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.

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.

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 $632, 692, 626, 992, 878, 908, 1112, 962, 998, 1412, 1202$ One would expect these numbers to have no small odd divisors (and each is a power of 2 times a prime except 988=21337) 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: $a_2=24$ BUT otherwise $a_j$ for $1\le j \le 42$ are all $2 \mod 6$. The proportion which are 2 mod 6 (ignoring the one value 24) 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 the second is 24 and the rest are 2 mod 6 up until entries 43, 48, 70, 81, 88. Over the entire $5001$ values a single one is 0 mod 6, 3846 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.

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.

For even $n$, $g(n)$ 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 $632, 692, 626, 992, 878, 908, 1112, 962, 998, 1412, 1202$ One would expect these numbers to have no small odd divisors (and each is a power of 2 times a prime except 988=21337) 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: $a_2=24$ BUT otherwise $a_j$ for $1\le j \le 47$ are all $2 \mod 6$. The proportion which are 2 mod 6 (ignoring the one value 24) 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 the second is 24 and the rest are 2 mod 6 up until entries 43, 48, 70, 81, 88. Over the entire $5001$ values a single one is 0 mod 6, 3846 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.

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 $632, 692, 626, 992, 878, 908, 1112, 962, 998, 1412, 1202$ One would expect these numbers to have no small odd divisors (and each is a power of 2 times a prime except 988=21337) 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: $a_2=24$ BUT otherwise $a_j$ for $1\le j \le 42$ are all $2 \mod 6$. The proportion which are 2 mod 6 (ignoring the one value 24) 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 the second is 24 and the rest are 2 mod 6 up until entries 43, 48, 70, 81, 88. Over the entire $5001$ values a single one is 0 mod 6, 3846 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.