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%I #11 Feb 27 2015 09:23:10
%S 1,1,1,1,1,1,2,2,1,1,2,1,3,2,1,3,2,1,2,2,2,4,4,1,2,3,3,3,3,2,2,4,3,3,
%T 3,2,5,4,3,3,4,3,4,5,2,3,5,3,5,5,2,5,5,3,5,4,4,5,6,5,4,4,3,4,5,5,7,7,
%U 1,5,7,4,7,7,4,3,8,5,5,6,6,5,6,4,6,6,5,10,7,3,5,8,7,9,7,4,4,9,5,4,8
%N Number of ways to write n as a*(2a-1)+ b*(2b-1) + c*(2c+1) + d*(2d+1), where a,b,c,d are nonnegative integers with a <= b and c <= d.
%C Conjecture: (i) a(n) > 0 for all n. In other words, any nonnegative integer can be expressed as the sum of two hexagonal numbers and two second hexagonal numbers.
%C (ii) Each nonnegative integer can be written as the sum of two pentagonal numbers and two second pentagonal numbers.
%C We have verified parts (i) and (ii) of the conjecture for n up to 10^7 and 10^6 respectively.
%H Zhi-Wei Sun, <a href="/A255350/b255350.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://arxiv.org/1502.03056">On universal sums a*x^2+b*y^2+f(z), a*T_x+b*T_y+f(z) and a*T_x+b*y^2+f(z)</a>, arXiv:1502.03056 [math.NT], 2015.
%e a(23) = 1 since 23 = 1*(2*1-1) + 1*(2*1-1) + 0*(2*0+1) + 3*(2*3+1).
%e a(68) = 1 since 68 = 1*(2*1-1) + 4*(2*4-1) + 1*(2*1+1) + 4*(2*4+1).
%t HQ[n_]:=IntegerQ[Sqrt[8n+1]]&&Mod[Sqrt[8n+1],4]==1
%t Do[r=0;Do[If[HQ[n-x(2x-1)-y(2y-1)-z(2z+1)],r=r+1],{x,0,(Sqrt[4n+1]+1)/4},{y,x,(Sqrt[8(n-x(2x-1))+1]+1)/4},{z,0,(Sqrt[4(n-x(2x-1)-y(2y-1))+1]-1)/4}];
%t Print[n," ",r];Continue,{n,0,10000}]
%Y Cf. A000326, A000384, A005449, A014105.
%K nonn
%O 0,7
%A _Zhi-Wei Sun_, Feb 21 2015
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