%I #15 Dec 27 2023 21:22:00
%S 10,22,48,78,144,168,288,210,474,480,570,714,630,930,924,1302,1386,
%T 1380,1500,1590,1470,2160,1680,2838,2100,2772,3498,2640,2760,3060,
%U 4452,4284,4170,4110,3780,4650,5874,3990,4950,5550,8802,8706,5850,6480,7728,7812
%N 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.
%C A002375(a(n)/2) = p.
%H J. Stauduhar, <a href="/A216048/b216048.txt">Table of n, a(n) for n = 1..100</a>
%e 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.
%e a(3) = 48: With n = 3, prime(3) = 5 and 48 = {5+43, 7+41, 11+37, 17+31, 19+29}.
%t nn = 10^4; ps = Boole[PrimeQ[Range[1, 2*nn, 2]]]; lst =
%t Table[Sum[ps[[i]] ps[[n  i + 1]], {i, Ceiling[n/2]}], {n,
%t nn}]; t = {}; p = 0; While[p = NextPrime[p];
%t pos = Position[lst, p, 1, 1]; pos != {}, AppendTo[t, 2*pos[[1, 1]]]];
%t t (* Modified T. D. Noe's code from A216047 *) (* _J. Stauduhar_, Sep 01 2012 *)
%Y Cf. A002375.
%K nonn
%O 1,1
%A _J. Stauduhar_, Aug 31 2012
