or providing a proof you can see by inspection that if we take the uppermost slopes of $B_a$ and place them column-wise into a matrix:
$$B_3 = \text{ }\begin{matrix}
    5&7&9&11&13&\ldots         \\
    11&17&23&29&35&\ldots      \\
    20&32&44&56&92&\ldots      \\
    32&52&72&92&112&\ldots     \\
    47&77&107&137&167&\ldots   \\
    65&107&149&191&233&\ldots  \\
    85&142&254&326&310&\ldots  \\
    \vdots&\vdots&\vdots&\vdots&\vdots&\ddots
    \end{matrix}$$
then the rows of $B$ are in linear arithmetic progression. Trivially the top row of $B$ has infinitely many primes. But surprisingly, at least to me, if $B(r)$ is the $r^{th}-row$ of $B$ and $$r \equiv{\{1,2\}} \text{ modulo }4$$ then $B(r)$ has infinitely many primes and thier density is already known and computed. The formulas below allow you to compute the $k^{th}$ entry of any row. 
        $$\begin{matrix}
  r=1\text{; }&2k+1\\
  r=2\text{; }&6k+5\\
  r=3\text{; }&12k+8\\
  r=4\text{; }&20k+12\\
  r=5\text{; }&30k+17\\
  r=6\text{; }&42k+23\\
  r=7\text{; }&56k+30\\
  r=8\text{; }&72k+38\\
  r=9\text{; }&90k+47\\
  r=10\text{; }&109k+57\\
  \vdots&\vdots&\\
  \end{matrix}$$
