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%I #11 Mar 17 2016 19:05:38
%S 1,1,1,1,1,1,2,3,2,1,3,1,2,2,3,2,2,3,2,1,4,4,2,2,2,2,1,5,4,2,2,2,3,4,
%T 4,5,2,2,3,3,5,2,5,3,2,4,5,4,2,3,3,3,3,4,3,1,2,5,3,4,3,4,4,4,5,4,3,4,
%U 4,3,6,5,5,3,3,3,6,6,2,4
%N Number of ordered ways to write n = x^3 + y*(y+1) + z*(3*z+2), where x and y are nonnegative integers and z is a nonzero integer.
%C Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 3, 4, 5, 6, 10, 12, 20, 27, 56, 101.
%C (ii) Each nonnegative integer can be written as x^3 + P(y,z) with x >= 0 and y and z integers, provided that P(y,z) is among y^2+z(3z+1), y(2y+1)+z(3z+1), y(2y+1)+z(3z+2), y(2y+1)+z(5z+2), y(2y+1)+z(5z+3), y(2y+1)+2z(3z+1), y(2y+1)+ 2z(3z+2), y(2y+1)+z(6z+5), y(2y+1)+z(7z+2), y(2y+1)+z(7z+6), y(3y+1)+z(4z+1), y(3y+1)+z(7z+2), y(3y+2)+z(4z+1).
%C (iii) Every n = 0,1,2,... can be expressed as f(x,y,z) with x >= 0 and y and z integers, provided that f(x,y,z) is among 2x^3+y^2+z(3z+1), 2x^3+y(y+1)+z(3z+2), 3x^3+y(y+1)+z(3z+2), 2x^3+y(2y+1)+z(3z+1), 2x^3+y(2y+1)+z(3z+2), 2x^3+y(2y+1)+z(5z+4), 2x^3+y(3y+1)+z(3z+2), 2x^3+y(3y+2)+z(4z+3).
%C Note that those y(2y+1) with y integral are just triangular numbers.
%C See also A262813 for a similar conjecture.
%H Zhi-Wei Sun, <a href="/A270469/b270469.txt">Table of n, a(n) for n = 1..10000</a>
%e a(10) = 1 since 10 = 0^3 + 1*2 + (-2)(3*(-2)+2).
%e a(12) = 1 since 12 = 1^3 + 2*3 + 1*(3*1+2).
%e a(20) = 1 since 20 = 0^3 + 3*4 + (-2)*(3*(-2)+2).
%e a(27) = 1 since 27 = 0^3 + 2*3 + (-3)*(3*(-3)+2).
%e a(56) = 1 since 56 = 0^3 + 0*1 + 4*(3*4+2).
%e a(101) = 1 since 101 = 2^3 + 8*9 + (-3)*(3*(-3)+2).
%t pQ[x_]:=pQ[x]=x>0&&IntegerQ[Sqrt[3x+1]]
%t Do[r=0;Do[If[pQ[n-x^3-y(y+1)],r=r+1],{x,0,n^(1/3)},{y,0,(Sqrt[4(n-x^3)+1]-1)/2}];Print[n," ",r];Continue,{n,1,80}]
%Y Cf. A000217, A000290, A000578, A001082, A002378, A262813, A262815, A262816.
%K nonn
%O 1,7
%A _Zhi-Wei Sun_, Mar 17 2016
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