%I #14 Sep 29 2017 06:09:28
%S 276,703,861,1225,2850,3003,4560,5151,8128,10878,11781,12090,12720,
%T 13366,14706,15400,16110,18721,21115,22366,24090,24531,26796,29161,
%U 29646,31125,32131,33153,36315,38503,39621,40186,42486,45451,47895
%N Hexagonal numbers (A000384) which are sum of 2 other hexagonal numbers > 0.
%C This is to A136117 as A000384 is to A000326. Duke and Schulze-Pillot (1990) proved that every sufficiently large integer (and hence every sufficiently large hexagonal number) can be written as the sum of three hexagonal numbers.
%H Donovan Johnson, <a href="/A133215/b133215.txt">Table of n, a(n) for n = 1..1000</a>
%H W. Duke and R. Schulze-Pilot, <a href="http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002107066">Representation of integers by positive ternary quadratic forms and equidistribution of lattice points on ellipsoids</a>, Invent. Math. 99(1990), 49-57.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HexagonalNumber.html">Hexagonal Number</a>.
%F {x: x>0 and x in A000384 and x = A000384(i) + A000384(j) for i>0 and j>0}, where A000384 = {n*(2*n-1) for n > 0}.
%e hex(19) = 703 = 378 + 325 = hex(14) + hex(13).
%e hex(21) = 861 = 630 + 231 = hex(18) + hex(11).
%e hex(25) = 1225 = 1035 + 190 = hex(23) + hex(10).
%e hex(38) = 2850 = 2415 + 435 = hex(35) + hex(15).
%e hex(39) = 3003 = 2850 + 153 = hex(38) + hex(9) = 2415 + 435 + 153 = hex(35) + hex(15) + hex(9).
%e hex(48) = 4560 = 2415 + 2145 = hex(35) + hex(33).
%t With[{upto=60000},Select[Union[Total/@Subsets[Table[n(2n-1),{n, Ceiling[ (1+Sqrt[1+8upto])/4]}],{2}]],IntegerQ[(1+Sqrt[1+8#])/4]&&#<=upto&]] (* _Harvey P. Dale_, Jul 24 2011 *)
%Y Cf. A000384, A136117.
%K nonn
%O 1,1
%A _Jonathan Vos Post_, Dec 18 2007
%E Added missing term 276 and a(8)-a(35) from _Donovan Johnson_, Sep 27 2008
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