%I #22 Apr 03 2017 14:24:12
%S 4,7,12,13,16,19,21,28,31,36,37,39,43,48,49,52,57,61,63,64,67,73,76,
%T 79,84,91,93,97,100,103,108,109,111,112,117,124,127,129,133,139,144,
%U 147,148,151,156,157,163,169,171,172,175,181,183,189,192,193,196,199
%N Numbers of the form x^2 + 3y^2 where x and y are positive integers.
%C Superset of primes of the form 6n+1 (A002476).
%C It seems that all integer solutions of ((a+b)^3 - (a-b)^3) / (2*b) = c^3 have c = x^2 + 3*y^2. - Juergen Buchmueller (pullmoll(AT)t-online.de), Apr 04 2008
%C To prove the case of cubes in Fermat's last theorem, Euler considered numbers of the form a^2 + 3b^2. In the equation x^3 + y^3 = z^3, Euler specified that x = a - b and y = a + b. - _Alonso del Arte_, Jul 19 2012
%C All terms == 0,1,3,4, or 7 (mod 9). - _Robert Israel_, Apr 03 2017
%D Paulo Ribenboim, 13 Lectures on Fermat's Last Theorem. New York: Springer-Verlag (1979): 4.
%H Robert Israel, <a href="/A092572/b092572.txt">Table of n, a(n) for n = 1..10000</a>
%H E. Akhtarkavan, M. F. M. Salleh and O. Sidek, <a href="https://www.idosi.org/wasj/wasj21(2)13/2.pdf">Multiple Descriptions Video Coding Using Coinciding Lattice Vector Quantizer for H.264/AVC and Motion JPEG2000</a>, World Applied Sciences Journal 21 (2): 157-169, 2013. - From _N. J. A. Sloane_, Feb 11 2013
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Eulers6nPlus1Theorem.html">Eulers 6n Plus 1 Theorem</a>
%e 7 is of the specified form, since 2^2 + 3 * 1^2 = 7.
%e So is 12, since 3^2 + 3 * 1^2 = 12, and 13, with 1^2 + 3 * 2^2 = 13.
%p N:= 1000: # to get all terms <= N
%p S:= {seq(seq(x^2 + 3*y^2, x = 1 .. floor(sqrt(N - 3*y^2))),
%p y=1..floor(sqrt(N/3-1)))}:
%p sort(convert(S,list)); # _Robert Israel_, Apr 03 2017
%t Union[Flatten[Table[a^2 + 3b^2, {a, 20}, {b, Ceiling[Sqrt[(400 - a^2)/3]]}]]] (* _Alonso del Arte_, Jul 19 2012 *)
%Y Cf. A002476, A092573, A092575, A158937 (similar definition but with duplicates left in).
%K nonn
%O 1,1
%A _Eric W. Weisstein_, Feb 28 2004