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{2*h^2, h >=1} union {3*k^2, k >=1}, in increasing order.
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%I #6 Oct 30 2014 18:37:01

%S 2,3,8,12,18,27,32,48,50,72,75,98,108,128,147,162,192,200,242,243,288,

%T 300,338,363,392,432,450,507,512,578,588,648,675,722,768,800,867,882,

%U 968,972,1058,1083,1152,1200,1250,1323,1352,1452,1458,1568,1587,1682

%N {2*h^2, h >=1} union {3*k^2, k >=1}, in increasing order.

%C Let S = {2*h^2, h >=1} and T = {3*k^2, k >=1}. Then S and T are disjoint. The position of 2*n^2 in (S union T) is A184808(n), and the position of 3*n^2 is A184809(n).

%H Clark Kimberling, <a href="/A249096/b249096.txt">Table of n, a(n) for n = 1..1000</a>

%e {2*h^2, h >=1} = {2, 8, 18, 32, 50, 72, 98, 128, 162, 200, ...};

%e {3*k^2, k >=1} = {3, 12, 27, 48, 75, 108, 147, 192, 243, ...};

%e so the union is {2, 3, 8, 12, 18, 27, 32, 48, 50, 72, 75, ...}

%t z = 120; s = Table[2 h^2, {h, 1, z}]; t = Table[3 k^2, {k, 1, z}]; v = Sort[Union[s, t]]

%Y Cf. A184808, A184809.

%Y A249367 is essentially the same sequence.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Oct 20 2014