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Maximal k <= n^2 + 1 such that every Goldbach representation of 2*k = p+q contains at least one prime from the set {prime(1), prime(2), ..., prime(n)}, or a(n)=0 if there is no such k.
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%I #45 Jul 15 2018 12:33:09

%S 2,4,8,10,14,22,22,28,32,38,46,49,49,58,58,68,74,74,82,82,87,94,94,98,

%T 112,116,121,128,136,146,146,146,155,155,164,166,184,184,184,200,206,

%U 206,221,221,224,238,244,265,265,268,268,268,286,286,286,286,344

%N Maximal k <= n^2 + 1 such that every Goldbach representation of 2*k = p+q contains at least one prime from the set {prime(1), prime(2), ..., prime(n)}, or a(n)=0 if there is no such k.

%C The restriction a(n) <= n^2 + 1 allows one to make the sequence computable for n >= 1 and, at the same time, to somewhat agree with heuristic arguments for large n.

%C The sequence is based on a conjecture stronger than Goldbach's Conjecture: for arbitrarily large N there exists a number m(N) such that, for k > m(N), the number of unordered Goldbach representations (A002375) of 2*k is greater than N.

%C Heuristic arguments would imply that m(N) ~ N*log^2(2*N). Then, conjecturally, for n >= 3, a(n) < n*log^2(2*n).

%C The existence of a(n) for arbitrary n says that, if we remove from the sequence of primes an arbitrarily large number M of the first terms, the Goldbach representations remain for all sufficiently large even numbers.

%e Let n=3. Then the set is {2,3,5}. The Goldbach representations of 2*k=16 are 3+13 and 5+11. Each of them contains a prime from {2,3,5}. So a(3) >= 8. Since, by definition, a(3) <= 10, consider also 2*k=18 and 20. We have 18=7+11, 20=7+13. These representations contain none of the primes 2,3,5. Therefore a(3)=8.

%Y Cf. A002375, A152451, A156284, A156537.

%K nonn

%O 1,1

%A _Vladimir Shevelev_, May 09 2014

%E More terms from _Peter J. C. Moses_, May 10 2014