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A254631 Number of ways to write n as x*(x+1)/2 + y*(3*y+2) + z*(3*z-2) with x,y,z nonnegative integers.

%I A254631 #7 Feb 03 2015 12:54:45
%S A254631 1,2,1,1,1,1,3,2,2,2,1,3,1,1,2,2,4,2,2,2,2,3,3,3,2,1,3,5,2,3,1,2,2,2,
%T A254631 5,1,5,4,2,2,3,5,3,3,4,4,3,3,2,3,2,3,3,2,3,5,4,5,3,2,5,4,6,2,2,3,6,3,
%U A254631 3,4,3,7,3,4,3,2,4,4,4,6,3,3,4,4,4,5,5,4,3,2,3,5,8,3,3,3,7,3,3,8,4
%N A254631 Number of ways to write n as x*(x+1)/2 + y*(3*y+2) + z*(3*z-2) with x,y,z nonnegative integers.
%C A254631 Conjecture: a(n) > 0 for all n, and a(n) > 1 for all n > 35.
%C A254631 We have proved that any nonnegative integer n can be written as x*(x+1)/2 + y*(3*y+2) + z*(3*z-2) with x,y,z integers.
%H A254631 Zhi-Wei Sun, <a href="/A254631/b254631.txt">Table of n, a(n) for n = 0..10000</a>
%H A254631 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.
%e A254631 a(13) = 1 since 13 = 0*1/2 + 1*(3*1+2) + 2*(3*2-2).
%e A254631 a(30) = 1 since 30 = 3*4/2 + 2*(3*2+2) + 2*(3*2-2).
%e A254631 a(35) = 1 since 35 = 1*2/2 + 3*(3*3+2) + 1*(3*1-2).
%t A254631 TQ[n_]:=IntegerQ[Sqrt[8n+1]]
%t A254631 Do[r=0;Do[If[TQ[n-y(3y+2)-z(3z-2)],r=r+1],{y,0,(Sqrt[3n+1]-1)/3},{z,0,(Sqrt[3(n-y(3y+2))+1]+1)/3}];
%t A254631 Print[n," ",r];Continue,{n,0,100}]
%Y A254631 Cf. A000217, A000567, A045944, A254573, A254574.
%K A254631 nonn
%O A254631 0,2
%A A254631 _Zhi-Wei Sun_, Feb 03 2015