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%I #22 Feb 12 2019 13:31:43
%S 1,0,0,1,0,0,1,0,0,1,1,0,0,1,0,0,1,0,0,0,1,1,0,1,1,0,0,1,0,0,1,1,0,0,
%T 1,0,1,0,0,1,0,1,2,0,0,1,1,0,0,1,0,0,1,0,0,1,1,1,1,0,1,1,0,1,0,1,0,1,
%U 1,0,0,0,1,0,0,2,1,0,2,1,0,1,1,0,1,0,1,0,1,0,0,2,0,1,1,0,0,1,1,1,0
%N Number of ways to write n as x*(2x+1) + y*(2y+1) + z*(2z+1), where x,y,z are nonnegative integers with x <= y <= z.
%C Conjecture 1: a(n) > 0 for any integer n > 138158.
%C We have verified this for n up to 2*10^6. Note that n*(2n+1) (n = 0,1,...) are the second hexagonal numbers (A014105).
%C Conjecture 2: Any integer n > 146858 can be written as the sum of three hexagonal numbers (A000384).
%C Conjecture 3: Any integer n > 33066 can be written as the sum of three pentagonal numbers (A000326).
%C Conjecture 4: Any integer n > 24036 can be written as the sum of three second pentagonal numbers (A005449).
%C Conjecture 5: Let N(1) = 114862, N(-1) = 166897, N(3) = 196987 and N(-3) = 273118. Then, for any r among 1, -1, 3 and -3, each integer n > N(r) can be written as x*(5x+r)/2 + y*(5y+r)/2 + z*(5z+r)/2 with x,y,z nonnegative integers.
%C We have verified Conjectures 2-5 for n up to 10^6.
%H Zhi-Wei Sun, <a href="/A306383/b306383.txt">Table of n, a(n) for n = 0..10000</a>
%H Zhi-Wei Sun, <a href="https://doi.org/10.1007/s11425-017-9354-4">Universal sums of three quadratic polynomials</a>, Sci. China Math., in press.
%e a(223595) = 1 with 223595 = 95*(2*95+1) + 200*(2*200+1) + 250*(2*250+1).
%e a(290660) = 1 with 290660 = 136*(2*136+1) + 149*(2*149+1) + 323*(2*323+1).
%t QQ[n_]:=QQ[n]=IntegerQ[Sqrt[8n+1]]&&Mod[Sqrt[8n+1],4]==1;
%t tab={};Do[r=0;Do[If[QQ[n-x(2x+1)-y(2y+1)],r=r+1],{x,0,(Sqrt[8n/3+1]-1)/4},{y,x,(Sqrt[4(n-x(2x+1))+1]-1)/4}];tab=Append[tab,r],{n,0,100}];Print[tab]
%Y Cf. A000217, A000326, A000384, A000566, A005449, A005475, A005476, A008443, A014105, A147875, A306382.
%K nonn
%O 0,43
%A _Zhi-Wei Sun_, Feb 11 2019
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