%I #18 Apr 14 2020 18:33:28
%S 1,2,2,1,1,2,2,1,1,2,1,2,1,1,1,2,6,3,2,2,2,4,2,3,3,2,5,3,2,1,3,6,2,1,
%T 1,2,4,3,4,2,3,5,3,2,2,2,2,2,1,2,3,7,3,2,2,3,4,2,3,2,3,4,4,2,5,5,9,3,
%U 1,4,3,8,2,4,2,4,9,3,2,5,2
%N Number of ways to write n as x^2 + 15*y^2 + z*(3z+1)/2, where x and y are nonnegative integers and z is an integer.
%C Conjecture: a(n) > 0 for any nonnegative integer n, and a(n) = 1 only for n = 0, 3, 4, 7, 8, 10, 12, 13, 14, 29, 33, 34, 48, 68, 113, 129, 220.
%C Let a,b,c,d,e,f be nonnegative integers with a > b, c > d, e > f, a == b (mod 2), c == d (mod 2), e == f (mod 2), a >= c >= e >= 2, b >= d if a = c, and d >= f if c = e. We have shown in arXiv:1502.03056 that if the ordered tuple (a,b,c,d,e,f) is universal, i.e., each n = 0,1,2,... can be written as x(ax+b)/2 + y(cy+d)/2 + z(ez+f)/2 with x,y,z integers, then (a,b,c,d,e,f) must be among the 12082 tuples listed in the linked a-file. We also conjecture that all the listed tuples not yet proved to be universal are indeed universal. Note that those numbers x(4x+2)/2 with x integral coincide with triangular numbers.
%H Zhi-Wei Sun, <a href="/A286944/b286944.txt">Table of n, a(n) for n = 0..10000</a>
%H Zhi-Wei Sun, <a href="/A286944/a286944_1.txt">List of those tuples (a,b,c,d,e,f) for which each nonnegative integer should be represented by x(ax+b)/2 + y(cy+d)/2 + z(ez+f)/2 with x,y,z integers</a>
%H Zhi-Wei Sun, <a href="http://math.scichina.com:8081/sciAe/EN/abstract/abstract517007.shtml">On universal sums of polygonal numbers</a>, Sci. China Math. 58(2015), no. 7, 1367-1396.
%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2016.07.024">On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d)</a>, J. Number Theory 171(2017), 275-283.
%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.
%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/RepresentationRiddle.pdf">Riddles of Representations of Integers</a>, presentation to Nanjing Normal Univ. (China, 2019).
%H Hai-Liang Wu and Zhi-Wei Sun, <a href="http://arxiv.org/abs/1707.06223">Some universal quadratic sums over the integers</a>, arXiv:1707.06223 [math.NT], 2017.
%e a(33) = 1 since 33 = 4^2 + 15*1^2 + 1*(3*1+1)/2.
%e a(34) = 1 since 34 = 2^2 + 15*1^2 + 3*(3*3+1)/2.
%e a(48) = 1 since 48 = 6^2 + 15*0^2 + (-3)*(3*(-3)+1)/2.
%e a(68) = 1 since 68 = 1^2 + 15*2^2 + 2*(2*3+1)/2.
%e a(113) = 1 since 113 = 6^2 + 15*0^2 + 7*(3*7+1)/2.
%e a(129) = 1 since 129 = 8^2 + 15*2^2 + (-2)*(3*(-2)+1)/2.
%e a(220) = 1 since 220 = 13^2 + 15*0^2 + 6*(3*6+1)/2.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];Do[r=0;Do[If[SQ[24(n-x^2-15y^2)+1],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[(n-x^2)/15]}];Print[n," ",r],{n,0,80}]
%Y Cf. A000290, A001318, A287616.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, May 16 2017