Why are Goldbach laggards biased towards $2 \bmod 6$?

Question

For even $n$, let $g(n)$ be the number of ways to write $n$ as a sum of two primes $n=p+q$ with $p \le q$. Define $a_k$ to be the largest $n$ with $g(n)=k$. I would bet money that no-one will disprove (in the next 10 years) that $a_k$ for $10 \le k \le 19$ are $632, 692, 626, 992, 878, 908, 1112, 998, 1412, 1202$. One would expect these numbers to have no small odd divisors (and each is a power of 2 times a prime) but I would expect them to be about evenly split between $2 \bmod 6$ and $4 \bmod 6$. Yet all of these are $2 \bmod 6$. Is there a model which accounts for this? (Update: $a_j$ for $1\le j \le 42$ are all $2 \bmod 6$. The proportion which are $2 \bmod 6$ never goes below 76% for $k \le 5001$. So it is not as blatant as I first thought but still quite pronounced.)

Is there a (heuristic) reason for this bias in favor of $2 \bmod 6$?

Discussion (revised thanks to Gerry and David): Of course we do not know if $a_k$ is well defined although we suspect it is. Data exists for $g(n)$ up to $2.5 \cdot 10^8$. The estimates are based on the size of n and the set of odd primes dividing it. This would predict that from some point on we never have $g(n+m)<g(n)$ with $n+m$ a multiple of 3 but $n$ not a multiple of 3. That seems to be true (I think). The OEIS has a list of Conjecturally largest even integer which is an unordered sum of two primes in exactly n ways and a link from there gives the first 5001 values. Of these all are $2 \bmod 6$ up until entries $43, 48, 70, 81, 88$. Over the entire $5001$ values, $3847$ are $2 \bmod 6$ and $1154$ are $4 \bmod 6$. The proportion which are $2 \bmod 6$ looks like it might converge to $\frac{3}{4}$.

The proportions which are in various congruence classes are \begin{gather*} [[0, 1], [2, 3846], [4, 1154]] \bmod 6 \\ [[2, 1536], [4, 911], [6, 874], [8, 1680]] \bmod 10 \\ [[2, 711], [4, 656], [6, 944], [8, 644], [10, 1033], [12, 1013]] \bmod 14 \\ [[0, 663], [2, 636], [4, 637], [6, 583], [8, 647], [10, 606], [12, 613], [14, 616]] \bmod 16. \end{gather*}

David suggests that it might relate to prime races and gives a good reference which in turn suggests a relation to congruence classes of squares. I suppose that even a slight advantage can bias the location of the extreme cases, but I'm not sure.

Answer 1

I had requested help finding references on this topic but I think that is the answer to this question.

One can use this heuristic procedure to explain why Goldbach's laggards are biased towards $ 2 \bmod 6$:

using these two examples in which they are represented the graphs of possible $p$ and $q $ pairs

case $ n=96 \equiv 0 \bmod 6$

$g(96)=7$

case $ n=98 \equiv 2 \bmod 6$

note the symmetry with respect to $n/2$

$g(98)=3$ also considering that $98=3+95$

it is evident that for $n =p+q $

• if $ n \equiv 0 \bmod 6$ then $p \equiv -1 \bmod 6 $ and $ q \equiv 1 \bmod 6$

• if $ n \equiv 2 \bmod 6$ then $p \equiv 1 \bmod 6$ and $ q \equiv 1 \bmod 6$

, or $p=3$ and $q \equiv -1 \bmod 6$

• and if $ n \equiv 4 \bmod 6$ then $p \equiv -1 \bmod 6 $ and $ q \equiv -1 \bmod 6$ , or $p=3$ and $q \equiv 1 \bmod 6$

as you can see some examples in the case of $n \equiv 0 \bmod 6$ the function $g(n)$ can assume a maximum value equal to double compared to the other two cases $n \equiv 2 \bmod6$ and $n \equiv 4 \bmod6$ why ignoring the case in which $p=3$ in these last two cases the possible pairs $p$ and $q $ are symmetrical respect to $n/2$, therefore these two cases reach the value $k$ for $n$ greater.

As regards the prevalence of $2 \bmod 6$ cases, this is due to the fact that the composite numbers $-1 \bmod 6$ are of the type $(6a-1)(6b+1)$ instead those $1 \bmod 6$ are of the type $(6a+1)(6b+1)$ or $(6a-1)(6b-1)$, the latter in addition to having a different distribution and fixed $n$ are slightly higher in number.