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A237638
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a(n) is the number of prime sets such that each set contains enough prime numbers to decompose every even number from 6 to 2n into the sum of two of its elements (reuse allowed), while none of the sets is a subset of another such set.
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2
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1, 1, 1, 1, 1, 2, 2, 3, 4, 4, 5, 6, 6, 9, 11, 11, 11, 13, 16, 23, 25, 31, 47, 57, 63, 70, 74, 79, 82, 122, 131, 129, 180, 215, 219, 323, 367, 446, 501, 531, 661, 867, 897, 1311, 1471, 1691, 1695, 2130, 2288, 2833, 3363, 3891, 5435, 8068, 8867, 13476, 15451, 15897
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OFFSET
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3,6
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LINKS
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EXAMPLE
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n=4, 2n=8. There is only one set of primes {3,5} such that 6=3+3, 8=3+5. So a(4)=1.
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n=8, 2n=16. We can find two sets, {3,5,7,11} and {3,5,7,13} that have such features. So a(8)=2. Here any set with more primes either contains an unused prime number or one of these two sets is a subset of them, like {3,5,7,11,13}, and thus is not considered. So a(8)=2.
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n=13, 2n=26. Five such sets are found: {3,5,7,11,13}, {3,5,7,13,17},{3,5,7,13,19}, {3,5,7,11,17,19}, {3,5,7,11,17,23}. So a(13)=5.
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MATHEMATICA
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a = {{{3}}}; Table[n2 = 2*n; na = {}; la = Last[a]; lo = Length[la]; Do[ok = 0; Do[p1 = la[[i, j]]; p2 = n2 - p1; If[MemberQ[la[[i]], p2], ok = 1], {j, 1, Length[la[[i]]]}];
If[ok == 1, na = Sort[Append[na, la[[i]]]], Do[p1 = la[[i, j]]; p2 = n2 - p1; If[PrimeQ[p2], ng = Sort[Append[la[[i]], p2]]; big = 0; If[Length[na] > 0, Do[If[Intersection[na[[k]], ng] == na[[k]], big = 1], {k, 1, Length[na]}]]; If[big == 0, na = Sort[Append[na, ng]]]], {j, 1, Length[la[[i]]]}]], {i, 1, lo}]; AppendTo[a, na]; Length[na], {n, 4, 60}](* Program lists the 4th item and beyond *)
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CROSSREFS
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KEYWORD
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nonn,hard
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AUTHOR
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STATUS
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approved
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