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a(n) is the number of decompositions of 2n into an unordered sum of two odd primes p and q such that at least one number among p-6 and q-6 is also prime, and both p and q are greater than 3.
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%I #12 Sep 28 2018 21:38:35

%S 0,0,0,0,0,0,0,1,2,1,2,3,2,2,3,1,3,3,1,2,4,2,2,4,3,3,5,2,4,6,1,4,5,1,

%T 4,5,4,3,6,4,3,8,4,3,8,2,5,7,3,5,5,3,4,7,5,5,9,5,6,12,3,4,10,2,6,6,4,

%U 4,6,6,5,9,5,5,12,3,5,9,2,7,8,5,4,11,7,3

%N a(n) is the number of decompositions of 2n into an unordered sum of two odd primes p and q such that at least one number among p-6 and q-6 is also prime, and both p and q are greater than 3.

%C It is conjectured that a(n) > 0 for n >= 8. (Conjecture 1)

%C If we categorize primes into the following four sets:

%C a) Even prime: 2

%C b) 3-multiple prime: 3

%C c) primes of the form 6m-1, i.e., 5, 11, 17, 23, 29, 41, ...

%C d) primes of the form 6m+1, i.e., 7, 13, 19, 31, 37, 43, ...

%C then the decomposition of an even number can also be categorized accordingly:

%C 1) cannot do so: 2;

%C 2) contains 2: only 4;

%C 3) contains 3: 6=3+3, 8=3+5, 10=3+7, 14=3+11, ... Obviously not all even numbers belong to this category;

%C 4) contains only primes from prime category c): 10=5+5, 16=5+11, 22=5+17, ... It is conjectured that all even numbers of the form 6m+4, m >= 1 belong to this category (Conjecture 2);

%C 5) contains one prime from category c) and one prime from category d): 12=5+7, 18=5+13, 24=5+19, ... It is conjectured that all even numbers of the form 6m, m > 1 belong to this category (Conjecture 3);

%C 6) contains two primes from prime category d): 14=7+7, 20=7+13, 26=7+19, ... It is conjectured that all even numbers of the form 6m+2, m > 1 belong to this category (Conjecture 4).

%C It is obvious that categories 4), 5), 6) have no elements in common.

%C For even numbers 2k in category 4), 5), and 6), if there exists a pair of dual prime composition 2k=p+q such that p-6 or q-6 is prime too, at least one pair of 2k's decomposition p+q can be generated from 2k-6's decomposition p-6,q or p,q-6.

%C Thus, if Conjecture 1 is true, the Goldbach conjecture is true.

%e For n=8, 2n=16, 16=5+11, and 11-6=5 is also prime. So a(8)=1;

%e ...

%e For n=18, 2n=36, 36=5+31,17+19,23+13,29+7, where in number pair (5,31), both 5-6=-1 and 31-6=25 are not prime, and at least one number in other pairs less 6 are prime: 17-6=11, 19-6=13, 23-6=17, 13-6=7, 29-6=23. So a(18)=3.

%t Table[m = 2*n; ct = 0; p = 3; While[While[p = NextPrime[p]; cp = m - p; ! PrimeQ[cp]]; (2*p) <= m, If[PrimeQ[p - 6] || PrimeQ[cp - 6], ct++]]; ct, {n, 1, 86}]

%Y Cf. A000040, A002375, A007582, A002476, A023201.

%K nonn,easy

%O 1,9

%A _Lei Zhou_, Sep 12 2018