Is every sufficiently large positive integer of the form $ab + ac + bc + 1$?

Question

Is every sufficiently large positive integer $A$ of the form $ab + ac + bc + 1$

where $a,b,c$ are some positive integers larger than some given positive integer $d$ ?

How large is sufficiently large , in other words how do we know our number $A$ is sufficiently large ( as a function of $d$ ) ?

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Update :

I read the following about positive integers that are not of the form $ ab + ac + bc $:

https://oeis.org/A000926

And I quote about that sequence above : " Note that for n in this sequence, n+1 is either a prime, twice a prime, the square of a prime, 8 or 16 "

This is why I considered the weird +1 in $ab + ac + bc + 1$

So composites larger than 16 not a prime , twice a prime or the square of a prime are of the form $ab + ac + bc + 1$.

And I wonder why !

I guess that relates strongly to the Original posted problem. Maybe I should have mentioned that before , sorry.

Will Jagy's answer gave the same OEIS link which I already knew. This also explains some of the comments.

Im no expert in genus theory or Galois theory but I assume this has a simple answer.

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Answer 1

Euler's "numerus idoneus" (or "numeri idonei", or idoneal, or suitable, or convenient numbers).

https://oeis.org/A000926

It is conjectured that the list given here is complete. Chowla showed that the list is finite and Weinberger showed that there is at most one further term.

(6) n is not of the form ab+ac+bc with 0 < a < b < c. [Rains]

Answer 2

Then equation: $$XY+XZ+YZ=N$$

If we ask what ever number: $p$

That the following sum can always be factored: $p^2+N=ks$

Solutions can be written. $$X=p$$ $$Y=s-p$$ $$Z=k-p$$

I like this formula the solutions of this equation. Number of solution for $xy +yz + zx = N$