%I #12 Oct 03 2016 03:11:10
%S 1,2,2,2,2,1,2,3,3,3,2,2,3,1,1,3,5,6,2,3,1,2,4,2,4,3,4,3,3,2,4,7,4,4,
%T 2,2,4,3,3,4,3,5,5,3,6,3,5,4,2,4,4,6,5,3,2,6,5,7,4,3,2,4,4,4,7,3,8,4,
%U 5,3,5,6,8,3,2,3,4,9,2,8,3,7,7,4,5,5,4,4,4,6,5,4,6,7,9,2,8,4,3,4,3
%N Number of ways to write n as the sum of a square, a second pentagonal number, and a hexagonal number.
%C Conjecture: a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 5, 13, 14, 20, 112, 125.
%C Compare this conjecture with the conjecture in A160324.
%C The conjecture that a(n) > 0 for all n = 0,1,2,... appeared in Conjecture 1.2(ii) of the author's JNT paper in the links. - _Zhi-Wei Sun_, Oct 03 2016
%H Zhi-Wei Sun, <a href="/A254668/b254668.txt">Table of n, a(n) for n = 0..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/0905.0635">On universal sums of polygonal numbers</a>, arXiv:0905.0635 [math.NT], 2009-2015.
%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.
%e a(20) = 1 since 20 = 2^2 + 3*(3*3+1)/2 + 1*(2*1-1).
%e a(112) = 1 since 112 = 7^2 + 6*(3*6+1)/2 + 2*(2*2-1).
%e a(125) = 1 since 125 = 5^2 + 8*(3*8+1)/2 + 0*(2*0-1).
%t SQ[n_]:=IntegerQ[Sqrt[n]]
%t Do[r=0;Do[If[SQ[n-y(3y+1)/2-z(2z-1)],r=r+1],{y,0,(Sqrt[24n+1]-1)/6},{z,0,(Sqrt[8(n-y(3y+1)/2)+1]+1)/4}];
%t Print[n," ",r];Continue,{n,0,100}]
%Y Cf. A000290, A000384, A005449, A160324, A160325, A254661.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Feb 04 2015