OFFSET
1,9
COMMENTS
It is conjectured that a(n) > 0 for n >= 8. (Conjecture 1)
If we categorize primes into the following four sets:
a) Even prime: 2
b) 3-multiple prime: 3
c) primes of the form 6m-1, i.e., 5, 11, 17, 23, 29, 41, ...
d) primes of the form 6m+1, i.e., 7, 13, 19, 31, 37, 43, ...
then the decomposition of an even number can also be categorized accordingly:
1) cannot do so: 2;
2) contains 2: only 4;
3) contains 3: 6=3+3, 8=3+5, 10=3+7, 14=3+11, ... Obviously not all even numbers belong to this category;
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);
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);
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).
It is obvious that categories 4), 5), 6) have no elements in common.
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.
Thus, if Conjecture 1 is true, the Goldbach conjecture is true.
EXAMPLE
For n=8, 2n=16, 16=5+11, and 11-6=5 is also prime. So a(8)=1;
...
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.
MATHEMATICA
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}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Lei Zhou, Sep 12 2018
STATUS
approved