An infinite family of prime-free quadratic sequences from the transposed triangular grid

Background

The triangular grid places integer $T(r-1)+c$ at row $r$, column $c$, where $T(n)=n(n+1)/2$. Transposing this grid, reading along SE diagonals of the triangular grid as columns, yields a new array whose column $d$ has values $$f_d(n) = T(n+d-2)+n = \frac{n^2+(2d-1)n+(d-1)(d-2)/2 + ... }{...}$$ More precisely, column $d$ of the transposed triangular grid has the quadratic polynomial $$f_d(n) = \tfrac{1}{2}n^2 + \left(d - \tfrac{1}{2}\right)n + \tfrac{(d-1)(d-2)}{2}+1,\quad n=1,2,3,\ldots$$

Example of the transposed triangular grid

The transposed triangular grid (first 12 columns, 12 rows), where composite values are marked *:

n\d |    1    2    3    4    5    6    7    8    9   10   11   12
----|------------------------------------------------------------
  1 |   *1    2   *4    7   11  *16  *22   29   37  *46  *56   67
  2 |    3    5   *8  *12   17   23  *30  *38   47  *57  *68  *80
  3 |   *6   *9   13  *18  *24   31  *39  *48  *58  *69  *81  *94
  4 |  *10  *14   19  *25  *32  *40  *49   59  *70  *82  *95  109
  5 |  *15  *20  *26  *33   41  *50  *60   71   83  *96 *110 *125
  6 |  *21  *27  *34  *42  *51   61  *72  *84   97 *111 *126 *142
  7 |  *28  *35   43  *52  *62   73  *85  *98 *112  127 *143 *160
  8 |  *36  *44   53  *63  *74  *86  *99  113 *128 *144 *161  179
  9 |  *45  *54  *64  *75  *87 *100 *114 *129 *145 *162 *180  199
 10 |  *55  *65  *76  *88  101 *115 *130 *146  163  181 *200 *220
 11 |  *66  *77   89 *102 *116  131 *147 *164 *182 *201 *221 *242
 12 |  *78  *90  103 *117 *132 *148 *165 *183 *202 *222 *243 *265

Columns $d=7$ and $d=11$ (i.e. $d=T(3)+1$ and $d=T(4)+1$) are entirely composite — every entry marked *. Note that column $d=10$ also appears composite in the first 6 rows but shows primes at $n=7$ (value 127) and $n=10$ (value 181), confirming it is not prime-free.

The factorisations for the two prime-free columns are: $$\mathrm{cell}(n,7) = \frac{(n+3)(n+10)}{2}, \qquad \mathrm{cell}(n,11) = \frac{(n+6)(n+15)}{2}$$ In both cases both factors exceed 1 for all $n\ge1$, guaranteeing compositeness.

Note: The prime-free property holds for $j\ge3$ (i.e. $d=7,11,16,22,\ldots$), where both factors $n+T(j-1)$ and $n+T(j+1)$ exceed 1 for all $n\ge1$. For $j=1,2$ ($d=2,4$) the smallest factor can equal 1 for small $n$, so those columns are not entirely composite.

The result

Column $d$ is entirely composite (prime-free) for every $d$ of the form $d = T(j)+1$ where $T(j)=j(j+1)/2$ is a triangular number, i.e. $d \in \{2, 4, 7, 11, 16, 22, 29, 37, \ldots\}$.

For these values of $d$, the polynomial $f_d(n)$ factors over $\mathbb{Z}$ as $$f_d(n) = \frac{(n + T(j-1))(n + T(j+1))}{\text{(integer denominator)}},$$ yielding an unconditional $O(1)$ factorisation: given any value $f_d(n)$, a non-trivial factor is recovered immediately from the closed-form expression without any search.

The infinite family

Since there are infinitely many triangular numbers $T(j)$, there are infinitely many entirely composite columns in the transposed triangular grid. These form an infinite family of prime-free quadratic sequences.

The discriminant of $f_d(n)$ is $\Delta = (2d-1)^2 - 4(d^2-3d+3) = 8d - 11$. The factorisation occurs precisely when $\Delta$ is a perfect square, which happens exactly at $d = T(j)+1$.

Questions

1. Is this infinite family of prime-free quadratic sequences known?
2. The columns not of the form $T(j)+1$ (e.g. $d=3,5,6,8,\ldots$) appear to contain infinitely many primes - is this provable, or does it follow from Bunyakovsky's conjecture?