A270469 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.

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, 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, 4, 3, 6, 5, 5, 3, 3, 3, 6, 6, 2, 4

Offset: 1

Author

Zhi-Wei Sun, Mar 17 2016

Keywords

• nonn

Comments

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.

(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).

(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).

Note that those y(2y+1) with y integral are just triangular numbers.

See also A262813 for a similar conjecture.

Examples

			a(10) = 1 since 10 = 0^3 + 1*2 + (-2)(3*(-2)+2).
a(12) = 1 since 12 = 1^3 + 2*3 + 1*(3*1+2).
a(20) = 1 since 20 = 0^3 + 3*4 + (-2)*(3*(-2)+2).
a(27) = 1 since 27 = 0^3 + 2*3 + (-3)*(3*(-3)+2).
a(56) = 1 since 56 = 0^3 + 0*1 + 4*(3*4+2).
a(101) = 1 since 101 = 2^3 + 8*9 + (-3)*(3*(-3)+2).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000217, A000290, A000578, A001082, A002378, A262813, A262815, A262816.

Programs

Mathematica

pQ[x_]:=pQ[x]=x>0&&IntegerQ[Sqrt[3x+1]]
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}]