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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 $2 \mod 6$ and $4 \mod 6$. Yet all of these are $2 \mod 6$. Is there a model which accounts for this? (update: $a_j$ for $1\le j \le 42$ are all $2 \mod 6$. The proportion which are 2 mod 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 (hueristic) reason for this bias in favor of $2 \mod 6$?

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 mod 6 up until entries 43, 48, 70, 81, 88. Over the entire $5001$ values, 3847 are 2 mod 6 and 1154 are 4 mod 6. The proportion which are $2 \mod 6$ looks like it might converge to $\frac{3}{4}$

The proportions which are in various congruence classes are

$$ [[0, 1], [2, 3846], [4, 1154]] \mod 6$$ $$ [[2, 1536], [4, 911], [6, 874], [8, 1680]] \mod 10$$ $$ [[2, 711], [4, 656], [6, 944], [8, 644], [10, 1033], [12, 1013]] \mod 14$$ $$ [[0, 663], [2, 636], [4, 637], [6, 583], [8, 647], [10, 606], [12, 613], [14, 616]] \mod 16$$

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.

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Multiplication by 2 yields the sequence at oeis.org/A000954 but that site doesn't discuss the problem raised here. –  Gerry Myerson Feb 3 '11 at 23:28
Numerical data suggests that, for almost all $x$, the number of primes less than $x$ which are $2 \mod 3$ is greater than those which are $1 \mod 3$. The term to search on is "prime race"; here is a friendly introduction mathdl.maa.org/images/upload_library/22/Ford/granville1.pdf Is this effect strong enough to explain your observations? –  David Speyer Feb 4 '11 at 0:21
If primes were as randomly distributed as I think they are, then since for $n\equiv 1\pmod{3}$ you are picking pairs from $\binom{P(2n;1,3)}{2}$, for $n\equiv 2\pmod{3}$ you are picking pairs from $\binom{P(2n;2,3)}{2}$, and for $n\equiv 0\pmod{3}$ you pick pairs from $P(2n;1,3)\times P(2n;2,3)$, you will likely have $g(3n)>g(3n+2)>g(3n+1)$. Now combining this with the standard growth conjectures on $g(n)$ (meaning, it is roughly increasing :)) this gives some "explanation". –  Gjergji Zaimi Feb 4 '11 at 5:55
To expand on David Speyer's comment: suppose that up to $N$ there are 100 primes congruent 2 mod 3, and only 80 congruent 1 mod 3. Then if $N$ is 4 mod 6 it could have as many as 50 Goldbach representations; if 0 mod 6, it could have as many as 80; if 2 mod 6, it can't have more than 40. This could well be why more laggards are 2 mod 6, and no laggards (except 24) are 0 mod 6. You have to go up to 608,981,813,029 before you get more primes 1 than 2 mod 3, which seems to be beyond the range of the Goldbach tables, so testing the hypothesis won't be easy. –  Gerry Myerson Feb 4 '11 at 6:03
Let the minutes record that Gjergji's comment wasn't yet visible to me when I posted mine. –  Gerry Myerson Feb 4 '11 at 6:05

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