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For even $n$, 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=2*13*37) 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$42$ 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.

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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=2*13*37) 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 $\frac{3}{4} \frac{3}{4}$

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.

2 updated information

Let

For even $n$, $g(n)$ be the number of ways to write $2n$ n$as a sum of two primes$2n=p+q$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 $316, 346632, 692, 313626, 496992, 439878, 454908, 5561112, 481962, 499998, 7061412, 601$ 1202$One would expect these numbers to have no small odd divisors (and each is a power of 2 times a prime except 499=13*37988=2*13*37) but I would expect them to be a about evenly split between$1 2 \mod 3$6$ and $2 4 \mod 3$6$. Yet all of these are$1 2 \mod 3$6$. 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: of (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 $g(n)$, For example decompositions of 2n into an unordered sum of two odd primes links to a table for $1 \le n \le 20000$) and the numbers have been computed 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 predicit 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). Question: Is 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 any reason 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 expect that$\frac{3}{4}

The proportions which are in various congruence class classes are

$$[[0, 1], [2, 3846], [4, 1154]] \mod 3 is correlated with g(n)? Has this been observed6$$$$[[2, 1536], [4, 911], [6, 874], [8, 1680]] \mod 10$$$$[[2, 711], [4, 656], [6, 944], [8, 644], [10, 1033], [12, 1013]] \mod 14$$$$[[0, 663], [2, 636], [4, 637], [6, 583], [8, 647], [10, 606], [12, 613], [14, 616]] \mod 16$$

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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