A237638 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.
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
Offset: 3
Examples
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. ... 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. ... 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.
Links
- Lei Zhou, Table of n, a(n) for n = 3..79
Programs
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Mathematica
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 *)
Comments