%I #13 Jul 28 2017 13:27:16
%S 0,0,0,1,1,1,1,1,1,3,1,1,4,2,2,2,1,2,4,3,2,4,2,4,4,3,1,4,5,2,5,1,3,6,
%T 5,2,3,6,3,9,3,1,6,3,5,4,4,6,7,3,2,5,3,6,9,6,3,7,6,2,8,5,4,8,6,3,4,6,
%U 3,12,2
%N Number of ways to write n as x^2 + 2*y^2 + z*(z+1)/2, where x is a nonnegative integer, and y and z are positive integers.
%C Conjecture: a(n) > 0 for all n > 2. In other words, each n = 0,1,2,... can be written as x^2 + 2y*(y+2) + z*(z+3)/2 with x,y,z nonnegative integers.
%C As pointed out by Sun in his 2007 paper in Acta Arith., a result of Jones and Pall implies that every n = 0,1,2,... can be written as x^2 + 2*(2y)^2 + z*(z+1)/2 with x,y,z nonnegative integers.
%C Let a,c,e be positive integers, and let b,d,f be nonnegative integers with a-b, c-d, e-f all even. Suppose that a|b, c|d and e|f. The author studied in arXiv:1502.03056 when each nonnegative integer can be written as x*(a*x+b)/2 + y*(c*y+d)/2 + z*(e*z+f)/2 with x,y,z nonnegative integers, and conjectured that the answer is positive if (a,b,c,d,e,f) is among the following ten tuples (4,0,2,0,1,3), (4,0,2,0,1,5), (4,0,2,6,1,1), (4,0,2,6,2,0), (4,4,2,0,1,3), (4,8,2,0,1,1), (4,8,2,0,1,3), (4,12,2,0,1,1), (6,0,2,0,1,3), (6,6,2,0,1,3).
%C See also A287616 and A286944 for related comments.
%H Zhi-Wei Sun, <a href="/A290342/b290342.txt">Table of n, a(n) for n = 0..10000</a>
%H B. W. Jones and G. Pall, <a href="https://doi.org/10.1007/BF02547347">Regular and semi-regular positive ternary quadratic forms</a>, Acta Math. 70 (1939), 165-191.
%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.4064/aa127-2-1">Mixed sums of squares and triangular numbers</a>, Acta Arith. 127(2007), 103-113.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1502.03056">On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2</a>, arXiv:1502.03056 [math.NT], 2015-2017.
%e a(10) = 1 since 10 = 1^2 + 2*2^2 + 1*2/2.
%e a(11) = 1 since 11 = 0^2 + 2*2^2 + 2*3/2.
%e a(16) = 1 since 16 = 2^2 + 2*1^2 + 4*5/2.
%e a(26) = 1 since 26 = 3^2 + 2*1^2 + 5*6/2.
%e a(31) = 1 since 31 = 1^2 + 2*1^2 + 7*8/2.
%e a(41) = 1 since 41 = 6^2 + 2*1^2 + 2*3/2.
%t TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
%t Do[r=0;Do[If[TQ[n-x^2-2y^2],r=r+1],{x,0,Sqrt[n]},{y,1,Sqrt[(n-x^2)/2]}];Print[n," ",r],{n,0,70}]
%Y Cf. A000217, A000290, A286944, A287616.
%K nonn
%O 0,10
%A _Zhi-Wei Sun_, Jul 27 2017