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A242247
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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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0
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2, 4, 8, 10, 14, 22, 22, 28, 32, 38, 46, 49, 49, 58, 58, 68, 74, 74, 82, 82, 87, 94, 94, 98, 112, 116, 121, 128, 136, 146, 146, 146, 155, 155, 164, 166, 184, 184, 184, 200, 206, 206, 221, 221, 224, 238, 244, 265, 265, 268, 268, 268, 286, 286, 286, 286, 344
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OFFSET
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1,1
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COMMENTS
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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.
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.
Heuristic arguments would imply that m(N) ~ N*log^2(2*N). Then, conjecturally, for n >= 3, a(n) < n*log^2(2*n).
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.
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LINKS
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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