

A216048


Consider the unordered Goldbach partitions of the even numbers m. Then a(n) is the least m which contains prime(n) such partitions composed of odd primes.


1



10, 22, 48, 78, 144, 168, 288, 210, 474, 480, 570, 714, 630, 930, 924, 1302, 1386, 1380, 1500, 1590, 1470, 2160, 1680, 2838, 2100, 2772, 3498, 2640, 2760, 3060, 4452, 4284, 4170, 4110, 3780, 4650, 5874, 3990, 4950, 5550, 8802, 8706, 5850, 6480, 7728, 7812
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OFFSET

1,1


COMMENTS



LINKS



EXAMPLE

With n = 1: prime(1) = 2, so we want the least m that has 2 such partitions. For m = 6, 6 has one such partition, {3+3}, but 1 is not a prime number. For m = 8, 8 has one such partition, {3+5}, but 1 is not a prime number. For m = 10, 10 has two such partitions, {3+7, 5+5}, so a(1) = 10.
a(3) = 48: With n = 3, prime(3) = 5 and 48 = {5+43, 7+41, 11+37, 17+31, 19+29}.


MATHEMATICA

nn = 10^4; ps = Boole[PrimeQ[Range[1, 2*nn, 2]]]; lst =
Table[Sum[ps[[i]] ps[[n  i + 1]], {i, Ceiling[n/2]}], {n,
nn}]; t = {}; p = 0; While[p = NextPrime[p];
pos = Position[lst, p, 1, 1]; pos != {}, AppendTo[t, 2*pos[[1, 1]]]];


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



